GSEB Class 12 Physics Solutions Chapter 3 Current Electricity

GSEB Solutions Class 12 Physics Chapter 3 Current Electricity

Question 1. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Answer: The car battery has an electromotive force (EMF) of 12 volts and an internal resistance of 0.4 Ω. The highest possible current flows when there is no external resistance in the circuit, effectively a short circuit. Using Ohm's law (Current = EMF / Resistance), the maximum current that can be drawn is \( \frac{12 \, \text{V}}{0.4 \, \Omega} = 30 \, \text{A} \).
In simple words: A car battery with 12V and 0.4Ω internal resistance can give a maximum of 30 Amperes of current.

🎯 Exam Tip: Remember that maximum current from a battery is obtained when the external resistance is zero (short-circuit condition), and it is calculated using the battery's EMF and internal resistance.

 

Question 2. A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer: We have a battery with an EMF of 10 V and an internal resistance of 3 Ω. The current flowing through the circuit is 0.5 A. First, we find the total resistance in the circuit using the formula for EMF: \( E = I(R + r) \), where R is the external resistance and r is the internal resistance. \( 10 \, \text{V} = 0.5 \, \text{A} \times (R + 3 \, \Omega) \) \( \frac{10 \, \text{V}}{0.5 \, \text{A}} = R + 3 \, \Omega \) \( 20 \, \Omega = R + 3 \, \Omega \) So, the resistance of the resistor R is \( 20 \, \Omega - 3 \, \Omega = 17 \, \Omega \). Next, we calculate the terminal voltage (V), which is the voltage across the external resistor. Using Ohm's law, \( V = I \times R \). \( V = 0.5 \, \text{A} \times 17 \, \Omega = 8.5 \, \text{V} \).
In simple words: A 10V battery with 3Ω internal resistance pushes 0.5A through a circuit. The external resistor's resistance is 17Ω, and the voltage across it (terminal voltage) is 8.5V.

🎯 Exam Tip: When a battery supplies current, its terminal voltage is less than its EMF due to the voltage drop across its internal resistance. Use \( V = E - Ir \) or \( V = IR \) for the external circuit.

 

Question 3.(a) Three resistors 1Ω, 2Ω, and 3Ω are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Answer:(a) When resistors are connected in series, their total resistance is simply the sum of individual resistances. So, the total resistance \( R \) of the combination is \( R = 1 \, \Omega + 2 \, \Omega + 3 \, \Omega = 6 \, \Omega \).
(b) The combination is connected to a 12 V battery with negligible internal resistance. First, calculate the total current \( I \) flowing through the circuit. Using Ohm's law, \( I = \frac{\text{EMF}}{\text{Total Resistance}} = \frac{12 \, \text{V}}{6 \, \Omega} = 2 \, \text{A} \). Since the resistors are in series, the same current (2 A) flows through each resistor. Now, calculate the potential drop across each resistor using \( V = IR \): Potential drop across the 1Ω resistor: \( V_1 = 2 \, \text{A} \times 1 \, \Omega = 2 \, \text{V} \). Potential drop across the 2Ω resistor: \( V_2 = 2 \, \text{A} \times 2 \, \Omega = 4 \, \text{V} \). Potential drop across the 3Ω resistor: \( V_3 = 2 \, \text{A} \times 3 \, \Omega = 6 \, \text{V} \).
In simple words: (a) Three resistors (1Ω, 2Ω, 3Ω) in series add up to 6Ω total resistance. (b) With a 12V battery, 2A of current flows. This means 2V drops across the 1Ω resistor, 4V across the 2Ω, and 6V across the 3Ω.

🎯 Exam Tip: For series combinations, current is the same through all components, while potential drops add up to the total voltage. For parallel combinations, voltage is the same across all components, while currents add up to the total current.

 

Question 4.(a) Three resistors 2Ω, 4Ω, and 5Ω are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery.

Answer:(a) For resistors in parallel, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances. \( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{2 \, \Omega} + \frac{1}{4 \, \Omega} + \frac{1}{5 \, \Omega} \) To add these fractions, find a common denominator, which is 20: \( \frac{1}{R_{\text{total}}} = \frac{10}{20 \, \Omega} + \frac{5}{20 \, \Omega} + \frac{4}{20 \, \Omega} = \frac{10 + 5 + 4}{20 \, \Omega} = \frac{19}{20 \, \Omega} \) So, the total resistance is \( R_{\text{total}} = \frac{20}{19} \, \Omega \).
(b) The parallel combination is connected to a 20 V battery with negligible internal resistance. In a parallel circuit, the voltage across each resistor is the same as the battery's EMF (20 V). The current through each resistor can be found using Ohm's law \( I = \frac{V}{R} \): Current through the 2Ω resistor: \( I_1 = \frac{20 \, \text{V}}{2 \, \Omega} = 10 \, \text{A} \). Current through the 4Ω resistor: \( I_2 = \frac{20 \, \text{V}}{4 \, \Omega} = 5 \, \text{A} \). Current through the 5Ω resistor: \( I_3 = \frac{20 \, \text{V}}{5 \, \Omega} = 4 \, \text{A} \). The total current drawn from the battery is the sum of the currents through individual resistors: \( I_{\text{total}} = I_1 + I_2 + I_3 = 10 \, \text{A} + 5 \, \text{A} + 4 \, \text{A} = 19 \, \text{A} \). Alternatively, using the total resistance: \( I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{20 \, \text{V}}{(20/19) \, \Omega} = 19 \, \text{A} \).
In simple words: (a) When 2Ω, 4Ω, and 5Ω resistors are in parallel, their combined resistance is 20/19 Ω. (b) With a 20V battery, the 2Ω resistor gets 10A, the 4Ω gets 5A, and the 5Ω gets 4A. The total current from the battery is 19A.

🎯 Exam Tip: Always remember that the voltage across parallel components is identical, while the current splits. For total current, sum individual branch currents or use the total equivalent resistance with Ohm's law.

 

Question 5. At room temperature (27.0°C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω. Given that the temperature coefficient of the material of the resistor is 1.70 x 10^-4°C^-1.

Answer: We are given the initial temperature \( T_1 = 27.0^\circ\text{C} \) and initial resistance \( R_1 = 100 \, \Omega \). The final resistance is \( R_2 = 117 \, \Omega \). The temperature coefficient of resistance is \( \alpha = 1.70 \times 10^{-4} \, ^\circ\text{C}^{-1} \). The relationship between resistance and temperature is \( R_2 = R_1 (1 + \alpha \Delta T) \), where \( \Delta T = T_2 - T_1 \). We can rearrange this to find \( \Delta T \): \( \frac{R_2}{R_1} = 1 + \alpha \Delta T \) \( \alpha \Delta T = \frac{R_2}{R_1} - 1 = \frac{R_2 - R_1}{R_1} \) \( \Delta T = \frac{R_2 - R_1}{R_1 \alpha} \) Substitute the given values: \( \Delta T = \frac{117 \, \Omega - 100 \, \Omega}{100 \, \Omega \times (1.70 \times 10^{-4} \, ^\circ\text{C}^{-1})} = \frac{17 \, \Omega}{17 \times 10^{-3} \, ^\circ\text{C}^{-1}} = \frac{1}{10^{-3}} \, ^\circ\text{C} = 1000 \, ^\circ\text{C} \). The final temperature \( T_2 \) is \( T_2 = T_1 + \Delta T = 27^\circ\text{C} + 1000^\circ\text{C} = 1027^\circ\text{C} \).
In simple words: A heating element's resistance increases from 100Ω to 117Ω as it gets hotter. Given its temperature coefficient, this change means its temperature rose by 1000°C. Starting at 27°C, the new temperature is 1027°C.

🎯 Exam Tip: Remember the formula \( R_2 = R_1 (1 + \alpha \Delta T) \) for temperature dependence of resistance. Be careful with units for temperature (often °C or K) and \( \alpha \).

 

Question 6. A negligibly small current is passed through a wire of length 15m and uniform cross-section 6.0 x 10^-7 m² and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

Answer: We are given the resistance \( R = 5.0 \, \Omega \), length \( l = 15 \, \text{m} \), and cross-sectional area \( A = 6.0 \times 10^{-7} \, \text{m}^2 \). The resistivity \( \rho \) of a material is related to its resistance, length, and cross-sectional area by the formula: \( R = \rho \frac{l}{A} \). To find resistivity, we can rearrange the formula: \( \rho = \frac{RA}{l} \). Substitute the given values: \( \rho = \frac{(5.0 \, \Omega) \times (6.0 \times 10^{-7} \, \text{m}^2)}{15 \, \text{m}} = \frac{30.0 \times 10^{-7} \, \Omega \cdot \text{m}}{15} \) \( \rho = 2 \times 10^{-7} \, \Omega \cdot \text{m} \).
In simple words: A wire 15m long with an area of 6.0 x 10^-7 m² has a resistance of 5.0Ω. Its material's resistivity is calculated as 2 x 10^-7 Ωm.

🎯 Exam Tip: The formula \( R = \rho \frac{l}{A} \) is fundamental for relating resistance to material properties and dimensions. Ensure consistent units (SI units are usually best) for length, area, and resistance to get resistivity in Ωm.

 

Question 7. A silver wire has a resistance of 2.1 Ω at 27.5°C and a resistance of 2.7 Ω at 100°C. Determine the temperature coefficient of resistivity of silver.

Answer: We have the initial resistance \( R_1 = 2.1 \, \Omega \) at initial temperature \( T_1 = 27.5^\circ\text{C} \). The final resistance is \( R_2 = 2.7 \, \Omega \) at final temperature \( T_2 = 100^\circ\text{C} \). The change in temperature is \( \Delta T = T_2 - T_1 = 100^\circ\text{C} - 27.5^\circ\text{C} = 72.5^\circ\text{C} \). The temperature coefficient of resistivity \( \alpha \) is given by the formula: \( \alpha = \frac{R_2 - R_1}{R_1 \Delta T} \) Substitute the given values: \( \alpha = \frac{2.7 \, \Omega - 2.1 \, \Omega}{2.1 \, \Omega \times 72.5^\circ\text{C}} = \frac{0.6 \, \Omega}{152.25 \, \Omega \cdot ^\circ\text{C}} \) \( \alpha \approx 0.0039 \, ^\circ\text{C}^{-1} \).
In simple words: A silver wire changes resistance from 2.1Ω to 2.7Ω when its temperature goes from 27.5°C to 100°C. This means its temperature coefficient of resistivity is about 0.0039 per degree Celsius.

🎯 Exam Tip: The temperature coefficient of resistivity \( \alpha \) describes how much a material's resistance changes per degree Celsius (or Kelvin). Its unit is typically °C^-1 or K^-1.

 

Question 8. A heating element using nichrome connected to a 230V supply draws an initial current of 3.2 A which settles after a few seconds to steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0°C? The temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 x 10^-4°C^-1.

Answer: The supply voltage is 230 V. Room temperature \( T_1 = 27.0^\circ\text{C} \). The temperature coefficient of resistance \( \alpha = 1.70 \times 10^{-4} \, ^\circ\text{C}^{-1} \). First, calculate the initial resistance \( R_1 \) when the current is 3.2 A: \( R_1 = \frac{V}{I_1} = \frac{230 \, \text{V}}{3.2 \, \text{A}} = 71.88 \, \Omega \). Next, calculate the steady resistance \( R_2 \) when the current settles to 2.8 A: \( R_2 = \frac{V}{I_2} = \frac{230 \, \text{V}}{2.8 \, \text{A}} = 82.14 \, \Omega \). Now, use the formula relating resistance and temperature: \( R_2 = R_1 (1 + \alpha \Delta T) \). Rearrange to find the temperature change \( \Delta T \): \( \Delta T = \frac{R_2 - R_1}{R_1 \alpha} = \frac{82.14 \, \Omega - 71.88 \, \Omega}{71.88 \, \Omega \times (1.70 \times 10^{-4} \, ^\circ\text{C}^{-1})} = \frac{10.26 \, \Omega}{0.0122196} \, ^\circ\text{C} \) \( \Delta T \approx 840 \, ^\circ\text{C} \). The steady temperature \( T_2 \) is \( T_2 = T_1 + \Delta T = 27^\circ\text{C} + 840^\circ\text{C} = 867^\circ\text{C} \).
In simple words: A nichrome heater drawing 3.2A initially then 2.8A steadily from a 230V supply shows its resistance changing with temperature. Knowing its temperature coefficient and starting at 27°C, the final stable temperature is 867°C.

🎯 Exam Tip: When dealing with heating elements, remember that resistance typically increases with temperature. Calculate resistance at different states (initial/final) using Ohm's law, then apply the temperature-resistance formula.

 

Question 9. Determine the current in each branch of the network shown in Fig.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक व्हीटस्टोन ब्रिज नेटवर्क को दर्शाता है जिसमें कई प्रतिरोधक और एक ईएमएफ स्रोत जुड़े हुए हैं। नेटवर्क में AB, BC, AD, DC, BD और DE शाखाएं शामिल हैं, जिनमें से प्रत्येक में प्रतिरोधक लगे हैं। एक 10V बैटरी बिंदु E और C के बीच जुड़ी है, जिससे परिपथ में धारा प्रवाहित होती है।
Answer: To find the current in each branch, we apply Kirchhoff's laws to the given network. Let \( I \) be the total current from the 10V battery, \( I_1 \) be the current through branch AB, and \( I_g \) be the current through the galvanometer branch BD. The current distribution is shown in the figure (not provided, but implicitly understood from the equations). Using Kirchhoff's Voltage Law (KVL) for different meshes: For the mesh ABDA: \( 10I_1 + 5I_g - 5(I - I_1) = 0 \) Simplifying, \( 15I_1 + 5I_g - 5I = 0 \), which further reduces to \( 3I_1 + I_g - I = 0 \) ... (1) For the mesh BDCB: \( 5I_g + 10(I - I_1 + I_g) - 5(I_1 - I_g) = 0 \) Simplifying, \( -15I_1 + 20I_g + 10I = 0 \), which reduces to \( 3I_1 - 4I_g - 2I = 0 \) ... (2) From external considerations or another mesh (e.g., ABCEA), if we assume the provided solution's intermediate steps: Let's use the provided equations from the OCR as a system: \( 3I_1 + I_g - I = 0 \) ... (1) \( 3I_1 - 4I_g - 2I = 0 \) ... (2) The solution then implies a third relation that simplifies to \( 6I_1 + I = 2 \) (4) and \( 15I_1 - 6I = 0 \) (5). Solving equations (4) and (5): From (5), \( 15I_1 = 6I \implies I = \frac{15}{6}I_1 = \frac{5}{2}I_1 \). Substitute \( I \) into (4): \( 6I_1 + \frac{5}{2}I_1 = 2 \) \( \frac{12I_1 + 5I_1}{2} = 2 \) \( 17I_1 = 4 \implies I_1 = \frac{4}{17} \, \text{A} \). Now find \( I \): \( I = \frac{5}{2}I_1 = \frac{5}{2} \times \frac{4}{17} = \frac{10}{17} \, \text{A} \). Now find \( I_g \) from (1): \( 3(\frac{4}{17}) + I_g - \frac{10}{17} = 0 \) \( \frac{12}{17} + I_g - \frac{10}{17} = 0 \) \( \frac{2}{17} + I_g = 0 \implies I_g = -\frac{2}{17} \, \text{A} \). The negative sign for \( I_g \) indicates that the actual direction of current in branch BD is opposite to what was assumed in the figure. It flows from D to B. Now we can determine the current in each branch: Current through AB: \( I_{\text{AB}} = I_1 = \frac{4}{17} \, \text{A} \). Current through AD: \( I_{\text{AD}} = I - I_1 = \frac{10}{17} \, \text{A} - \frac{4}{17} \, \text{A} = \frac{6}{17} \, \text{A} \). Current through BC: \( I_{\text{BC}} = I_1 - I_g = \frac{4}{17} \, \text{A} - (-\frac{2}{17} \, \text{A}) = \frac{4+2}{17} \, \text{A} = \frac{6}{17} \, \text{A} \). Current through DC: \( I_{\text{DC}} = (I - I_1) + I_g = \frac{6}{17} \, \text{A} + (-\frac{2}{17} \, \text{A}) = \frac{6-2}{17} \, \text{A} = \frac{4}{17} \, \text{A} \). Current through BD: \( I_{\text{BD}} = I_g = -\frac{2}{17} \, \text{A} \). This means 2/17 A flows from D to B.
In simple words: By using Kirchhoff's rules for the network, we find the currents in each part. Current through AB is 4/17 A, AD is 6/17 A, BC is 6/17 A, DC is 4/17 A, and BD is 2/17 A, flowing from D to B.

🎯 Exam Tip: For complex networks, Kirchhoff's current and voltage laws are essential. Always draw a clear diagram with assumed current directions. A negative result for a current means it flows in the opposite direction to the assumption.

 

Question 10.(a) In a metre bridge the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
(b) Determine the balance point of the bridge if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक मीटर ब्रिज व्यवस्था को दिखाता है, जो अज्ञात प्रतिरोधों को मापने के लिए एक व्हीटस्टोन ब्रिज का प्रकार है। इसमें प्रतिरोधक X और Y के लिए दो गैप हैं, एक 1-मीटर लंबा तार जिसमें एक स्लाइडिंग संपर्क और एक गैल्वेनोमीटर जुड़ा हुआ है।
Answer:(a) In a meter bridge, at the balance point, the ratio of resistances is equal to the ratio of the lengths of the wire sections. Let \( l_1 \) be the balance length from end A, so \( l_1 = 39.5 \, \text{cm} \). The other length is \( l_2 = 100 - l_1 = 100 - 39.5 = 60.5 \, \text{cm} \). The known resistance is \( Y = 12.5 \, \Omega \). The unknown resistance is \( X \). The balance condition is \( \frac{X}{Y} = \frac{l_1}{l_2} \). So, \( X = Y \times \frac{l_1}{l_2} = 12.5 \, \Omega \times \frac{39.5 \, \text{cm}}{60.5 \, \text{cm}} \). \( X = 12.5 \times 0.65289 \approx 8.16 \, \Omega \). Thick copper strips are used for connections to ensure their resistance is very small (negligible). This prevents them from affecting the measurements and ensures that the resistance of the connecting wires does not introduce errors into the bridge formula.
(b) If resistors X and Y are interchanged, the new balance point \( l' \) will be obtained when the ratio is reversed: \( \frac{Y}{X} = \frac{l'}{100 - l'} \) \( \frac{12.5 \, \Omega}{8.16 \, \Omega} = \frac{l'}{100 - l'} \) \( 1.5318 \approx \frac{l'}{100 - l'} \) \( 1.5318 \times (100 - l') = l' \) \( 153.18 - 1.5318 l' = l' \) \( 153.18 = 2.5318 l' \) \( l' = \frac{153.18}{2.5318} \approx 60.5 \, \text{cm} \). So, the balance point shifts to 60.5 cm from end A (or 39.5 cm from the other end, which was the original length \( l_1 \)).
(c) At the balance point, there is no potential difference between the points where the galvanometer is connected. Therefore, no current flows through the galvanometer. If the galvanometer and the cell (EMF source) are interchanged at this balanced state, the bridge will still remain balanced, and the galvanometer will continue to show no deflection (zero current). This is due to the principle of reciprocity in balanced Wheatstone bridge circuits.
In simple words: (a) With a meter bridge, if Y is 12.5Ω and the balance point is at 39.5cm, then X is about 8.16Ω. Thick copper strips are used to avoid unwanted resistance in the connections. (b) If X and Y are swapped, the balance point will move to 60.5cm. (c) At the balance point, swapping the galvanometer and battery won't make the galvanometer show current, as the bridge remains balanced.

🎯 Exam Tip: For meter bridge questions, remember the balance condition \( \frac{R_1}{R_2} = \frac{l_1}{l_2} \). The use of thick copper strips is to minimize contact resistance, a common practical aspect. The interchange of cell and galvanometer at the balance point is a key concept illustrating bridge reciprocity.

 

Question 11. A storage battery, of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Answer: We have a storage battery with EMF \( E_b = 8.0 \, \text{V} \) and internal resistance \( r = 0.5 \, \Omega \). It is being charged by a DC supply of \( E_s = 120 \, \text{V} \) through a series resistor \( R = 15.5 \, \Omega \). During charging, the supply voltage drives current against the battery's EMF. The net EMF in the circuit is \( E_{\text{net}} = E_s - E_b = 120 \, \text{V} - 8.0 \, \text{V} = 112 \, \text{V} \). The total resistance in the circuit is \( R_{\text{total}} = R + r = 15.5 \, \Omega + 0.5 \, \Omega = 16.0 \, \Omega \). The charging current \( I \) is \( I = \frac{E_{\text{net}}}{R_{\text{total}}} = \frac{112 \, \text{V}}{16.0 \, \Omega} = 7 \, \text{A} \). The terminal voltage of the battery during charging is given by \( V_t = E_b + Ir \). (Note the plus sign because voltage is supplied to the battery during charging, increasing its terminal voltage). \( V_t = 8.0 \, \text{V} + (7 \, \text{A} \times 0.5 \, \Omega) = 8.0 \, \text{V} + 3.5 \, \text{V} = 11.5 \, \text{V} \). The purpose of having a series resistor in the charging circuit is to limit the charging current. Without this resistor, a very large current would flow into the battery, which could overheat and damage the battery or the charging supply. It ensures safe and controlled charging.
In simple words: A 8V battery with 0.5Ω internal resistance is charged by a 120V supply using a 15.5Ω series resistor. The charging current is 7A, and the battery's terminal voltage during charging is 11.5V. The series resistor protects the battery from excessive current.

🎯 Exam Tip: For charging circuits, the net EMF is the difference between the supply voltage and the battery's EMF. Terminal voltage during charging is \( E + Ir \). Always remember the safety function of a series resistor in charging circuits.

 

Question 12. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

Answer: In a potentiometer, the EMF of a cell is directly proportional to the balance length. Let \( E_1 = 1.25 \, \text{V} \) be the EMF of the first cell, and \( l_1 = 35.0 \, \text{cm} \) be its balance length. Let \( E_2 \) be the EMF of the second cell, and \( l_2 = 63.0 \, \text{cm} \) be its balance length. According to the principle of potentiometer: \( \frac{E_1}{E_2} = \frac{l_1}{l_2} \) We need to find \( E_2 \): \( E_2 = E_1 \times \frac{l_2}{l_1} \) Substitute the given values: \( E_2 = 1.25 \, \text{V} \times \frac{63.0 \, \text{cm}}{35.0 \, \text{cm}} = 1.25 \, \text{V} \times 1.8 = 2.25 \, \text{V} \).
In simple words: A potentiometer measures cell voltages. If a 1.25V cell balances at 35cm, another cell balancing at 63cm must have an EMF of 2.25V.

🎯 Exam Tip: The core principle of a potentiometer is that EMF is directly proportional to the balance length (\( E \propto l \)). This is a common and straightforward application in problems.

 

Question 13. The number density of free electrons in a copper conductor is 8.5 x 10²⁸ m^-3. How long does an electron take to drift from one end of a wire of 3.0 m long to its other end? The area of cross-section of the wire is 2.0 x 10^-6 m² and it is carrying a current of 3.0 A.

Answer: We are given: Number density of free electrons, \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \). Length of the wire, \( l = 3.0 \, \text{m} \). Area of cross-section, \( A = 2.0 \times 10^{-6} \, \text{m}^2 \). Current, \( I = 3.0 \, \text{A} \). Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \). First, calculate the drift velocity \( v_d \) using the relation \( I = nAe v_d \): \( v_d = \frac{I}{nAe} \) Substitute the values: \( v_d = \frac{3.0 \, \text{A}}{(8.5 \times 10^{28} \, \text{m}^{-3}) \times (2.0 \times 10^{-6} \, \text{m}^2) \times (1.6 \times 10^{-19} \, \text{C})} \) \( v_d = \frac{3.0}{8.5 \times 2.0 \times 1.6 \times 10^{28 - 6 - 19}} = \frac{3.0}{27.2 \times 10^3} = \frac{3.0}{27200} \) \( v_d \approx 1.1 \times 10^{-4} \, \text{m/s} \). Now, calculate the time \( t \) an electron takes to drift the length \( l \) of the wire: \( t = \frac{l}{v_d} \) Substitute the values: \( t = \frac{3.0 \, \text{m}}{1.1 \times 10^{-4} \, \text{m/s}} \approx 2.72 \times 10^4 \, \text{seconds} \).
In simple words: In a 3-meter copper wire carrying 3A of current, electrons move very slowly. Given the number of electrons and the wire's size, it would take an electron about 27,200 seconds (over 7.5 hours) to drift from one end to the other.

🎯 Exam Tip: Remember the relationship \( I = nAe v_d \) for drift velocity. Drift velocity is typically very small. The time taken for an electron to traverse a wire's length is \( t = l/v_d \).

 

Question 14. The earth's surface has a negative surface charge of 10^-9 Cm^-2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth's surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 x 10⁶m.)

Answer: We are given: Surface charge density \( \sigma = 10^{-9} \, \text{C/m}^2 \). Current flowing over the globe, \( I = 1800 \, \text{A} \). Radius of Earth, \( R_e = 6.37 \times 10^6 \, \text{m} \). First, calculate the total charge \( Q \) on the Earth's surface. The surface area of the Earth is \( 4\pi R_e^2 \). \( Q = \sigma \times \text{Area} = \sigma \times 4\pi R_e^2 \) \( Q = (10^{-9} \, \text{C/m}^2) \times 4 \times 3.14 \times (6.37 \times 10^6 \, \text{m})^2 \) \( Q = 10^{-9} \times 4 \times 3.14 \times (40.5769 \times 10^{12}) \, \text{C} \) \( Q \approx 509.6 \times 10^3 \, \text{C} = 5.096 \times 10^5 \, \text{C} \). Now, calculate the time \( t \) required to neutralize this charge, using the formula \( I = \frac{Q}{t} \), so \( t = \frac{Q}{I} \). \( t = \frac{5.096 \times 10^5 \, \text{C}}{1800 \, \text{A}} \) \( t \approx 283.1 \, \text{seconds} \). So, roughly 283 seconds would be required to neutralize the Earth's surface charge if there were no replenishment mechanism.
In simple words: Earth's surface has a negative charge density, and a current of 1800A flows globally. Given Earth's radius, the total charge is large. If this charge wasn't constantly replenished by phenomena like thunderstorms, it would take only about 283 seconds for the current to neutralize it.

🎯 Exam Tip: This problem combines concepts of surface charge density, current, and global dimensions. Remember to calculate total charge from surface density and area, then use the definition of current (\( I = Q/t \)) to find time. Always identify given parameters and derived quantities clearly.

 

Question 15.(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What is the current drawn from the supply and its terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Answer:(a) For six cells connected in series: Total EMF \( E_{\text{total}} = 6 \times (\text{EMF of one cell}) = 6 \times 2.0 \, \text{V} = 12.0 \, \text{V} \). Total internal resistance \( r_{\text{total}} = 6 \times (\text{internal resistance of one cell}) = 6 \times 0.015 \, \Omega = 0.09 \, \Omega \). The external resistance is \( R = 8.5 \, \Omega \). The total resistance in the circuit is \( R_{\text{circuit}} = R + r_{\text{total}} = 8.5 \, \Omega + 0.09 \, \Omega = 8.59 \, \Omega \). The current drawn from the supply is \( I = \frac{E_{\text{total}}}{R_{\text{circuit}}} = \frac{12.0 \, \text{V}}{8.59 \, \Omega} \approx 1.397 \, \text{A} \). The terminal voltage of the battery is the voltage across the external resistor \( V_t = IR \): \( V_t = 1.397 \, \text{A} \times 8.5 \, \Omega \approx 11.88 \, \text{V} \).
(b) For a secondary cell after long use: EMF \( E = 1.9 \, \text{V} \). Internal resistance \( r = 380 \, \Omega \). Maximum current is drawn when the external resistance is zero (short-circuit condition): \( I_{\text{max}} = \frac{E}{r} = \frac{1.9 \, \text{V}}{380 \, \Omega} = 0.005 \, \text{A} \). A car's starting motor requires a very large current, typically hundreds of amperes, to operate. A current of only 0.005 A is extremely small and completely insufficient to drive a car's starting motor. Therefore, this cell cannot drive the starting motor of a car.
In simple words: (a) Six 2V cells with 0.015Ω internal resistance each, when connected in series to an 8.5Ω resistor, draw about 1.397A of current, and the terminal voltage is 11.88V. (b) A worn-out cell (1.9V, 380Ω) can only provide a tiny maximum current of 0.005A, which is too weak to start a car.

🎯 Exam Tip: Remember how series combinations affect total EMF and internal resistance. Terminal voltage is \( V = E - Ir \) when delivering current. Max current is \( E/r \). Be aware of the typical current requirements for practical devices like car motors.

 

Question 16. Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (\(\rho_{Al}\) = 2.63 x 10^-8 Ωm, \(\rho_{Cu}\) = 1.72 x 10^-8 Ωm, Relative density of Al = 2.7 and that of Cu = 8.9.)

Answer: Let both wires have equal length \( l \) and equal resistance \( R \). The resistance \( R \) of a wire is given by \( R = \rho \frac{l}{A} \), where \( \rho \) is resistivity and \( A \) is the cross-sectional area. From this, the area \( A = \rho \frac{l}{R} \). The mass \( M \) of a wire is given by \( M = \text{density} \times \text{Volume} = \text{density} \times A \times l \). Substitute the expression for \( A \): \( M = \text{density} \times \left( \rho \frac{l}{R} \right) \times l = \frac{\text{density} \times \rho \times l^2}{R} \). For copper (Cu) and aluminium (Al), since \( l \) and \( R \) are the same for both wires, the ratio of their masses depends on the product of their density and resistivity: \( \frac{M_{Cu}}{M_{Al}} = \frac{\text{density}_{Cu} \times \rho_{Cu}}{\text{density}_{Al} \times \rho_{Al}} \) Given values: \( \rho_{Al} = 2.63 \times 10^{-8} \, \Omega \text{m} \) \( \rho_{Cu} = 1.72 \times 10^{-8} \, \Omega \text{m} \) Relative density of Al = 2.7 (so density_Al = \( 2.7 \times 10^3 \, \text{kg/m}^3 \)) Relative density of Cu = 8.9 (so density_Cu = \( 8.9 \times 10^3 \, \text{kg/m}^3 \)) Let's use relative densities for the ratio, as the \( 10^3 \) factor cancels out. \( \frac{M_{Cu}}{M_{Al}} = \frac{8.9 \times (1.72 \times 10^{-8})}{2.7 \times (2.63 \times 10^{-8})} = \frac{8.9 \times 1.72}{2.7 \times 2.63} = \frac{15.308}{7.101} \approx 2.15 \) Since \( \frac{M_{Cu}}{M_{Al}} \approx 2.15 \), this means \( M_{Cu} \approx 2.15 \times M_{Al} \). Therefore, the aluminium wire is lighter than the copper wire for the same length and resistance. Aluminium wires are preferred for overhead power cables because: 1. **Lighter Weight:** For the same resistance, aluminium wires are significantly lighter than copper wires. This reduces the strain on supporting structures (poles and towers) and makes installation easier and cheaper. 2. **Cost-Effective:** Aluminium is generally cheaper than copper. 3. **Good Conductivity:** While copper is a better conductor, aluminium still has good conductivity sufficient for power transmission.
In simple words: Even though copper conducts electricity better, an aluminum wire with the same length and resistance is lighter than a copper one. Because aluminum is lighter and cheaper, it is chosen for long overhead power cables.

🎯 Exam Tip: This problem requires understanding how resistance, density, and material properties relate to mass. Overhead cables prioritize lightweight materials for reduced structural load, even if conductivity isn't superior.

 

Question 17. What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Current A	Voltage V	Current A	Voltage V
0.2	3.94	3.0	59.2
0.4	7.87	4.0	78.8
0.6	11.8	5.0	98.6
0.8	15.7	6.0	118.5
1.0	19.7	7.0	138.2
2.0	39.4	8.0	158.0

Answer: To draw conclusions, we calculate the resistance \( R = \frac{V}{I} \) for various data points: For I = 0.2 A, V = 3.94 V, \( R = \frac{3.94}{0.2} = 19.7 \, \Omega \). For I = 0.4 A, V = 7.87 V, \( R = \frac{7.87}{0.4} = 19.675 \, \Omega \). For I = 0.6 A, V = 11.8 V, \( R = \frac{11.8}{0.6} \approx 19.67 \, \Omega \). For I = 0.8 A, V = 15.7 V, \( R = \frac{15.7}{0.8} = 19.625 \, \Omega \). For I = 1.0 A, V = 19.7 V, \( R = \frac{19.7}{1.0} = 19.7 \, \Omega \). For I = 2.0 A, V = 39.4 V, \( R = \frac{39.4}{2.0} = 19.7 \, \Omega \). For I = 3.0 A, V = 59.2 V, \( R = \frac{59.2}{3.0} \approx 19.73 \, \Omega \). And so on. **Conclusions:** 1. **Ohm's Law:** The ratio of voltage to current (\( V/I \)) remains approximately constant (around 19.7 Ω) over a wide range of current values. This indicates that manganin is an ohmic material and obeys Ohm's law. 2. **Resistivity and Temperature:** Since the resistance stays nearly constant despite the varying current (which would cause varying heating and thus temperature changes in the resistor), it implies that the resistivity of manganin is almost independent of temperature. This makes manganin a suitable material for standard resistors.
In simple words: The table shows that for a manganin resistor, the voltage divided by the current (resistance) is almost always the same. This means manganin follows Ohm's Law, and its resistance doesn't change much even when its temperature changes due to current.

🎯 Exam Tip: When analyzing V-I data, calculate the ratio V/I to determine resistance. If the resistance is constant, the material is ohmic. A material whose resistance is stable despite changes in current (and thus temperature) indicates a low temperature coefficient of resistance, making it suitable for precision applications.

 

Question 18. Answer the following questions.
(a) A steady current flows in a metallic conductor of the non-uniform cross-section. Which of these quantities is constant along the conductor - current, current density, electric field, drift speed?
(b) Is Ohm's law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm's law.
(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

Answer:(a) When a steady current flows through a metallic conductor of non-uniform cross-section, the **current** remains constant throughout the conductor. This is because charge is conserved, meaning the rate of flow of charge must be the same at all points. However, current density (\( J = I/A \)), electric field (\( E = J/\sigma \)), and drift speed (\( v_d = J/(ne) \)) will vary with the cross-sectional area \( A \).
(b) No, Ohm's law is not universally applicable to all conducting elements. It is an empirical law that holds true for a class of materials called ohmic conductors (like metals at constant temperature). Materials that do not obey Ohm's law are called non-ohmic conductors. Examples include vacuum diode tubes, semiconductor diodes, and electrolytes.
(c) For a voltage supply, the maximum current \( I_{max} \) that can be drawn from it is inversely proportional to its internal resistance \( r \) (i.e., \( I_{max} = E/r \), assuming external resistance is negligible). Therefore, if a low voltage supply needs to provide high currents, its internal resistance must be very low to allow a large current flow.
(d) A high tension (HT) supply, such as one providing 6 kV, must have a very large internal resistance for safety reasons. If the circuit containing the HT supply accidentally gets short-circuited, a large internal resistance will limit the current to a safe value (\( I = E/r \), where r is large). This prevents damage to the power supply and reduces the risk of electrical hazards.
In simple words: (a) In a wire that changes thickness, only the total current stays the same; current density, electric field, and electron drift speed change. (b) Ohm's law doesn't work for everything, like diodes. (c) To get lots of current from a small voltage, the power source needs very low internal resistance. (d) High-voltage power sources need high internal resistance to limit current during a short circuit, keeping them safe.

🎯 Exam Tip: Understand the difference between current and current density, and why current is conserved. Be able to distinguish between ohmic and non-ohmic materials. The concept of internal resistance is crucial for both low-voltage/high-current applications (where it should be low) and high-voltage safety (where it should be high).

 

Question 19. Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficient of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10²²/10³).

Answer:(a) Alloys of metals usually have **greater** resistivity than that of their constituent metals.
(b) Alloys usually have much **lower** temperature coefficient of resistance than pure metals.
(c) The resistivity of the alloy manganin is **nearly independent of** increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of **10²²**.
(For reference, insulators have resistivity around \( 10^{11} \) to \( 10^{19} \, \Omega\text{m} \), while metals have resistivity around \( 10^{-8} \, \Omega\text{m} \). The ratio would be \( 10^{11}/10^{-8} = 10^{19} \) or \( 10^{19}/10^{-8} = 10^{27} \). The option 10²² is closest to the typical order of magnitude differences.)
In simple words: (a) Alloys are generally better at resisting current than the pure metals they are made from. (b) Alloys don't change their resistance much with temperature, unlike pure metals. (c) Manganin's resistance stays almost the same even if it gets hotter. (d) Insulators block current vastly more than metals, by a factor of about 10²².

🎯 Exam Tip: These are key properties of alloys and different material types. Remember that alloys often combine properties to be useful (e.g., high resistivity and low temperature coefficient for standard resistors). Know the approximate orders of magnitude for resistivity of conductors, semiconductors, and insulators.

 

Question 20.(a) Given n resistors each of resistance R. How will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum effective resistance?
(b) Given the resistances of 1Ω, 2Ω, 3Ω. How will you combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5)Ω, (iii) 6Ω, (iv) (6/11)Ω?
(c) Determine the equivalent resistance of networks shown in the figure.

Answer:(a) (i) To get the maximum effective resistance, you should combine the resistors in **series**. Maximum effective resistance, \( R_{\text{max}} = R + R + ... + R \) (n times) \( = nR \).
(ii) To get the minimum effective resistance, you should combine the resistors in **parallel**. Minimum effective resistance, \( \frac{1}{R_{\text{min}}} = \frac{1}{R} + \frac{1}{R} + ... + \frac{1}{R} \) (n times) \( = \frac{n}{R} \). So, \( R_{\text{min}} = \frac{R}{n} \). The ratio of the maximum to minimum effective resistance is: \( \frac{R_{\text{max}}}{R_{\text{min}}} = \frac{nR}{R/n} = n^2 \).
(b) Given resistors are \( R_1 = 1 \, \Omega \), \( R_2 = 2 \, \Omega \), and \( R_3 = 3 \, \Omega \). (i) To get \( 11/3 \, \Omega \): Connect \( R_1 \) and \( R_2 \) in parallel, then connect this combination in series with \( R_3 \). Parallel combination of \( R_1 \) and \( R_2 \): \( R_{12p} = \frac{R_1 R_2}{R_1 + R_2} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \, \Omega \). Total equivalent resistance: \( R_{\text{eq}} = R_{12p} + R_3 = \frac{2}{3} \, \Omega + 3 \, \Omega = \frac{2+9}{3} \, \Omega = \frac{11}{3} \, \Omega \).
(ii) To get \( 11/5 \, \Omega \): Connect \( R_2 \) and \( R_3 \) in parallel, then connect this combination in series with \( R_1 \). Parallel combination of \( R_2 \) and \( R_3 \): \( R_{23p} = \frac{R_2 R_3}{R_2 + R_3} = \frac{2 \times 3}{2 + 3} = \frac{6}{5} \, \Omega \). Total equivalent resistance: \( R_{\text{eq}} = R_{23p} + R_1 = \frac{6}{5} \, \Omega + 1 \, \Omega = \frac{6+5}{5} \, \Omega = \frac{11}{5} \, \Omega \).
(iii) To get \( 6 \, \Omega \): Connect all three resistors in **series**. \( R_{\text{eq}} = R_1 + R_2 + R_3 = 1 \, \Omega + 2 \, \Omega + 3 \, \Omega = 6 \, \Omega \).
(iv) To get \( 6/11 \, \Omega \): Connect all three resistors in **parallel**. \( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6+3+2}{6} = \frac{11}{6} \). So, \( R_{\text{eq}} = \frac{6}{11} \, \Omega \).
(c) The figure shows a network made of four identical sections connected in series. Each section consists of a 2Ω effective resistance (formed by two 1Ω resistors in series) in parallel with a 4Ω effective resistance (formed by two 2Ω resistors in series). Let's find the equivalent resistance of one such section: For two 1Ω resistors in series: \( R_{1s} = 1 \, \Omega + 1 \, \Omega = 2 \, \Omega \). For two 2Ω resistors in series: \( R_{2s} = 2 \, \Omega + 2 \, \Omega = 4 \, \Omega \). These two series combinations are then connected in parallel. So, one section's resistance is: \( R_{\text{section}} = \frac{R_{1s} \times R_{2s}}{R_{1s} + R_{2s}} = \frac{2 \, \Omega \times 4 \, \Omega}{2 \, \Omega + 4 \, \Omega} = \frac{8 \, \Omega^2}{6 \, \Omega} = \frac{4}{3} \, \Omega \). Since there are four such sections connected in series, the total equivalent resistance of the network is: \( R_{\text{total}} = 4 \times R_{\text{section}} = 4 \times \frac{4}{3} \, \Omega = \frac{16}{3} \, \Omega \).
In simple words: (a) To get the biggest resistance, connect them in a line (series); for the smallest, connect them side-by-side (parallel). The ratio of max to min resistance is n squared. (b) For 1Ω, 2Ω, 3Ω: you can get 11/3Ω by R1 || R2 then series R3; 11/5Ω by R2 || R3 then series R1; 6Ω by all in series; and 6/11Ω by all in parallel. (c) The given network is four identical parts in a row, where each part has 2Ω and 4Ω in parallel. The total resistance of this network is 16/3Ω.

🎯 Exam Tip: Master the formulas for series and parallel combinations. For complex networks, break them down into smaller, simpler series/parallel parts. For designing specific resistances, think about which combinations add up or reduce resistance as needed.

 

Question 21. Determine the current drawn from a 12V supply with internal resistance 0.5 Ω by the infinite network shown in figure. Each resistor has 1Ω resistance.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक अनंत सीढ़ी नेटवर्क (infinite ladder network) को दर्शाता है जिसमें 1Ω प्रतिरोधक एक दोहराए जाने वाले पैटर्न में जुड़े हुए हैं। यह नेटवर्क एक 12V बिजली आपूर्ति से जुड़ा है जिसकी आंतरिक प्रतिरोध 0.5 Ω है, और हमें नेटवर्क से निकलने वाली कुल धारा ज्ञात करनी है।
Answer: The network shown is an infinite ladder network. To find its equivalent resistance, let's assume the equivalent resistance of the entire infinite network, from a certain point onwards, is \( Y \). If we add one more repeating unit to this network, its equivalent resistance remains \( Y \). The repeating unit consists of a 1Ω resistor, then a parallel combination of a 1Ω resistor and the infinite network \( Y \), and finally another 1Ω resistor. So, \( Y = 1 \, \Omega + (1 \, \Omega \, || \, Y) + 1 \, \Omega \) \( Y = 2 \, \Omega + \frac{1 \times Y}{1 + Y} \, \Omega \) Multiply by \( (1+Y) \) to clear the denominator: \( Y(1+Y) = 2(1+Y) + Y \) \( Y + Y^2 = 2 + 2Y + Y \) \( Y^2 - 2Y - 2 = 0 \) This is a quadratic equation for \( Y \). Using the quadratic formula \( Y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( Y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3} \, \Omega \). Since resistance cannot be negative, we take the positive root: \( Y = 1 + \sqrt{3} \, \Omega \approx 1 + 1.732 \, \Omega = 2.732 \, \Omega \). This is the equivalent resistance of the infinite network. The network is connected to a 12V supply with an internal resistance \( r = 0.5 \, \Omega \). The total resistance in the circuit is \( R_{\text{total}} = Y + r = 2.732 \, \Omega + 0.5 \, \Omega = 3.232 \, \Omega \). The current drawn from the supply is \( I = \frac{\text{EMF}}{R_{\text{total}}} = \frac{12 \, \text{V}}{3.232 \, \Omega} \approx 3.71 \, \text{A} \).
In simple words: For this never-ending (infinite) resistor network, we find its total resistance by using a trick where we assume the whole thing has a resistance 'Y'. After solving a simple equation, 'Y' comes out to be about 2.732Ω. When connected to a 12V supply with 0.5Ω internal resistance, the total current drawn is about 3.71A.

🎯 Exam Tip: Infinite ladder networks are solved by identifying the repeating unit and assuming the equivalent resistance of the entire infinite portion is \( Y \). Then, express the equivalent resistance of the first repeating unit plus the remaining infinite portion in terms of \( Y \), and solve the resulting equation for \( Y \). Discard negative resistance values.

 

Question 22. Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf and the balance point turns out to be at 82.3 cm length of the wire.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पोटेंशियोमीटर परिपथ को दर्शाता है जिसका उपयोग एक मानक सेल और एक अज्ञात सेल के ईएमएफ की तुलना करने के लिए किया जाता है। मुख्य परिपथ में एक ड्राइवर सेल (2V, 0.4Ω आंतरिक प्रतिरोध) और एक लंबी तार AB है। द्वितीयक परिपथ में एक सेल, गैल्वेनोमीटर और एक उच्च प्रतिरोधक (600 kΩ) शामिल हैं।
Answer:(a) The value of \( \varepsilon \) (unknown EMF): Let \( E_1 = 1.02 \, \text{V} \) be the EMF of the standard cell, and \( l_1 = 67.3 \, \text{cm} \) be its balance length. Let \( \varepsilon \) be the EMF of the unknown cell, and \( l_2 = 82.3 \, \text{cm} \) be its balance length. According to the principle of potentiometer: \( \frac{\varepsilon}{E_1} = \frac{l_2}{l_1} \). \( \varepsilon = E_1 \times \frac{l_2}{l_1} = 1.02 \, \text{V} \times \frac{82.3 \, \text{cm}}{67.3 \, \text{cm}} \) \( \varepsilon = 1.02 \times 1.22288 \approx 1.25 \, \text{V} \).
(b) The very high resistance (600 kΩ) is put in series with the standard cell and galvanometer to limit the current flowing through the galvanometer when the jockey is far from the balance point. This protects the sensitive galvanometer from damage due to large currents.
(c) No, the balance point is not affected by this high resistance. At the balance point, no current flows through the galvanometer circuit, so the presence of the high resistance has no effect on the null point position.
(d) The balance point is not directly affected by the internal resistance of the driver cell in terms of the ratio of EMFs to balance lengths. However, the internal resistance does affect the potential gradient across the potentiometer wire. If the internal resistance is very high, the potential gradient might be too low, making it difficult to obtain a balance point or reducing the sensitivity of the potentiometer.
(e) No, the method would not work if the driver cell's EMF was 1.0 V (instead of 2.0 V). For a balance point to be found, the EMF of the cell being measured must be less than the potential drop across the potentiometer wire (which is always less than or equal to the driver cell's EMF). If the driver cell's EMF is 1.0 V, it would be less than the standard cell's EMF (1.02 V) and the unknown cell's EMF (1.25 V), so no balance point could be achieved.
(f) The circuit in its current form is not suitable for accurately measuring extremely small EMFs (like a few mV). This is because for such small EMFs, the balance point would occur very close to end A (the zero point of the wire). Measuring such a short length accurately is difficult and leads to significant relative errors. To modify the circuit for very small EMFs, the potential gradient across the potentiometer wire needs to be significantly reduced. This can be done by adding a large resistance in series with the driver cell in the primary circuit.
In simple words: (a) A potentiometer setup measures an unknown cell's EMF (ε) as 1.25V, given a standard cell (1.02V) and their balance points. (b) A 600kΩ resistor protects the galvanometer by limiting current. (c) This high resistance doesn't change the balance point. (d) The driver cell's internal resistance affects the potential gradient but not the balance point itself. (e) If the driver cell was 1.0V, it couldn't measure cells with higher EMFs like 1.02V or 1.25V. (f) The current setup isn't good for tiny voltages because the balance point would be too close to one end, making measurements hard. To fix this, the potential drop along the wire must be made much smaller.

🎯 Exam Tip: Understand all components of a potentiometer: driver cell, potentiometer wire, standard cell, unknown cell, galvanometer, and series resistance. Know the conditions for obtaining a balance point (\( E_{\text{measured}} < E_{\text{driver}} \)) and the significance of potential gradient for sensitivity. The role of protective resistance in the galvanometer circuit is important.

 

Question 23. Figure shows a potentiometer circuit for comparison of two resistances. The balance point with resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ɛ?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पोटेंशियोमीटर परिपथ को दर्शाता है जिसका उपयोग दो प्रतिरोधों (R और X) की तुलना करने के लिए किया जाता है। एक सेल (ε) पोटेंशियोमीटर तार पर एक विभव पात बनाता है। द्वितीयक परिपथ में अज्ञात प्रतिरोध X, ज्ञात प्रतिरोध R और एक गैल्वेनोमीटर जुड़े हैं, और संतुलन बिंदु तार पर मापा जाता है।
Answer: In a potentiometer circuit used for comparing resistances, the ratio of the resistances is equal to the ratio of their corresponding balance lengths. Given: Known resistance \( R = 10.0 \, \Omega \). Balance length for R, \( l_1 = 58.3 \, \text{cm} \). Unknown resistance \( X \). Balance length for X, \( l_2 = 68.5 \, \text{cm} \). The relationship is \( \frac{X}{R} = \frac{l_2}{l_1} \). So, \( X = R \times \frac{l_2}{l_1} = 10.0 \, \Omega \times \frac{68.5 \, \text{cm}}{58.3 \, \text{cm}} \). \( X = 10.0 \times 1.174957 \approx 11.75 \, \Omega \). If you failed to find a balance point with the given cell of EMF \( \varepsilon \): This indicates that the potential difference across the resistance (R or X) in the secondary circuit is greater than the total potential drop across the potentiometer wire AB. To find a balance point, the potential drop across the length of the wire must be greater than the voltage being measured. You could: 1. **Increase the EMF of the driver cell:** Use a driver cell with a higher EMF in the primary circuit. 2. **Decrease the series resistance in the primary circuit:** If there's a rheostat or series resistor in the primary circuit (connected to the driver cell and potentiometer wire), decrease its value to increase the potential gradient along the potentiometer wire. 3. **Use a longer potentiometer wire:** A longer wire would provide a greater total potential drop for the same potential gradient or allow for a larger potential drop across the full length.
In simple words: For this potentiometer, if a 10.0Ω resistor balances at 58.3cm, then an unknown resistor that balances at 68.5cm has a resistance of 11.75Ω. If you can't find a balance point, it means the voltage you're trying to measure is too high for the potentiometer wire; you need to increase the voltage across the main wire or make the wire longer.

🎯 Exam Tip: Potentiometer problems for resistance comparison use the same principle as EMF comparison. If a balance point cannot be found, it implies that the potential to be measured is greater than the potential available across the potentiometer wire. Understanding the remedies for this situation (adjusting primary circuit or wire length) is important.

 

Question 24. Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in the open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पोटेंशियोमीटर व्यवस्था को दर्शाता है जिसका उपयोग 1.5V सेल के आंतरिक प्रतिरोध को निर्धारित करने के लिए किया जाता है। मुख्य परिपथ में एक 2.0V ड्राइवर सेल तार AB पर एक निरंतर विभव पात बनाए रखता है। द्वितीयक परिपथ में, 1.5V सेल एक गैल्वेनोमीटर और 9.5Ω के शंट प्रतिरोधक के साथ जुड़ा हुआ है।
Answer: We are given: EMF of the cell to be measured, \( E = 1.5 \, \text{V} \). (This value is not strictly needed for the internal resistance calculation if lengths are given, but it confirms the context.) Balance point in open circuit (when no current is drawn from the cell), \( l_1 = 76.3 \, \text{cm} \). This length corresponds to the EMF \( E \). External resistance connected in the secondary circuit, \( R = 9.5 \, \Omega \). Balance point when current is drawn from the cell (i.e., when \( R \) is connected), \( l_2 = 64.8 \, \text{cm} \). This length corresponds to the terminal voltage \( V \) across \( R \). The formula for internal resistance \( r \) using potentiometer readings is: \( r = R \left( \frac{l_1 - l_2}{l_2} \right) \) Substitute the given values: \( r = 9.5 \, \Omega \left( \frac{76.3 \, \text{cm} - 64.8 \, \text{cm}}{64.8 \, \text{cm}} \right) \) \( r = 9.5 \, \Omega \left( \frac{11.5}{64.8} \right) \) \( r = 9.5 \times 0.177469 \dots \) \( r \approx 1.686 \, \Omega \).
In simple words: Using a potentiometer, a cell gives a balance point at 76.3cm when open. When a 9.5Ω resistor is connected, the balance point moves to 64.8cm. Based on these lengths and the external resistor, the cell's internal resistance is found to be about 1.686Ω.

🎯 Exam Tip: For determining internal resistance with a potentiometer, remember the key formula \( r = R \left( \frac{l_1 - l_2}{l_2} \right) \). \( l_1 \) is the balance length for EMF (open circuit), and \( l_2 \) is for terminal voltage (closed circuit with external resistance \( R \)).

GSEB Class 12 Physics Current Electricity Additional Important Questions And Answers

 

Question 1. What do you mean by the steady current?
Answer: A steady current is an electric current where its strength or intensity remains constant over time.
In simple words: A steady current always has the same strength and does not change.
🎯 Exam Tip: Understanding the difference between steady and varying currents is fundamental for circuit analysis.

 

Question 2. What do you mean by the varying current?
Answer: A varying current is an electric current where its strength or intensity does not remain constant; it changes over time.
In simple words: A varying current changes its strength as time passes.
🎯 Exam Tip: Questions about varying currents often involve concepts like AC circuits or transient phenomena.

 

Question 3. In a steady flow, if charge q is flowing through a conductor for a time t, what is the charge flowing per unit time?
Answer: In a steady flow, the charge flowing per unit time is given by \( \frac{q}{t} \). This quantity represents the average current.
In simple words: When charge flows steadily, the amount of charge that moves each second is found by dividing total charge by total time.
🎯 Exam Tip: This is the basic definition of current, important for initial understanding of charge flow.

 

Question 4. In a varying flow, if ∆q is the charge flowing in a time ∆t, within the limit ∆t → 0, what is the electric current?
Answer: In a varying flow, the electric current is defined as the instantaneous rate of charge flow, which is given by \( \mathrm{Lt}_{\Delta t \to 0} \frac{\Delta q}{\Delta t} \).
In simple words: For a changing flow, electric current is how fast the charge moves at one exact moment, found by taking a very small change in charge over a very small change in time.
🎯 Exam Tip: This concept introduces calculus into physics, defining instantaneous current, which is crucial for understanding changing circuits.

 

Question 5. What is the unit of currency?
Answer: The unit of electric current is the ampere.
In simple words: Current is measured in amperes.
🎯 Exam Tip: Knowing the SI units for physical quantities is essential for all physics calculations and conceptual questions.

 

Question 6. Current through a conductor is said to be 1 C/s. What does it mean?
Answer: If the current through a conductor is 1 C/s, it means that one coulomb of charge flows through a cross-section of the conductor in one second.
In simple words: 1 C/s means 1 Coulomb of charge passes through the wire every second.
🎯 Exam Tip: This definition links the unit of current (Ampere) directly to its fundamental units of charge and time (Coulomb per second).

 

Question 7. Why does current become a scalar quantity?
Answer: Current is considered a scalar quantity because it does not follow the rules of vector addition. While current has both magnitude and direction, its total sum in a circuit always remains the same, regardless of the angle at which conductors meet or carry current.
In simple words: Current is a scalar because it does not add up like forces or velocities. Even if wires bend, the total current stays the same, so we treat it as a simple number.
🎯 Exam Tip: This is a common conceptual question distinguishing scalars from vectors, emphasizing that not all quantities with direction are vectors.

 

Question 8. If I is the current passing normal to an area A, what is current per unit area?
Answer: The current per unit area, when current I passes normally through area A, is called current density, denoted by \( \hat{j} \). It is given by \( \hat{j} = \frac{I}{\vec{A}} = \sigma \vec{E} \).
In simple words: Current per unit area means how much current flows through a small patch of the wire. We call this current density.
🎯 Exam Tip: Current density is a vector quantity that describes the flow of charge more precisely than total current, especially in non-uniform conductors.

 

Question 9. What do you mean by current density?
Answer: Current density refers to the amount of electric current flowing through a unit cross-sectional area of a conductor.
In simple words: Current density tells us how much electric current goes through a small piece of the conductor's surface.
🎯 Exam Tip: Current density is crucial for understanding microscopic current flow and is related to drift velocity.

 

Question 10. Why is current density a vector quantity?
Answer: Current density is a vector quantity because it has both magnitude and a specific direction. Its direction is the same as the direction of the electric field \( \vec{E} \) that causes the current to flow, making \( \hat{j} = \sigma \vec{E} \).
In simple words: Current density is a vector because it has a size and points in a certain direction, just like the electric field that makes current move.
🎯 Exam Tip: Distinguishing between current (scalar) and current density (vector) is a key conceptual point.

 

Question 11. The free electrons inside a metallic conductor are in random motion. Can we detect current in the conductor?
Answer: No, we cannot detect current in the conductor when free electrons are in random motion. This is because their average velocity is zero, meaning there is no net flow of charge in any particular direction.
In simple words: Even with many moving electrons, if they move randomly, there is no overall push in one way. So, no current is detected.
🎯 Exam Tip: Current requires a net, directed motion of charge carriers, not just random movement.

 

Question 12. What is the essential condition for current flow in a conductor?
Answer: The essential condition for current flow in a conductor is the application of a potential difference across it. This sets up an electric field, which then causes the electrons to move in a directed flow, resulting in an electric current.
In simple words: For current to flow, you need a voltage difference across the wire. This push makes electrons move in one direction.
🎯 Exam Tip: Potential difference is the driving force for current in a conductor, similar to pressure driving fluid flow.

 

Question 13. How do you produce a net flow of electric charges [electric current] in a conductor?
Answer: To produce a net flow of electric charges (electric current) in a conductor, a potential difference (p.d.) must be applied. This causes an electric field to act on the electrons, making them drift in a specific direction and creating a current.
In simple words: To make current flow, you put a voltage difference across a conductor. This makes the electrons move in a line, creating current.
🎯 Exam Tip: This re-emphasizes the fundamental requirement for current flow: a driving potential difference.

 

Question 14. When an electric field (E) is applied to the conductor, what is the force experienced by an electron?
Answer: When an electric field (E) is applied to a conductor, an electron experiences a force given by \( \text{Force} = -eE \). Here, 'e' is the magnitude of the charge of an electron, and the negative sign indicates that the force is in the opposite direction to the electric field.
In simple words: When an electric field is present, an electron feels a push that is opposite to the field's direction.
🎯 Exam Tip: This equation \( F = qE \) is basic to electromagnetism; the negative sign for electrons is important.

 

Question 15. If 'm' is the mass of an electron, what is its acceleration?
Answer: If 'm' is the mass of an electron, its acceleration 'a' when an electric field E is applied is given by \( a = \frac{-eE}{m} \). This formula comes from Newton's second law (\( F = ma \)), where \( F = -eE \).
In simple words: If an electron has mass 'm' and feels a push 'F', it will speed up or slow down with an acceleration 'a' equal to that push divided by its mass.
🎯 Exam Tip: This links electrical force to mechanical acceleration, demonstrating how electric fields make charges move.

 

Question 16. In the electric field, what happens to the motion of the electrons?
Answer: In an electric field, the electrons in a conductor begin to drift in a direction opposite to the electric field. This directed motion is what constitutes the electric current.
In simple words: When an electric field is turned on, electrons start moving in a line, but in the opposite direction to the field's push.
🎯 Exam Tip: The conventional current direction is opposite to the electron drift direction because electrons carry negative charge.

 

Question 17. The number density of conduction electrons in a conductor is 8.5 x 10²⁸m^-3. The area of cross-section is 10^-6m² and I = 3A.
(a) What is the drift velocity of electrons?
(b) Compare the drift velocity of electrons with thermal velocity.
(c) What is the time taken by an electron to drift from one end of the conductor 3m long to the other end?
(d) The time taken by an electron to travel in a conductor is very large. But when we switch on an electric circuit, there is no time delay to complete the circuit. Why?
Answer: Given values: Number density \( n = 8.5 \times 10^{28} \text{m}^{-3} \), Current \( I = 3\text{A} \), Area of cross-section \( A = 10^{-6} \text{m}^2 \), Charge of electron \( e = 1.6 \times 10^{-19} \text{C} \).
(a) The drift velocity of electrons \( V_d \) can be calculated using the formula \( I = n e A V_d \).
So, \( V_d = \frac{I}{neA} \)
\( V_d = \frac{3}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}} \)
\( V_d = 2.2 \times 10^{-6} \text{ms}^{-1} = 2.2 \, \mu \text{ms}^{-1} \).
(b) The thermal speed of electrons is typically around \( 10^5 \text{ms}^{-1} \). Comparing this with the drift velocity (\( 2.2 \times 10^{-6} \text{ms}^{-1} \)), it is clear that the drift velocity is extremely small.
(c) For a conductor 3m long, the time \( t \) taken by an electron to drift from one end to the other is \( t = \frac{\text{length}}{V_d} \).
\( t = \frac{3}{2.2 \times 10^{-6}} \)
\( t = 1.36 \times 10^6 \text{s} \).
(d) Even though individual electrons drift very slowly, the electric field that causes them to move propagates through the circuit at nearly the speed of light. Therefore, as soon as the circuit is switched on, the electric field is established almost instantly throughout the conductor, causing all electrons to start drifting simultaneously, resulting in immediate current flow.
In simple words: (a) Electrons move very slowly along the wire due to current. (b) This slow movement is tiny compared to their fast, random jiggling. (c) It would take a very long time for one electron to travel a few meters. (d) The circuit turns on fast because the electrical signal (the "push") moves at light speed, even if the electrons themselves are slow.
🎯 Exam Tip: This question highlights the difference between drift velocity and the speed of signal propagation, a common misconception. The calculation of drift velocity is a standard application of the \( I = neAV_d \) formula.

 

Question 18.
(a) How can you measure the voltage across resistance and current through the circuit?
(b) How can you modify the given circuit to measure voltage across resistance and current through the circuit?
Answer:
(a) To measure voltage across a resistance, a voltmeter is connected in parallel with the resistance. To measure current through a circuit, an ammeter is connected in series with the circuit.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक साधारण विद्युत परिपथ को दर्शाता है जिसमें एक बैटरी (E), एक कुंजी (k) और एक प्रतिरोधक (R) जुड़े हुए हैं। वोल्टेज और धारा को मापने के लिए, एक एमीटर को प्रतिरोधक के साथ श्रृंखला में और एक वोल्टमीटर को प्रतिरोधक के साथ समानांतर में जोड़ना होगा।
(b) To modify the circuit to measure voltage and current: An ammeter (A) should be placed in series with the resistance R to measure current. A voltmeter (V) should be placed in parallel across the resistance R to measure voltage.
In simple words: (a) We use a voltmeter to check voltage by connecting it across the part we want to measure, and an ammeter to check current by connecting it in the line of flow. (b) To measure, put the ammeter in the same line as the resistor and the voltmeter in a separate line across the resistor.
🎯 Exam Tip: Correct placement of ammeters (series, low resistance) and voltmeters (parallel, high resistance) is a critical practical skill in circuit experiments.

 

Question 19. Why is an ammeter connected in series and a voltmeter in parallel?
Answer: An ammeter is connected in series because it has a very low internal resistance. This allows almost the entire current to flow through it without significantly changing the circuit's total resistance, thus enabling accurate measurement of the current. A voltmeter is connected in parallel because it has a very high internal resistance. This ensures that it draws negligible current from the main circuit, allowing it to measure the potential difference across the component without altering the circuit's operation.
In simple words: An ammeter connects in line to catch all the current because it has low resistance. A voltmeter connects across a component to read its voltage without taking much current, because it has high resistance.
🎯 Exam Tip: This is a fundamental concept in circuit measurements. Remembering the ideal resistances (ammeter near zero, voltmeter near infinite) is key.

 

Question 20. Without replacing the source, how can you vary the potential difference across the resistance wire?
Answer: To vary the potential difference across the resistance wire without replacing the source, a rheostat (variable resistor) can be introduced into the circuit. By adjusting the rheostat, the total resistance of the circuit changes, which in turn alters the current and consequently the potential difference across the resistance wire.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक सरल परिपथ दिखाया गया है जिसमें एक बैटरी (E), एक कुंजी (k), एक प्रतिरोधक (R) और एक रियोस्टेट (Rh) जुड़े हुए हैं। रियोस्टेट का उपयोग परिपथ के कुल प्रतिरोध को बदलकर R पर वोल्टेज को नियंत्रित करने के लिए किया जाता है।
In simple words: You can change the voltage across a wire by adding a "rheostat" (a variable resistor) to the circuit. Adjusting the rheostat changes how much current flows, which then changes the voltage across the wire.
🎯 Exam Tip: Rheostats are essential components for controlling current and voltage in experimental setups and are often used in potentiometer circuits.

 

Question 21. How can you state the relation between V and I as a law?
Answer: The relation between voltage (V) across a conductor and current (I) flowing through it can be stated as Ohm's law, which mathematically is expressed as \( V = IR \). This law indicates that for a given resistance R, the voltage is directly proportional to the current.
In simple words: The relationship between voltage and current is called Ohm's law, which simply says voltage equals current times resistance.
🎯 Exam Tip: Ohm's law is a cornerstone of circuit analysis, but it's important to remember its limitations (e.g., it applies to ohmic materials).

 

Question 22. What is conductance?
Answer: Conductance is a measure of how easily electric current flows through a material. It is defined as the reciprocal of resistance.
In simple words: Conductance tells us how well electricity can flow through something. It is the opposite of resistance.
🎯 Exam Tip: Conductance (G = 1/R) is measured in Siemens (S) or mhos (\( \Omega^{-1} \)).

 

Question 23. What is the variation of resistance with length and area of cross-section of the conductor?
Answer: The resistance (R) of a conductor varies directly with its length (l) and inversely with its area of cross-section (A). This relationship can be expressed as \( R \propto l \) and \( R \propto \frac{1}{A} \). Combining these, \( R = \rho \frac{l}{A} \), where \( \rho \) is resistivity.
In simple words: A longer wire has more resistance, and a thicker wire has less resistance.
🎯 Exam Tip: This formula \( R = \rho \frac{l}{A} \) is crucial for calculating resistance and understanding how material properties and dimensions affect it.

 

Question 24. What is the relation between drift velocity and relaxation time?
Answer: The drift velocity (\( V_d \)) of electrons in a conductor is directly proportional to the relaxation time (\( \tau \)). The relation is given by \( V_d = \frac{eE\tau}{m} \), where e is the electron charge, E is the electric field, and m is the electron mass.
In simple words: How fast electrons drift depends on how long they can move freely before hitting something. This free-moving time is called relaxation time.
🎯 Exam Tip: Relaxation time represents the average time between collisions of electrons with the lattice ions in a conductor, influencing its conductivity.

 

Question 25. What is the relation between the electric field, length of the wire, and the potential difference between the conductor?
Answer: The electric field (E) inside a conductor is related to the potential difference (V) across its ends and its length (l) by the formula \( E = \frac{V}{l} \). This shows that the electric field is the potential gradient along the conductor. Furthermore, the current (I) is also related to the drift velocity (\( V_d \)) by \( I = neAV_d \), and current density is \( j = neV_d \). Substituting \( V_d = \frac{eE\tau}{m} \), we get \( j = \frac{ne^2E\tau}{m} \). Since \( j = \sigma E \), the conductivity \( \sigma = \frac{ne^2\tau}{m} \). The resistance is \( R = \rho \frac{l}{A} \) and resistivity \( \rho = \frac{m}{ne^2\tau} \). Combining these, \( V = IR \).
In simple words: The electric field in a wire is the voltage divided by its length. This field makes current flow. The current's speed depends on how long electrons travel before hitting things. All these parts are connected by Ohm's Law.
🎯 Exam Tip: This question connects the microscopic model of current (drift velocity, relaxation time) to the macroscopic Ohm's law and the concept of electric field as a potential gradient.

 

Question 26. What is the relation between resistivity and relaxation time?
Answer: Resistivity \( \rho \) is inversely proportional to the relaxation time \( \tau \). The relationship is given by \( \rho = \frac{m}{ne^2\tau} \), where m is the electron mass, n is the number density of free electrons, and e is the electron charge.
In simple words: Resistivity, which shows how much a material resists current, goes down when electrons can move freely for longer (longer relaxation time).
🎯 Exam Tip: A longer relaxation time means fewer collisions, leading to higher conductivity and lower resistivity. This is crucial for understanding the microscopic origins of resistance.

 

Question 27. When the temperature increases, what happens to relaxation time?
Answer: When the temperature of a conductor increases, the thermal vibrations of the lattice ions become more energetic. This leads to more frequent collisions between free electrons and the ions, causing the relaxation time \( \tau \) to decrease.
In simple words: When a wire gets hotter, the tiny particles inside shake more. This makes it harder for electrons to move smoothly, so the time they fly free between bumps gets shorter.
🎯 Exam Tip: This explains why the resistance of most conductors increases with temperature, as a shorter relaxation time leads to higher resistivity.

 

Question 28. When relaxation time decreases, what happens to resistivity?
Answer: When relaxation time decreases, the resistivity of the material increases. This is because resistivity is inversely proportional to relaxation time (\( \rho \propto \frac{1}{\tau} \)). A shorter relaxation time means more frequent collisions, hindering electron flow and thus increasing resistance.
In simple words: If electrons hit things more often (shorter relaxation time), the material becomes more resistant to electricity, meaning its resistivity goes up.
🎯 Exam Tip: This direct relationship between decreasing relaxation time and increasing resistivity is a key consequence of the microscopic model of conduction.

 

Question 29. What is the variation of resistivity with temperature?
Answer: For conductors, resistivity (\( \rho \)) generally increases with temperature. The relationship can be approximated by \( \rho = \rho_0(1 + \alpha \Delta t) \), where \( \rho_0 \) is the resistivity at a reference temperature, \( \alpha \) is the temperature coefficient of resistance, and \( \Delta t \) is the change in temperature. This is because increased temperature leads to increased thermal vibrations, reducing relaxation time and thus increasing resistivity.
In simple words: For most metals, when it gets hotter, the resistance to electricity (resistivity) goes up because electrons bump into atoms more often.
🎯 Exam Tip: This formula is important for understanding and predicting how material resistance changes with temperature, which is relevant for various applications like thermistors.

 

Question 30. What happens to the resistivity of a conductor when the temperature becomes absolute zero or 0°K?
Answer: When the temperature of a conductor approaches absolute zero (0°K), its resistivity ideally disappears. In some materials called superconductors, resistivity drops abruptly to zero below a certain critical temperature.
In simple words: If a conductor gets extremely cold, down to absolute zero, its ability to resist electricity goes away completely.
🎯 Exam Tip: This concept leads to superconductivity, a fascinating phenomenon with potential for lossless energy transmission.

 

Question 31. What is superconductivity?
Answer: Superconductivity is a phenomenon where certain metals and alloys completely lose their electrical resistance when cooled below a specific critical temperature. For example, in 1911, Dutch physicist Kamerlingh Onnes discovered that purified mercury loses all its resistivity abruptly at 4.2 K. Above this temperature, its resistivity is small but finite. This complete absence of resistance allows current to flow without any energy loss.
In simple words: Superconductivity is when special materials, made very cold, suddenly lose all their resistance to electricity, letting current flow forever without losing power.
🎯 Exam Tip: Superconductivity has significant technological implications, from powerful magnets to highly efficient electrical transmission.

 

Question 32. How can you define temperature coefficient of resistance?
Answer: The temperature coefficient of resistance (\( \alpha \)) is defined as the ratio of the increase in resistance per unit original resistance per degree rise in temperature. Mathematically, if \( R_1 \) is resistance at \( T_1 \) and \( R_2 \) at \( T_2 \), then \( \alpha = \frac{R_2 - R_1}{R_1(T_2 - T_1)} \).
In simple words: The temperature coefficient of resistance tells us how much a material's resistance changes for every degree its temperature goes up, compared to its original resistance.
🎯 Exam Tip: \( \alpha \) helps predict how resistance will vary with temperature. Conductors have a positive \( \alpha \), while semiconductors often have a negative \( \alpha \).

 

Question 33. When the current in the conductor is high, what happens to the conductor?
Answer: When the current in a conductor is high, according to Joule's law of heating (\( H = I^2Rt \)), a significant amount of heat is generated. This causes the conductor to heat up, potentially leading to a rise in its temperature.
In simple words: If too much electricity flows through a wire, the wire gets hot.
🎯 Exam Tip: Excessive heating due to high current can melt wires or damage components, which is why fuses and circuit breakers are used.

 

Question 34. What is an ohmic resistance?
Answer: An ohmic resistance is a resistance that obeys Ohm's law. For such resistances, the voltage (V) across them is directly proportional to the current (I) flowing through them, and their V-I characteristic graph is a straight line passing through the origin.
In simple words: An ohmic resistance is a material where the voltage and current always follow Ohm's law directly.
🎯 Exam Tip: Most metallic conductors behave as ohmic resistances under normal conditions.

 

Question 35. What is a non-ohmic resistance?
Answer: A non-ohmic resistance is a resistance that does not obey Ohm's law. For these materials, the relationship between voltage (V) and current (I) is not linear, meaning their V-I characteristic graph is not a straight line.
In simple words: A non-ohmic resistance is a material where voltage and current do not always follow Ohm's law, so their relationship is not a simple straight line.
🎯 Exam Tip: Examples include semiconductor diodes, vacuum tubes, and electrolytes, where resistance can change with voltage, current, or temperature.

 

Question 36. Ohm's law is not a universal law. Why?
Answer: Ohm's law is not considered a universal law because it is an empirical or experimental law that applies only to certain materials (ohmic conductors) under specific conditions. Many materials, such as semiconductors, vacuum tubes, and electrolytes, do not strictly obey Ohm's law, showing non-linear V-I characteristics.
In simple words: Ohm's law is not for everything because it's based on tests and only works for some materials, not all.
🎯 Exam Tip: While essential, it's crucial to understand the limitations of Ohm's law and the existence of non-ohmic devices.

 

Question 37. In parallel combination, what is the potential difference across each resistor?
Answer: In a parallel combination of resistors, the potential difference (voltage) across each resistor is the same. This is because all components in a parallel circuit are connected across the same two points, and thus experience the same voltage drop.
In simple words: When resistors are connected side-by-side, the voltage push across each one is exactly the same.
🎯 Exam Tip: This is a key characteristic of parallel circuits, while current divides among the branches.

 

Question 38.
(a) What is the effective resistance between A and C?
(b) How much current enters the circuit through A?
(c) What is the current through each resistor?
(d) What is the amount of current leaving point A?
(e) What is the potential difference between points A and C?
(f) How much emf is applied?
(g) What is the value of the potential at B?
(h) What is the value of potential at D?
(i) How much current flows through a galvanometer if it is connected between B and D?
(j) If the resistor between A and D is replaced by 2Ω, what happens to galvanometer deflection?
(k) Can you again reduce the deflection to zero by adjusting resistance between D and C?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वीटस्टोन ब्रिज परिपथ को दर्शाता है जिसमें चार प्रतिरोधक (1Ω प्रत्येक) एक चतुर्भुज के रूप में जुड़े हुए हैं। एक 2V की बैटरी A और C बिंदुओं के बीच जुड़ी है। यह एक संतुलित ब्रिज का उदाहरण है जहाँ B और D के बीच कोई धारा प्रवाहित नहीं होती है।
(a) For the given Wheatstone bridge-like structure with 1Ω resistors, the effective resistance between A and C can be found by treating it as two parallel branches, each with two 1Ω resistors in series. Each branch has \( 1\Omega + 1\Omega = 2\Omega \). So, the equivalent resistance \( R_{AC} = \frac{(1+1)(1+1)}{(1+1)+(1+1)} = \frac{2 \times 2}{2+2} = \frac{4}{4} = 1\Omega \).
(b) The current entering the circuit through A is \( I = \frac{V}{R_{AC}} = \frac{2\text{V}}{1\Omega} = 2\text{A} \).
(c) Since the branches ABC and ADC are symmetrical and have equal resistance (2Ω each), the current divides equally. Current through ABC (\( I_{ABC} \)) = Current through ADC (\( I_{ADC} \)) = \( \frac{2\text{A}}{2} = 1\text{A} \). Thus, current through each 1Ω resistor is 1A.
(d) The amount of current leaving point A is the total current entering, which is 2A.
(e) The potential difference between points A and C is equal to the supply voltage, which is 2V.
(f) The emf applied to the combination is 2V.
(g) The potential at B. Assuming potential at C is 0V and A is 2V. Potential drop across AB is \( I \times R = 1\text{A} \times 1\Omega = 1\text{V} \). So, \( V_B = V_A - 1\text{V} = 2\text{V} - 1\text{V} = 1\text{V} \).
(h) The potential at D. Potential drop across AD is \( I \times R = 1\text{A} \times 1\Omega = 1\text{V} \). So, \( V_D = V_A - 1\text{V} = 2\text{V} - 1\text{V} = 1\text{V} \).
(i) Since \( V_B = V_D = 1\text{V} \), there is no potential difference between B and D. Therefore, no current flows through a galvanometer connected between B and D.
(j) If the resistor between A and D is replaced by 2Ω, the bridge becomes unbalanced. This means \( \frac{R_{AB}}{R_{BC}} \neq \frac{R_{AD}}{R_{DC}} \). Consequently, there will be a potential difference between B and D, and the galvanometer will show a deflection.
(k) Yes, you can again reduce the deflection to zero. If \( R_{AD} \) is changed to 2Ω, the bridge is unbalanced. To re-balance it, the ratio \( \frac{R_{AB}}{R_{BC}} \) must equal \( \frac{R_{AD}}{R_{DC}} \). If \( R_{AB}=1\Omega \), \( R_{BC}=1\Omega \), \( R_{AD}=2\Omega \), then for balance \( \frac{1}{1} = \frac{2}{R_{DC}} \implies R_{DC} = 2\Omega \). So, by adjusting \( R_{DC} \) to 2Ω, the deflection can be brought to zero.
In simple words: (a) The total resistance of this shape of circuit is 1Ω. (b) 2 amperes of current enter the circuit. (c) 1 ampere flows through each of the 1Ω resistors. (d) 2 amperes leave point A. (e) The voltage difference across A and C is 2 volts. (f) The battery supplies 2 volts. (g) The voltage at point B is 1 volt. (h) The voltage at point D is also 1 volt. (i) Because B and D have the same voltage, no current flows through a meter between them. (j) If one resistor changes, the voltages at B and D become different, and the meter will show current. (k) Yes, you can make the meter read zero again by changing another resistor to match the new balance.
🎯 Exam Tip: This extensive problem tests a deep understanding of Wheatstone bridge principles, including series/parallel combinations, current division, potential calculations, and balancing conditions.

 

Question 39.
(a) What is the general value of the potential at B?
(b) What is general value of potential at D?
(c) What is the condition to obtain zero deflection in the galvanometer?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक संतुलित वीटस्टोन ब्रिज परिपथ को दर्शाता है जहाँ एक बैटरी (2V) F और E के बीच जुड़ी है। ब्रिज में चार प्रतिरोधक (P, Q, R, S) हैं और एक गैल्वेनोमीटर (G) B और D के बीच जुड़ा है। यह व्यवस्था ब्रिज के संतुलन की स्थिति को दर्शाती है।
(a) The general value of the potential at B (\( V_B \)) can be expressed as a fraction of the total voltage (V) across the branch, considering the resistors P and Q in series. So, \( V_B = \left(\frac{V}{P+Q}\right)Q \).
(b) Similarly, the general value of the potential at D (\( V_D \)) can be expressed as a fraction of the total voltage (V) across the other branch, considering resistors R and S in series. So, \( V_D = \left(\frac{V}{R+S}\right)S \).
(c) To obtain zero deflection in the galvanometer, the potentials at points B and D must be equal, i.e., \( V_B = V_D \). This leads to the condition for a balanced Wheatstone bridge: \( \frac{P}{R} = \frac{Q}{S} \).
In simple words: (a) The voltage at point B is part of the total voltage, depending on resistors P and Q. (b) The voltage at point D is also part of the total voltage, depending on resistors R and S. (c) For no current to flow through the middle meter, the voltages at B and D must be the same, which means the resistor ratios on both sides must be equal.
🎯 Exam Tip: This question summarizes the core principles of a Wheatstone bridge, especially the balancing condition, which is vital for precise resistance measurements.

 

Question 40. A circuit is shown with a cell of emf E and internal resistance r connected to an external resistance R.
(c) What is the potential difference across the terminals of the cell?
(d) Is it equal to the emf of the cell?
(e) Find the condition under which potential difference and emf are equal?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सेल (जिसका emf E और आंतरिक प्रतिरोध r है) को एक बाहरी प्रतिरोधक (R) से जुड़ा हुआ दिखाता है। यह सरल परिपथ सेल के टर्मिनल वोल्टेज और कुल प्रतिरोध की गणना के लिए आधार प्रदान करता है।
(a) The total resistance of the circuit is the sum of the external resistance (R) and the internal resistance (r) of the cell: Total resistance \( = R + r \).
(b) The current (I) flowing through the circuit is given by Ohm's law applied to the entire circuit: \( I = \frac{E}{R+r} \).
(c) The potential difference across the terminals of the cell (\( V_t \)), also known as terminal voltage, is the voltage available across the external resistance. It is given by \( V_t = IR = E - Ir \). Substituting I, \( V_t = \frac{ER}{R+r} \).
(d) No, the terminal potential difference (\( V_t \)) is generally not equal to the emf (E) of the cell. It is less than the emf due to the voltage drop across the internal resistance (\( Ir \)).
(e) The potential difference across the terminals will be equal to the emf of the cell if the internal resistance (r) of the cell is zero or negligible, or if no current is drawn from the cell (i.e., the circuit is open, \( I = 0 \)). In both cases, \( Ir = 0 \), so \( V_t = E \).
In simple words: (a) The total resistance in the circuit is the outside resistance plus the battery's inside resistance. (b) The current is the battery's total voltage divided by this total resistance. (c) The voltage you measure across the battery's ends (terminal voltage) is the total battery voltage minus the voltage lost inside the battery. (d) No, the terminal voltage is usually less than the battery's own voltage (emf). (e) They are only equal if the battery has no internal resistance or if no current is flowing.
🎯 Exam Tip: Understanding the difference between EMF and terminal voltage, and the role of internal resistance, is fundamental for battery circuits. The condition \( V_t = E \) is important for ideal sources.

 

Question 41. How can you compare the emf of two cells?
Answer: The electromotive forces (emfs) of two cells can be compared using a potentiometer. By finding the balance lengths (\( l_1 \) and \( l_2 \)) for each cell when connected to the potentiometer, their emfs (\( E_1 \) and \( E_2 \)) are related by \( \frac{E_1}{E_2} = \frac{l_1}{l_2} \).
In simple words: You can compare two batteries' strengths using a potentiometer. The lengths on the potentiometer wire where the meter shows zero current tell you which battery is stronger.
🎯 Exam Tip: Potentiometers offer a precise method for comparing emfs without drawing significant current, making them ideal for delicate measurements.

 

Question 42. What is meant by the internal resistance of a cell?
Answer: The internal resistance of a cell is the opposition offered by the electrolytes and electrodes within the cell to the flow of electric current. This resistance causes a drop in the terminal voltage when current is drawn from the cell. A freshly prepared primary cell typically has a low internal resistance.
In simple words: Internal resistance is the push-back from inside the battery itself against the flow of electricity. It makes the battery's actual output voltage a little less than its full potential when used.
🎯 Exam Tip: Internal resistance leads to power loss (heating) within the cell and reduces the terminal voltage when current is supplied.

 

Question 43. Pick the odd one out of the following.
(d) Gauss' law
(e) Faraday's law
Answer: The odd one out is (a) Ohm's law.
In simple words: Ohm's law is about how current flows in wires, while the others are bigger laws about electric and magnetic fields.
🎯 Exam Tip: Ohm's law relates to resistance and current in circuits, while Gauss' law, Ampere's law, and Faraday's law are fundamental laws of electromagnetism (Maxwell's equations).

 

Question. Assertion: In a simple battery circuit, the point at the lowest potential is the positive terminal of the battery. Reason: The electron flows from higher potential to lower potential.
Answer: (d) Both the assertion and reason are false.
In simple words: The statement that the positive terminal is the lowest potential is wrong, and the reason given about electron flow is also wrong.
🎯 Exam Tip: Conventionally, current flows from higher (positive) to lower (negative) potential, while electrons, being negatively charged, flow from lower (negative) to higher (positive) potential.

 

Question 45. Match the following
Answer:

Device	Material	Reason for use
Connecting wire	Copper	Good conductor
Standard resistance	Constantan	Very small temperature coefficient of resistance
Fuse wire	Alloy	Low melting point and high resistivity

In simple words: Connecting wires use copper because it conducts well. Standard resistors use constantan because its resistance does not change much with heat. Fuse wires use an alloy that melts easily and has high resistance to protect circuits.
🎯 Exam Tip: This question tests knowledge of material properties and their practical applications in electrical components.

 

Question 46. Fill in the blanks using the words given in the brackets.
[Free electrons, free holes, drift speed, relaxation time, Ohm's law, critical temperature, ure, r.m.s speed, super temperature, negative, Coulomb's law, time constant]
In a metallic conductor electric current is due to the movement of .......(i)..... Metallic conductors obey ....(ii)..... In superconductors, the resistance drops suddenly to zero at a sufficiently low temperature called ..(iii)... Metals have ....(iv)... temperature coefficient of resistance.
Answer:
(i) In a metallic conductor electric current is due to the movement of **free electrons**.
(ii) Metallic conductors obey **Ohm's law**.
(iii) In superconductors, the resistance drops suddenly to zero at a sufficiently low temperature called **critical temperature**.
(iv) Metals have **positive** temperature coefficient of resistance.
In simple words: (i) Current in metals comes from free electrons moving. (ii) Metals follow Ohm's law. (iii) Superconductors lose resistance below a critical temperature. (iv) Metals get more resistant when hotter (positive temperature coefficient).
🎯 Exam Tip: This fill-in-the-blanks question covers fundamental concepts of conduction, Ohm's law, superconductivity, and temperature dependence of resistance.

 

Question 47. Match the following
Answer:

A	B	C
Drift speed	Electric field	10-3
Relaxation time	Temperature	10-7
Electron mobility in metal	Proportionality constant	104

In simple words: This table matches terms like drift speed and relaxation time with what they are related to or their typical values.
🎯 Exam Tip: Matching questions test your ability to associate concepts with their definitions, factors, or typical magnitudes.

 

Question 48.
(a) S.I unit of elective current is ampere. Then what do you mean by saying current is one ampere?
(b) Give the relation between electric current and current density and their respective units?
Answer:
(a) Saying current is one ampere means that one coulomb of charge drifts through any cross-section of the conductor per second.
(b) Electric current (I) is the total charge flowing per unit time. Current density (\( \hat{j} \)) is the current per unit area, i.e., \( \hat{j} = \frac{I}{A} \). The unit of current is ampere (A), and the unit of current density is ampere per square meter (\( \text{A/m}^2 \)).
In simple words: (a) One ampere means one coulomb of charge passes through a wire every second. (b) Electric current is the total flow, measured in amperes. Current density is how much current flows through a small patch, measured in amperes per square meter.
🎯 Exam Tip: Clearly distinguishing between scalar current and vector current density is important for describing charge flow comprehensively.

 

Question 49.
(a) Name the factors on which resistance of a conductor depends on.
(b) For a given conductor and at a given temperature how the resistance depends on
i. its length
ii. its area of cross-section
iii. From the above derive an expression for resistivity.
Answer:
(a) The resistance of a conductor depends on four main factors: the nature of the material, its length, its area of cross-section, and its temperature.
(b)
i. For a given conductor at a given temperature, resistance (R) is directly proportional to its length (l): \( R \propto l \). This means a longer wire has more resistance.
ii. Resistance (R) is inversely proportional to its area of cross-section (A): \( R \propto \frac{1}{A} \). This means a thicker wire has less resistance.
iii. Combining these two relationships, we get \( R \propto \frac{l}{A} \). To convert this proportionality into an equation, we introduce a constant of proportionality, \( \rho \) (rho), which is called resistivity. Thus, \( R = \rho \frac{l}{A} \). From this, we can derive the expression for resistivity: \( \rho = \frac{RA}{l} \).
In simple words: (a) How much a wire resists electricity depends on what it's made of, how long it is, how thick it is, and how hot it is. (b) (i) Longer wires resist more. (ii) Thicker wires resist less. (iii) Resistivity is a material's inherent ability to resist, calculated from its resistance, area, and length.
🎯 Exam Tip: The formula \( R = \rho \frac{l}{A} \) is one of the most fundamental in current electricity, crucial for calculations and conceptual understanding of resistance.

 

Question 50.
(a) Is Ohm's law applicable to all elements? If not give example.
(b) Is e.m.f. a force or some other physical quantity?
(c) Define e.m.f
(d) What is the source of e.m.f. in a primary cell?
Answer:
(a) No, Ohm's law is not universally applicable to all elements. It strictly holds true only for ohmic conductors (like most metals) under specific conditions. Many elements and devices, such as semiconductors (e.g., semiconductor diodes), vacuum tubes, and some electrolytes, do not obey Ohm's law. These are called non-Ohmic conductors, and their current-voltage (V-I) relationship is non-linear.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक ओमिय प्रतिरोधक (ऊपर) और एक अर्धचालक डायोड (नीचे) के I-V विशेषताओं को दर्शाता है। ओमिय प्रतिरोधक के लिए, I-V ग्राफ एक सीधी रेखा है, जबकि अर्धचालक डायोड के लिए, यह गैर-रेखीय है, जो दर्शाता है कि यह ओम के नियम का पालन नहीं करता है।
For non-Ohmic conductors:
- The relation between V and I is not linear.
- The relation between V and I depends on the sign of V.
- The relation between V and I is not unique; I may have more than one value for a given value of V.
(b) EMF (electromotive force) is not a force in the mechanical sense. Instead, it is a form of energy per unit charge, making it a physical quantity representing potential difference or voltage. It is typically measured in volts.
(c) EMF is defined as the energy required by a source to circulate a unit charge once completely around an electric circuit. It is also equal to the maximum potential difference available across the electrodes of a cell when no current is drawn from it (i.e., in an open circuit).
(d) The source of EMF in a primary cell is chemical energy, which is converted into electrical energy through electrochemical reactions.
In simple words: (a) Ohm's law doesn't work for everything; for example, it doesn't work for simple electronic parts like diodes. (b) EMF is not a push force, but rather the energy given to each unit of charge, like a voltage. (c) EMF is the total push a battery gives to make electricity flow, measured when nothing is connected. (d) A basic battery makes electricity from chemical reactions.
🎯 Exam Tip: This comprehensive question covers the scope of Ohm's law, the definition and nature of EMF, and its origin in primary cells, which are all crucial concepts in current electricity. Understanding the V-I characteristics of ohmic and non-ohmic devices is essential.

 

Question 51. A cell with an electromotive force (emf) of 1.5 V and an internal resistance of 0.5 Ω is connected to a non-linear conductor. The V-I graph for this conductor is shown in the figure. Graphically determine the current drawn from the cell and its terminal voltage.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक V-I ग्राफ दिखाता है जिसमें x-अक्ष पर वोल्टेज (V) और y-अक्ष पर धारा (I) है। इसमें एक वक्र है जो एक नॉन-लीनियर कंडक्टर का प्रतिनिधित्व करता है और एक सीधी रेखा है जो सेल की विशेषताओं को दर्शाती है। इन दोनों रेखाओं का प्रतिच्छेदन बिंदु सेल से खींची गई धारा और उसके टर्मिनल वोल्टेज को दर्शाता है।
Answer:We are given the cell's emf \( \epsilon = 1.5 \, V \) and internal resistance \( r = 0.5 \, \Omega \). Let the current be I. The terminal voltage V across the cell is given by the formula: \( \epsilon - IR = V \) Therefore, \( \epsilon - V = Ir \) Substituting the given values: \( 1.5 - V = I \times 0.5 \) This equation \( 1.5 - V = 0.5I \) represents a straight line for the cell. To find the operating point, we plot this straight line on the given V-I graph of the non-linear conductor. From the provided graph information, when V = 1.5 V, I = 0 A. And when V = 0 V, I = 3 A. By plotting this line on the conductor's V-I graph, the point where the line intersects the V-I curve will give the operating voltage and current. Graphically, the intersection point indicates an operating voltage of approximately 0.9 V and a current of 1.2 A.In simple words: We draw a line on the conductor's graph that shows how the cell's voltage changes with current. Where this line crosses the conductor's curve, we find the amount of current flowing and the voltage across the cell.

🎯 Exam Tip: When solving graphical problems, accurately plotting the load line (cell's V-I characteristic) on the device's V-I characteristic curve is crucial. The intersection point is the operating point and must be read carefully from the graph.

 

Question 52. The resistivity of copper, constantan and silver are \( 1.7 \times 10^{-6} \, \Omega \, \text{cm} \), \( 39.1 \times 10^{-6} \, \Omega \, \text{cm} \) and \( 10^{-6} \, \Omega \, \text{cm} \) respectively.
(a) Which is the best conductor?
(b) Give reason for your answer.
(c) Define resistivity.
(d) Which material is used for potentiometer wire? Why?
Answer:(a) Silver (b) Silver is the best conductor because it has the lowest resistivity (symbolized by \( \rho \)) among the given materials. A lower resistivity means less opposition to current flow. (c) Resistivity is a property of a material that shows how strongly it resists electric current. It is defined as the resistance of a conductor made from that material, having unit length and unit cross-sectional area. (d) Constantan is typically used for potentiometer wires. The reason for this choice is that constantan has a very small temperature coefficient of resistance. This means its resistivity changes very little with variations in temperature, making it reliable for accurate measurements.In simple words: Silver is best for carrying electricity because it resists current the least. Resistivity tells us how much a material fights against electricity. Constantan is good for specific uses like potentiometers because its resistance stays almost the same even when the temperature changes.

🎯 Exam Tip: Remember that materials with lower resistivity are better conductors. For precision instruments like potentiometers, materials with a low temperature coefficient of resistance are preferred to ensure stable readings unaffected by temperature fluctuations.

 

Question 53.
1. If a person touches a live wire, will he get stuck to it?
2. Why do we use copper wires as connecting wires?
3. A wire is carrying a current. Is it charged? If not, why?
Answer:1. Sometimes, a person might get stuck to a live wire. Even very small electric currents can disrupt the body's nervous system. This can temporarily stop a person from being able to control their muscles, making it hard to let go of the high-voltage point. 2. Copper wires are used as connecting wires because copper is an excellent conductor of electricity, meaning it has low resistivity. Additionally, it can be easily shaped into various sizes and forms, making it practical for wiring. 3. No, a wire carrying current is not charged. The net charge in the wire remains zero even when current flows. This is because current is the movement of free electrons, and for every electron that enters one end of the wire, another electron leaves the other end, keeping the total number of charges balanced.In simple words: Touching a live wire can sometimes make you unable to let go because electricity affects your nerves. Copper wires are used for connections because they let electricity flow easily and can be shaped. A wire with current is not charged because as electrons flow in, other electrons flow out, keeping it balanced.

🎯 Exam Tip: Understand the difference between electric current (flow of charge) and net charge. While current involves moving charges, a conductor maintains electrical neutrality overall.

 

Question 54. Light from a bathroom bulb gets dimmer for a moment when a geyser is switched on. Why?
Answer: When a geyser is switched on, the light from a bathroom bulb dims temporarily. This happens because a geyser typically has much lower resistance than a bulb. When the geyser is turned on, it draws a significantly larger amount of current from the main power supply compared to the bulb. This sudden increase in current causes a larger voltage drop across the internal resistance of the household wiring (and the supply itself), leading to a momentary decrease in the voltage available to the bulb, thus making it dimmer.In simple words: When you turn on a geyser, it uses a lot of electricity. This causes the total voltage supplied to your house to drop a little for a moment, making the light bulb connected to the same circuit appear dimmer.

🎯 Exam Tip: This phenomenon illustrates the concept of voltage drop due to internal resistance in a circuit. When a high-power appliance is connected, it draws more current, increasing the voltage drop across the supply's internal resistance, and reducing the terminal voltage for other parallel components.

 

Question 55. The figure above shows four colored bands (1, 2, 3, and 4) on a resistor.
(a) What is the meaning of each number (band)?
(b) Why is such a system used?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक रेज़िस्टर दिखाता है जिस पर 1, 2, 3 और 4 क्रमांक के चार रंगीन बैंड अंकित हैं। ये बैंड रेज़िस्टर के प्रतिरोध मान और उसकी सहनशीलता (टॉलरेंस) को दर्शाने के लिए उपयोग किए जाते हैं, जो कि इलेक्ट्रॉनिक घटकों के लिए एक मानक रंग कोडिंग प्रणाली का हिस्सा है।
Answer:(a) The meaning of each colored band on the resistor is as follows:
(1) The first colored band indicates the first significant figure of the resistance value.
(2) The second colored band indicates the second significant figure of the resistance value.
(3) The third colored band represents the multiplier, which is a power of ten to be multiplied by the significant figures.
(4) The fourth colored band indicates the tolerance, which specifies how much the actual resistance value can vary from the marked value. If there is no fourth band, the tolerance is usually considered to be \( \pm 20\% \). (b) This color coding system is used for several reasons: Firstly, resistors can be very small, making it difficult to print their resistance values directly on them in clear numbers. Secondly, there is a need for a universally accepted standard method for marking and measuring resistance values, ensuring consistency across different manufacturers and regions. Therefore, for convenience and clarity, the color coding system is employed.In simple words: The colored bands on a resistor tell us its value: the first two bands give the numbers, the third band is a multiplier, and the fourth band shows how accurate the resistor is. This system is used because resistors are often too small to print numbers on them clearly and to have a standard way to show their values.

🎯 Exam Tip: Memorize the color code chart for resistors (Black=0, Brown=1, Red=2, Orange=3, Yellow=4, Green=5, Blue=6, Violet=7, Gray=8, White=9) and their respective multipliers and tolerance values (Gold=\( \pm 5\% \), Silver=\( \pm 10\% \), No band=\( \pm 20\% \)).

 

Question 56. Two potentiometers are shown in figures (P1) and (P2). The first one has a 6 m length and the second one 10 m.
(a) Which one do you prefer while finding the internal resistance of a primary cell? Why?
(b) Will the balance point be affected by the internal resistance of the given cell?
(c) Suppose you connect the two potentiometers in series, will the combination give you a better accuracy?
(d) What important precaution should you take while using the modified connection mentioned in (c)?
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में दो अलग-अलग पोटेंशियोमीटर तारों को दर्शाया गया है, जिन्हें (P1) और (P2) नाम दिया गया है। पहले पोटेंशियोमीटर की तार की लंबाई 6 मीटर है जबकि दूसरे की लंबाई 10 मीटर है। ये पोटेंशियोमीटर बैटरी के ईएमएफ और आंतरिक प्रतिरोध को मापने के लिए उपयोग किए जाते हैं, और तार की लंबाई उनकी संवेदनशीलता को प्रभावित करती है।
Answer:(a) To find the internal resistance of a primary cell, you should prefer the potentiometer with a length of 10 m. This is because a longer potentiometer wire makes the instrument more sensitive, allowing for more precise balance point measurements. (b) Yes, the balance point will be affected by the internal resistance of the given cell. The balance point is achieved when the potential drop across a segment of the potentiometer wire equals the terminal voltage of the cell, which itself is affected by the internal resistance. (c) If you connect the two potentiometers in series, the total length of the potentiometer wire increases (16 m). A longer wire generally leads to a smaller potential gradient along the wire, which in turn results in higher sensitivity and thus better accuracy for measurements. (d) When using the modified connection (potentiometers in series), an important precaution is to ensure that both potentiometers are made of the same material and have a uniform resistance per unit length. This consistency is crucial for maintaining a linear potential gradient across the entire combined length and for accurate measurements.In simple words: For better measurements, a longer potentiometer wire is better because it is more sensitive. The internal resistance of the cell changes where the balance point is found. Connecting two potentiometers together can make measurements more accurate, but you must make sure they are similar and have the same resistance all along their length.

🎯 Exam Tip: Potentiometer sensitivity is directly proportional to its length. A longer wire allows for a smaller potential gradient, leading to a more precise determination of the balance point.

 

Question 57. What is conventional current?
Answer: Conventional current is defined as the flow of positive charge carriers. It flows from the positive terminal to the negative terminal of a cell in the external circuit.In simple words: Conventional current is imagined to be the flow of positive charge from the positive side to the negative side of a battery in a circuit.

🎯 Exam Tip: Always remember that conventional current direction is opposite to the actual flow of electrons in a metallic conductor.

 

Question 58. What is the significance of the direction of the current?
Answer: The direction of conventional current is opposite to the direction in which electrons actually flow. In simple terms, conventional current flows from the positive terminal to the negative terminal. On the other hand, the direction of electronic current (actual electron flow) is from the negative terminal to the positive terminal. The significance lies in establishing a standard for circuit analysis, even though the charge carriers in metals are negative electrons.In simple words: The direction of current is important because it sets a standard for how we talk about electricity. We usually say current flows from positive to negative, even though tiny electrons actually move from negative to positive.

🎯 Exam Tip: Distinguish between conventional current (positive charge flow) and electronic current (electron flow), understanding that they are in opposite directions. This distinction is fundamental in circuit theory.

 

Question 59. How does drift velocity of electrons in a metallic conductor vary with temperature?
Answer: The drift velocity of electrons in a metallic conductor decreases as the temperature increases. This is because, at higher temperatures, the atoms in the conductor vibrate more vigorously. These increased vibrations cause more frequent collisions between the free electrons and the vibrating atoms, which reduces the average time an electron can move freely before colliding (relaxation time). A shorter relaxation time leads to a lower average drift velocity.In simple words: When a metal gets hotter, electrons move slower on average because they bump into vibrating atoms more often. So, hotter means slower electron drift.

🎯 Exam Tip: Understand that increased temperature leads to higher atomic vibrations, reducing the mean free path and relaxation time of electrons, which in turn decreases drift velocity and increases resistance.

 

Question 60. How is p.d (V) related to drift velocity (vd)?
Answer: The potential difference (V) is related to the drift velocity (vd) through the following equations: The drift velocity is given by \( v_d = \frac{eE\tau}{m} \), where e is the electron charge, E is the electric field, \( \tau \) is the relaxation time, and m is the electron mass. The electric field E across a conductor of length l with a potential difference V is given by \( E = \frac{V}{l} \). Substituting E into the drift velocity equation: \( v_d = \frac{e(V/l)\tau}{m} \)
\( \implies v_d = \frac{eV\tau}{ml} \) From this, we can see that the drift velocity \( v_d \) is directly proportional to the potential difference V, assuming other factors are constant: \( v_d \propto V \).In simple words: The speed at which electrons drift in a wire is directly linked to the voltage applied across it. If you increase the voltage, the electrons will drift faster.

🎯 Exam Tip: Remember the direct relationship between drift velocity and electric field (or potential difference). Also, recall the inverse relationship with mass and the direct relationship with charge and relaxation time.

 

Question 61. Define resistivity.
Answer: Resistivity is a basic property of a material that shows how much it opposes the flow of electric current. It is specifically defined as the resistance of a conductor made from that material, having a unit length (e.g., 1 meter) and a unit cross-sectional area (e.g., 1 square meter).In simple words: Resistivity tells us how much a specific material resists electricity. Imagine a small cube of that material; its resistance is its resistivity.

🎯 Exam Tip: Resistivity is an intrinsic property of the material, independent of its dimensions, unlike resistance which depends on the length and cross-sectional area of the conductor.

 

Question 62. Why are connecting wires made of copper?
Answer: Connecting wires are commonly made of copper for several important reasons. Firstly, copper has very low resistivity, which means it is an excellent conductor and allows electric current to flow with minimal resistance and energy loss. Secondly, copper is a diamagnetic material. This property means it is not significantly magnetized when current flows through it, which is desirable in many electrical applications.In simple words: Copper is used for wires because it lets electricity flow very easily and does not become magnetic when current passes through it.

🎯 Exam Tip: Key properties for good connecting wires are high electrical conductivity (low resistivity) and ductility (ability to be drawn into thin wires). Copper excels in both.

 

Question 63. What is a thermistor?
Answer: A thermistor is a special type of resistor whose electrical resistance changes significantly and predictably with changes in temperature. They are commonly made from semiconductor materials and can have either a positive (resistance increases with temperature) or negative (resistance decreases with temperature) temperature coefficient.In simple words: A thermistor is a resistor whose resistance changes a lot when its temperature changes.

🎯 Exam Tip: Thermistors are vital components in temperature sensing, measurement, and control applications due to their high sensitivity to temperature changes.

 

Question 64. A wire is carrying current. Is it charged? Give reason.
Answer: No, a wire carrying current is not charged. The reason is that current is the flow of free electrons within the conductor. For every electron that enters the wire from the battery, another electron leaves the wire to the battery at any given instant. This means that the total number of negative charges (electrons) within the wire remains balanced by the total number of positive charges (atomic nuclei). Therefore, the net electric charge of the wire is always zero.In simple words: A wire carrying electricity is not charged. This is because for every electron that moves into the wire, another electron moves out, keeping the total number of positive and negative charges perfectly balanced inside the wire.

🎯 Exam Tip: Understand that current is a movement of charge, not an accumulation of net charge. Conductors remain electrically neutral even while current flows.

 

Question 65. Two wires of equal lengths, one of copper and the other of manganin have equal resistance values. Which wire is thicker? Explain.
Answer: We are given two wires of equal length (l) and equal resistance (R), one made of copper and the other of manganin. Let's denote the resistivity of copper as \( \rho_1 \) and its cross-sectional area as \( A_1 \). Let's denote the resistivity of manganin as \( \rho_2 \) and its cross-sectional area as \( A_2 \). The resistance of a wire is given by \( R = \rho \frac{l}{A} \). Since the resistances are equal: \( R_1 = R_2 = R \)
So, \( \rho_1 \frac{l}{A_1} = \rho_2 \frac{l}{A_2} \) Since the lengths (l) are also equal, we can simplify this to: \( \frac{\rho_1}{A_1} = \frac{\rho_2}{A_2} \)
This implies \( \frac{A_2}{A_1} = \frac{\rho_2}{\rho_1} \) We know that copper has a lower resistivity than manganin ( \( \rho_1 < \rho_2 \) ). Since \( \rho_1 < \rho_2 \), it must be that \( A_1 < A_2 \). Therefore, the manganin wire must be thicker than the copper wire to have the same resistance, given its higher resistivity.In simple words: Copper resists electricity less than manganin. If both wires are the same length and have the same total resistance, the manganin wire must be thicker to allow electricity to flow as easily as through the thinner copper wire.

🎯 Exam Tip: Remember the relationship \( R = \rho \frac{l}{A} \). For a fixed length and resistance, a material with higher resistivity must have a larger cross-sectional area (be thicker).

 

Question 66. Give the color code of a resistor with a value of 240 MΩ and a tolerance of 10%.
Answer: We need to determine the color bands for a resistor with a value of 240 MΩ \( \pm 10\% \). First, convert 240 MΩ to ohms: \( 240 \, \text{M}\Omega = 240 \times 10^6 \, \Omega = 24 \times 10^7 \, \Omega \). Now, let's break down the value for the color bands:
- The first significant figure is 2, which corresponds to the color Red.
- The second significant figure is 4, which corresponds to the color Yellow.
- The multiplier is \( 10^7 \), which corresponds to the color Violet.
- The tolerance is \( \pm 10\% \), which corresponds to the color Silver. So, the color code for a 240 MΩ \( \pm 10\% \) resistor is Red, Yellow, Violet, Silver.In simple words: To find the resistor's color code for 240 MΩ with 10% tolerance, we look up the colors for 2 (Red), 4 (Yellow), and \( 10^7 \) (Violet) for the value, and Silver for the 10% accuracy.

🎯 Exam Tip: Mastering the resistor color code is essential. Practice with various resistance values and tolerances to quickly identify the corresponding color bands.

 

Question 67. Two students X and Y perform an experiment on a potentiometer separately using the circuit diagram shown here. Keeping other things unchanged:
(i) Student X increases the value of resistance R.
(ii) Student Y decreases the value of resistance S in the setup. How would these changes affect the position of a null point in each case and why?
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक पोटेंशियोमीटर परिपथ दिखाया गया है जिसमें एक मुख्य सेल (E) और एक बाहरी प्रतिरोध (R) लगा है। एक दूसरा सेल (E') एक गैल्वेनोमीटर (G) और एक स्लाइडिंग संपर्क के साथ पोटेंशियोमीटर तार पर जुड़ा हुआ है, जिससे नल बिंदु की स्थिति का पता लगाया जाता है। आंतरिक प्रतिरोध (S) भी परिपथ में दर्शाया गया है।
Answer:(i) When student X increases the value of resistance R: If resistance R in the main circuit (driver cell circuit) is increased, the total current flowing from the driver cell decreases. Consequently, the potential gradient (potential drop per unit length) across the potentiometer wire also decreases. A smaller potential gradient means that a longer length of the potentiometer wire will be needed to balance the emf of the secondary cell. Therefore, the balance point will shift towards the right end of the potentiometer wire. (ii) When student Y decreases the value of resistance S: The resistance S mentioned here refers to the internal resistance of the cell whose emf is being measured. If this refers to a series resistance in the galvanometer circuit (as usually done with a shunt resistance for safety or internal resistance of the source cell), decreasing it would not typically shift the balance point itself, which is determined by the EMF of the cell and potential gradient of the wire. However, if S refers to a series resistor in the main circuit with the driver cell, decreasing S would increase the main current, thereby increasing the potential gradient and shifting the balance point to the left. Assuming S is the internal resistance of the cell being measured, it affects the terminal voltage but the null point is for EMF in open circuit. If S refers to a protective resistance *in series with the galvanometer*, decreasing it would increase the sensitivity of the galvanometer, making it easier to find the exact null point, but not shifting its position. If S is an external variable resistance used in series with the cell being measured *to determine internal resistance*, then decreasing S will cause a larger current to flow through the cell, reducing its terminal voltage, which in turn would shift the balance point to the left if measuring terminal voltage. Without a clear diagram for 'S', interpreting it as a resistance in the driver circuit is common, which would increase the main current and increase potential gradient, thus shifting the balance point towards the left.In simple words: If you increase resistance R, less current flows, the voltage drop per length of wire decreases, and the balance point moves to the right. If you decrease resistance S (assuming it's in the main circuit), more current flows, the voltage drop per length increases, and the balance point moves to the left.

🎯 Exam Tip: The balance point in a potentiometer depends on the potential gradient across the wire. Any change that alters this gradient (like changing the main circuit resistance) will shift the balance point. Internal resistance of the measured cell affects its terminal voltage, but the null point for EMF is generally in open circuit.

 

Question 68. A voltage of 30V is applied across a resistor with first, second, and third rings of blue, black, and yellow colors respectively. Find the value of current through the resistor.
Answer:First, we need to determine the resistance value from its color bands:
- First band: Blue (corresponds to digit 6)
- Second band: Black (corresponds to digit 0)
- Third band: Yellow (corresponds to multiplier \( 10^4 \)) So, the resistance R = \( 60 \times 10^4 \, \Omega = 600,000 \, \Omega = 6 \times 10^5 \, \Omega \). The applied voltage V = 30 V. Using Ohm's Law, I = \( \frac{V}{R} \):
\( I = \frac{30 \, \text{V}}{60 \times 10^4 \, \Omega} \)
\( I = \frac{30}{600000} \, \text{A} \)
\( I = \frac{1}{20000} \, \text{A} \)
\( I = 0.00005 \, \text{A} \)
\( I = 0.5 \times 10^{-4} \, \text{A} \) Therefore, the current through the resistor is \( 0.5 \times 10^{-4} \, \text{A} \).In simple words: First, we use the color bands (Blue, Black, Yellow) to find the resistor's value, which is 600,000 ohms. Then, using Ohm's law (Current = Voltage / Resistance) with the given voltage of 30V, we calculate the current flowing through it.

🎯 Exam Tip: Always remember to correctly interpret the resistor color code (significant figures, multiplier, and tolerance) and apply Ohm's Law accurately for calculations.

 

Question 69. One end of an aluminum wire whose diameter is 3 mm is welded to one end of a copper wire of diameter 2 mm. The composite wire carries a steady current of 20 mA. What is the current density in each wire?
Answer:Given data: Total steady current, I = 20 mA = \( 20 \times 10^{-3} \, \text{A} \). For the Aluminum (Al) wire: Diameter = 3 mm, so radius \( r_1 = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m} \). Cross-sectional area \( A_1 = \pi r_1^2 = 3.14 \times (1.5 \times 10^{-3})^2 \, \text{m}^2 = 3.14 \times 2.25 \times 10^{-6} \, \text{m}^2 = 7.065 \times 10^{-6} \, \text{m}^2 \). Current density in Al wire, \( J_1 = \frac{I}{A_1} = \frac{20 \times 10^{-3} \, \text{A}}{7.065 \times 10^{-6} \, \text{m}^2} \).
\( J_1 \approx 2.83 \times 10^3 \, \text{A/m}^2 \). For the Copper (Cu) wire: Diameter = 2 mm, so radius \( r_2 = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \). Cross-sectional area \( A_2 = \pi r_2^2 = 3.14 \times (1 \times 10^{-3})^2 \, \text{m}^2 = 3.14 \times 1 \times 10^{-6} \, \text{m}^2 = 3.14 \times 10^{-6} \, \text{m}^2 \). Current density in Cu wire, \( J_2 = \frac{I}{A_2} = \frac{20 \times 10^{-3} \, \text{A}}{3.14 \times 10^{-6} \, \text{m}^2} \).
\( J_2 \approx 6.37 \times 10^3 \, \text{A/m}^2 \).In simple words: We first find the cross-sectional area for each wire using its diameter. Since the same total current flows through both wires, we divide the current by each wire's area to find its current density. The thinner copper wire will have a higher current density than the thicker aluminum wire.

🎯 Exam Tip: Remember that when wires are connected in series, the current flowing through them is the same, but the current density will vary inversely with the cross-sectional area. Always convert units to SI (meters, amperes) for consistent calculations.

 

Question 70. Twelve equal resistors, each of resistance 12 Ω, form the edges of a cube as shown. A battery of emf 24 volts is connected across the diagonally opposite corners of this cube. Determine the equivalent resistance and current through each edge.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक घन के आकार में जुड़े हुए 12 रेज़िस्टर दिखाता है, जहाँ प्रत्येक रेज़िस्टर घन के एक किनारे को दर्शाता है। एक बैटरी घन के दो विपरीत कोनों (मान लीजिए A और E) के बीच जुड़ी हुई है। यह व्यवस्था कुल समतुल्य प्रतिरोध और प्रत्येक किनारे से बहने वाली धारा को निर्धारित करने के लिए है, जिसमें धारा के वितरण की समरूपता का उपयोग किया गया है।
Answer:Given: Resistance of each resistor, R = 12 Ω. Emf of the battery = 24 V. When a battery is connected across diagonally opposite corners of the cube (e.g., A and E), due to the symmetry of the cube, the current distributes itself in a specific pattern.
- Current through the paths originating from A (AB, AD, AK) is equal. Let this be I/3, where I is the total current.
- Current through the paths that are one step away from A (BC, BL, DC, DN, KL, KN) is equal. Let this be I/6.
- Current through the paths converging to E (CM, LM, NM) is equal. Let this also be I/3. Let \( R_{eq} \) be the equivalent resistance of the entire cube. Using Kirchhoff's voltage law for the mesh ABCMEA (or any symmetrical path from A to E):
\( (\frac{I}{3})R + (\frac{I}{6})R + (\frac{I}{3})R = E \)
\( I R (\frac{1}{3} + \frac{1}{6} + \frac{1}{3}) = E \)
\( I R (\frac{2+1+2}{6}) = E \)
\( I R (\frac{5}{6}) = E \) So, \( \frac{5IR}{6} = E \) The equivalent resistance \( R_{eq} = \frac{E}{I} \). From the equation, \( R_{eq} = \frac{5R}{6} \). Given R = 12 Ω,
\( R_{eq} = \frac{5}{6} \times 12 \, \Omega = 10 \, \Omega \). The total current drawn from the battery, I = \( \frac{E}{R_{eq}} = \frac{24 \, \text{V}}{10 \, \Omega} = 2.4 \, \text{A} \). Now, determine the current through each edge:
- Current through branches AB, AD, AK: \( I_1 = \frac{I}{3} = \frac{2.4 \, \text{A}}{3} = 0.8 \, \text{A} \).
- Current through branches BC, BL, DC, DN, KL, KN: \( I_2 = \frac{I}{6} = \frac{2.4 \, \text{A}}{6} = 0.4 \, \text{A} \).
- Current through branches CM, LM, NM: \( I_3 = \frac{I}{3} = \frac{2.4 \, \text{A}}{3} = 0.8 \, \text{A} \).In simple words: We use the cube's symmetry to figure out how the total current splits into different paths. By applying Kirchhoff's law, we find that the total resistance of the cube is \( \frac{5}{6} \) of one resistor's value. Then, we calculate the total current and finally, the current through each set of identical edges.

🎯 Exam Tip: For symmetrical circuits like a cube, identifying symmetrical current paths simplifies calculations significantly. Kirchhoff's laws are fundamental, but symmetry can provide shortcuts for current distribution.

 

Question 71. 12 cells, each having the same emf \( \epsilon \), are connected in series and kept in a closed box. Some of the cells are wrongly connected. The battery is in series with an ammeter and two cells of identical nature to others. The current is 3A when external cells aid the battery and 2A when they oppose. How many cells are wrongly connected in the battery?
Answer:Let N = 12 be the total number of cells. Let n be the number of cells wrongly connected. When a cell is wrongly connected in series, it opposes the emf of the other cells, effectively reducing the net emf by \( 2\epsilon \) for each wrongly connected cell. So, the effective emf of the 12 cells in the box = \( (12 - n) \epsilon - n \epsilon = (12 - 2n) \epsilon \). Let \( r_c \) be the internal resistance of each cell. The total internal resistance of the 12 cells is \( 12r_c \). Case 1: Two external cells aid the battery. Total cells = 12 (from box) + 2 (external) = 14 cells. Total emf = \( (12 - 2n)\epsilon + 2\epsilon = (14 - 2n)\epsilon \). Total internal resistance = \( 12r_c + 2r_c = 14r_c \). Let R be the external resistance in the circuit. Current \( I_1 = 3 \, \text{A} \). Using Ohm's law: \( I_1 = \frac{(14 - 2n)\epsilon}{R + 14r_c} \)
\( 3 = \frac{(14 - 2n)\epsilon}{R_{total}} \) ... (1) (Here, \( R_{total} \) represents \( R + 14r_c \)) Case 2: Two external cells oppose the battery. Total cells = 12 (from box) + 2 (external) = 14 cells. Total emf = \( (12 - 2n)\epsilon - 2\epsilon = (10 - 2n)\epsilon \). Total internal resistance = \( 12r_c + 2r_c = 14r_c \). Current \( I_2 = 2 \, \text{A} \). Using Ohm's law: \( I_2 = \frac{(10 - 2n)\epsilon}{R + 14r_c} \)
\( 2 = \frac{(10 - 2n)\epsilon}{R_{total}} \) ... (2) Now, divide equation (1) by equation (2):
\( \frac{3}{2} = \frac{(14 - 2n)\epsilon / R_{total}}{(10 - 2n)\epsilon / R_{total}} \)
\( \frac{3}{2} = \frac{14 - 2n}{10 - 2n} \) Cross-multiply:
\( 3(10 - 2n) = 2(14 - 2n) \)
\( 30 - 6n = 28 - 4n \) Rearrange terms to solve for n:
\( 30 - 28 = 6n - 4n \)
\( 2 = 2n \)
\( n = 1 \) Therefore, there is 1 cell wrongly connected in the battery.In simple words: We have 12 cells, some facing the wrong way. We add two more cells in series, first helping and then opposing. By comparing the current in both situations, we can calculate that only one cell out of the original 12 was connected incorrectly.

🎯 Exam Tip: For cells in series, remember that wrongly connected cells reduce the net emf by \( 2\epsilon \) and that the total internal resistance simply adds up. Apply Ohm's law to form simultaneous equations.

 

Question 72. A cell of emf E and internal resistance r is connected to an external resistance R. Show that at maximum power transfer, R = r.
Answer:Consider a cell with emf E and internal resistance r connected to an external resistance R. The total resistance in the circuit is \( R_{total} = R + r \). The current flowing in the circuit is \( I = \frac{E}{R + r} \). The power (P) delivered to the external resistance R is given by \( P = I^2 R \). Substitute the expression for I:
\( P = \left(\frac{E}{R + r}\right)^2 R \)
\( P = \frac{E^2 R}{(R + r)^2} \) To find the condition for maximum power transfer, we need to differentiate P with respect to R and set the derivative to zero: \( \frac{dP}{dR} = 0 \).
\( \frac{dP}{dR} = E^2 \frac{d}{dR} \left( \frac{R}{(R + r)^2} \right) \) Using the quotient rule \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u = R \) and \( v = (R+r)^2 \).
\( u' = 1 \)
\( v' = 2(R+r) \times 1 = 2(R+r) \) So, \( \frac{dP}{dR} = E^2 \frac{1 \cdot (R + r)^2 - R \cdot 2(R + r)}{(R + r)^4} \) Set \( \frac{dP}{dR} = 0 \):
\( \implies \frac{(R + r)^2 - 2R(R + r)}{(R + r)^4} = 0 \) Since \( (R+r)^4 \) cannot be zero (unless E is zero, in which case power is zero), the numerator must be zero:
\( (R + r)^2 - 2R(R + r) = 0 \) Factor out \( (R + r) \):
\( (R + r) [ (R + r) - 2R ] = 0 \) Since \( R+r \neq 0 \) for a practical circuit, we must have:
\( (R + r) - 2R = 0 \)
\( R + r - 2R = 0 \)
\( r - R = 0 \)
\( R = r \) Thus, maximum power is transferred to the external resistance when the external resistance (R) is equal to the internal resistance (r) of the cell.In simple words: For a cell to deliver the most power to an outside resistor, the resistance of that outside resistor must be the same as the cell's own internal resistance.

🎯 Exam Tip: The maximum power transfer theorem (R=r) is a key concept in electrical engineering. Be able to derive it using differentiation, and understand its implications for circuit design and efficiency.

 

Question 73. One section AB of a circuit absorbs 53.0 W of power when a current i = 1.20 A passes through it in the direction shown.
(a) Find the potential difference between A and B.
(b) If element C does not have an internal resistance, what is its emf?
(c) Which terminal left or right is positive?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक परिपथ खंड AB को दर्शाता है जिसमें एक प्रतिरोध (19.0 Ω) और एक सेल (C) जुड़ा हुआ है। धारा i (1.20 A) बिंदु A से बिंदु B की ओर प्रवाहित होती है। इस खंड द्वारा अवशोषित शक्ति 53.0 W है।
Answer:Given: Power absorbed, P = 53.0 W. Current, i = 1.20 A. The section AB consists of a resistance R = 19.0 Ω and an element C, which we can assume is a cell with internal resistance 'r' if it's absorbing power and behaving like a charging battery or a resistor with an internal emf opposing current. Let's assume element C is a cell with an emf \( \epsilon \) and internal resistance 'r'. The power absorbed by the section AB is the sum of power dissipated in the resistance and power absorbed by the cell if it's being charged, or the difference if it's supplying power. Given it *absorbs* power, it acts like a load. The total power absorbed in section AB is given by \( P = i^2 (R + r) \) if C is just a resistance, or \( P = iV_{AB} \). The problem statement implies C is a cell, so the total power absorbed is \( P = i^2R + i\epsilon_c \), where \( \epsilon_c \) is the emf of cell C. From the provided solution, it seems that C is modeled as an internal resistance 'r' along with the given 19.0 Ω. The solution implies 'r' is the internal resistance of 'C'. So, let the total effective resistance of element C be r. Total resistance in the path = \( 19.0 \, \Omega + r \). Power absorbed P = 53.0 W. Current i = 1.20 A. From the solution's steps: \( P = i^2 (R_{external} + r_{internal}) \) \( 53.0 = (1.20)^2 (19.0 + r) \) \( 53.0 = 1.44 (19.0 + r) \) \( \frac{53.0}{1.44} = 19.0 + r \) \( 36.8055 \approx 19.0 + r \) \( r = 36.8055 - 19.0 \) \( r \approx 17.8 \, \Omega \) (This value for 'r' is calculated based on the assumption that C is another resistance, making R_total = 19.0 + r). (a) Potential difference between A and B: \( V_{AB} = i (R + r) \) (where R is 19.0 Ω and r is the effective resistance calculated above for C). \( V_{AB} = 1.2 (19.0 + 17.8) \) \( V_{AB} = 1.2 (36.8) \) \( V_{AB} = 44.16 \, \text{V} \approx 44.2 \, \text{V} \) (b) If element C does not have an internal resistance, what is its emf? In this case, element C would solely be an emf \( \epsilon \). The power absorbed would be \( P = i^2 R_{19.0\Omega} + i\epsilon \). \( 53.0 = (1.20)^2 (19.0) + 1.20 \epsilon \) \( 53.0 = 1.44 \times 19.0 + 1.20 \epsilon \) \( 53.0 = 27.36 + 1.20 \epsilon \) \( 1.20 \epsilon = 53.0 - 27.36 \) \( 1.20 \epsilon = 25.64 \) \( \epsilon = \frac{25.64}{1.20} \) \( \epsilon \approx 21.37 \, \text{V} \) (c) Which terminal left or right is positive? Since the current is directed from left (A) to right (B), and power is being absorbed by the section, this means that point A is at a higher potential than point B. Therefore, the left terminal (A) is positive relative to the right terminal (B).In simple words: We first find the total resistance of the section using the power absorbed and the current. Then, we calculate the voltage across the section. If the element 'C' has no internal resistance, we find its voltage (emf) using the same power formula. Since current flows from left to right, the left side (A) has a higher voltage, meaning it's the positive terminal.

🎯 Exam Tip: For power absorption, remember that \( P = iV \) for the entire section. If a cell is part of the absorbing load (e.g., charging), its emf contributes to the total power absorbed. Current always flows from higher to lower potential.