GSEB Class 10 Science Solutions Chapter 12 Electricity

Gujarat Board Class 10 Science Electricity InText Questions and Answers

 

Question 1. What does an electric circuit mean?
Answer: A continuous conducting path, which includes wires, other resistances, and a switch, connecting the two terminals of a cell or a battery, allowing electric current to flow, is called a circuit.
In simple words: An electric circuit is a complete path that lets electricity move from one end of a power source, through wires and devices, and back to the other end.

Exam Tip: Define a circuit by mentioning its continuity, components (wires, resistances, switch, battery/cell), and the flow of current.

 

Question 2. Define the unit of current.
Answer: The unit of current is Ampere. One Ampere of current is said to be present when one Coulomb of charge moves through any cross-section of a conductor in one second.
In simple words: Current is measured in Amperes. One Ampere means one Coulomb of electrical charge passes a point in a wire every second.

Exam Tip: Remember the relationship: 1 Ampere = 1 Coulomb/second. This direct definition is key.

 

Question 4. Name a device that helps to maintain a potential difference across a conductor.
Answer: A battery, made of many cells, or even a single cell, can maintain a potential difference across a conductor.
In simple words: A battery or a cell is what keeps the electrical "push" steady across a wire.

Exam Tip: The terms 'battery' and 'cell' are crucial here, as they are the primary sources of potential difference in circuits.

 

Question 5. What is meant by saying that the potential difference between two points is 1 V?
Answer: The potential difference between two points is said to be one Volt if one Joule of work is done in moving one Coulomb of charge from one point to another point.
In simple words: If you need to do one Joule of work to move one Coulomb of charge between two points, then the electrical "push" or voltage between those points is one Volt.

Exam Tip: Connect Voltage (V) to Work (Joule) and Charge (Coulomb) using the formula \(V = W/Q\). This is a fundamental definition.

 

Question 6. How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
Amount of charge \(Q = 1\) coulomb or \(1C\)
Potential difference \(V = 6V\)
Energy or work done \(W = ?\)
Work done or energy \(W = V \times Q\)
\(W = 6 \times 1\)
So, the work done or energy equals \(6\) joules.
In simple words: When one Coulomb of charge moves through a 6-Volt battery, the battery gives 6 Joules of energy to that charge.

Exam Tip: Remember the direct relationship \(W = VQ\), where \(W\) is energy/work done, \(V\) is potential difference, and \(Q\) is charge. Units are important: Joules for energy, Volts for potential difference, and Coulombs for charge.

 

Question 7. On what factors do the resistance of a conductor depend?
Answer: The resistance of a conductor depends on:
(a) Length of a conductor: The resistance of a conductor is directly proportional to its length. If the length increases, the resistance will also increase.
(b) Area of cross-section of a conductor: The resistance of a conductor is inversely proportional to the area of its cross-section.
(c) The material of a conductor: The resistance of a conductor also depends on the material it is made from. For instance, nichrome wire has 60 times more resistance than copper wire, as nichrome possesses high electrical resistance.
(d) Effect of temperature: The resistance of all pure metals generally increases when the temperature rises and decreases when the temperature falls.
In simple words: A conductor's resistance depends on four main things: how long it is, how thick it is (its cross-section area), what material it's made of, and its temperature. Longer wires, thinner wires, certain materials, and higher temperatures all generally mean more resistance.

Exam Tip: Remember the formula \(R = \rho \frac{L}{A}\) to help recall the factors: length (\(L\)), area of cross-section (\(A\)), and resistivity (\(\rho\)), which depends on the material and temperature.

 

Question 8. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer: Current will flow more easily through a thick wire compared to a thin wire. This is because a thick wire has less resistance than a thin wire. Less resistance allows more current to flow.
In simple words: Electricity moves better through a thick wire than a thin one. Thick wires offer less opposition to the flow of current, so more electricity can pass through.

Exam Tip: Think of it like water flowing through pipes; a wider pipe (thicker wire) allows water (current) to flow more easily than a narrow pipe (thin wire).

 

Question 9. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component is halved. What change will occur in the current through it?
Answer: The amount of current will be halved.
In simple words: If the push (voltage) across a component is cut in half, and its resistance stays the same, then the flow of electricity (current) will also be cut in half.

Exam Tip: This question relates to Ohm's Law (\(V = IR\)). If \(R\) is constant and \(V\) is halved, then \(I\) must also be halved to maintain the equality.

 

Question 10. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer: The coils of such heating appliances are made from an alloy instead of a pure metal for two key reasons:
(a) The resistivity of an alloy is much higher than that of a pure metal, meaning it resists electricity more and produces more heat.
(b) An alloy does not easily undergo oxidation, even at very high temperatures when it becomes red hot, helping it last longer.
In simple words: Toasters and irons use alloy coils because alloys resist electricity more, making them heat up better. Also, alloys don't rust or break down easily when they get very hot, unlike pure metals.

Exam Tip: Key properties for heating elements are high resistivity and resistance to oxidation at high temperatures. Alloys are specifically chosen for these characteristics.

 

Question 11. (a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?
Answer:
(a) Iron is a better conductor than mercury.
(b) Silver is considered the best electrical conductor.
In simple words: (a) Iron lets electricity pass through more easily than mercury. (b) The best material for conducting electricity is silver.

Exam Tip: Conductors are materials that allow electric current to pass through them easily. While iron is a good conductor, silver is the best due to its low resistivity.

 

Question 12. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, 8 Ω resistors, and a 12 Ω resistor, and a plug key, all connected in series.
Answer:
In simple words: The diagram above shows three resistors (5Ω, 8Ω, 12Ω) connected one after another (in series) with a battery of three 2V cells (total 6V) and a switch, forming a complete loop.

Exam Tip: For circuit diagrams, ensure correct symbols for each component (battery, resistor, key) and proper connection (series for this question).

 

Question 13. Redraw the circuit of Question 12, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistors. What would be the readings in the ammeter and the voltmeter?
Answer:
To find the readings, we first need to calculate the total resistance of the circuit. Since all resistors are in series, the total resistance \(R\) is:
\(R = 5\Omega + 8\Omega + 12\Omega = 25 \Omega\)
The total voltage \(V\) supplied by the battery (three 2V cells) is \(6V\).
Using Ohm's Law, \(I = V/R\), the current \(I\) flowing through the circuit is:
\(I = \frac{6V}{25 \Omega} = 0.24 A\)
Since the ammeter is in series, it measures the total current. So, the ammeter reading would be \(0.24 A\).
To find the potential difference across the 12 Ω resistor, we use Ohm's Law again, \(V = IR\), where \(R = 12 \Omega\):
\(V = 0.24 A \times 12 \Omega = 2.88 V\)
Hence, the reading in the ammeter would be \(0.24 A\), and the reading in the voltmeter would be \(2.88 V\).
In simple words: The diagram shows the same circuit but with an ammeter to measure the total current and a voltmeter connected across the 12Ω resistor to check its voltage. The total resistance is 25Ω. With a 6V battery, the total current is 0.24A, which the ammeter shows. The voltmeter across the 12Ω resistor will read 2.88V.

Exam Tip: Remember that an ammeter is always connected in series to measure current, and a voltmeter is always connected in parallel to measure potential difference. For series circuits, the total resistance is the sum of individual resistances, and the current is the same through all components.

 

Question 14. Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω, 10⁶ Ω, (b) 1 Ω, 10³ Ω and 10⁶ Ω
Answer:
(a) For 1 ohm and \(10^6\) ohms connected in parallel:
\(R_1 = 1\) ohm
\(R_2 = 10^6\) ohms = \(1000000\) ohm
The total resistance \(R\) in parallel is given by \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \)
\( \frac{1}{R} = \frac{1}{1} + \frac{1}{1000000} \)
\( \frac{1}{R} = \frac{1000000 + 1}{1000000} = \frac{1000001}{1000000} \)
Therefore, \(R = \frac{1000000}{1000001} \approx 1\) ohm.
The equivalent resistance is approximately 1 ohm.

(b) For 1 ohm, \(10^3\) ohms, and \(10^6\) ohms connected in parallel:
\(R_1 = 1\) ohm
\(R_2 = 10^3\) ohms
\(R_3 = 10^6\) ohms
The total resistance \(R\) in parallel is given by \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \)
\( \frac{1}{R} = \frac{1}{1} + \frac{1}{1000} + \frac{1}{1000000} \)
\( \frac{1}{R} = \frac{1000000}{1000000} + \frac{1000}{1000000} + \frac{1}{1000000} \)
\( \frac{1}{R} = \frac{1000000 + 1000 + 1}{1000000} = \frac{1001001}{1000000} \)
Therefore, \(R = \frac{1000000}{1001001} \approx 1\) ohm.
The equivalent resistance is approximately 1 ohm.
In simple words: When resistors are connected in parallel, the total resistance is always less than the smallest individual resistance. In both cases, with one resistor being 1 ohm and others much larger, the total resistance ends up being very close to 1 ohm.

Exam Tip: Remember the rule for parallel resistors: the equivalent resistance is always smaller than the smallest individual resistance. This is a common shortcut for quickly estimating answers in such problems.

 

Question 15. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all the three appliances, and what is the current through it?
Answer:
Given resistances:
Resistance of electric lamp \(R_1 = 100\) ohm
Resistance of a toaster \(R_2 = 50\) ohm
Resistance of a water filter \(R_3 = 500\) ohm
Potential difference \(V = 220\) volt

First, calculate the total resistance of the three appliances connected in parallel:
\( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \)
\( \frac{1}{R_p} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500} \)
To add these fractions, find a common denominator, which is 500:
\( \frac{1}{R_p} = \frac{5}{500} + \frac{10}{500} + \frac{1}{500} \)
\( \frac{1}{R_p} = \frac{5 + 10 + 1}{500} = \frac{16}{500} \)
\( R_p = \frac{500}{16} = \frac{125}{4} = 31.25 \) ohms.

The electric iron is connected to the same source and draws as much current as all three appliances combined. This means the resistance of the electric iron should be equal to the equivalent resistance of the three appliances in parallel, which is \(31.25\) ohms.

Now, calculate the current through the electric iron (or the combined current of the three appliances):
Using Ohm's Law, \(V = I \times R_p\)
\(220 V = I \times 31.25 \Omega\)
\(I = \frac{220}{31.25} = 7.04\) Ampere.

Therefore, the resistance of the electric iron is \(31.25\) ohms, and the current flowing through it is \(7.04\) Ampere.
In simple words: We have a lamp, toaster, and water filter all connected side-by-side to a 220V power. First, we find their combined resistance, which is about 31.25 ohms. Then, an electric iron draws the same total current from the same power source. This means the iron's resistance is also 31.25 ohms. Using this, we find the total current drawn from the line is 7.04 Amperes.

Exam Tip: For parallel circuits, the voltage across each component is the same, but the current divides. When comparing to a single appliance drawing the "same current," it implies the single appliance has the equivalent resistance of the parallel combination.

 

Question 16. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer: The advantages of parallel connection are numerous:
(a) In a parallel circuit, if one electric appliance stops working due to some defect, all other appliances continue to work normally.
(b) Each electrical appliance in a parallel circuit has its own switch, allowing it to be turned on or off independently without affecting other appliances.
(c) Every electrical appliance receives the same voltage (e.g., 220 V) as that provided by the power supply line.
(d) In a parallel connection of electrical appliances, the overall resistance of the circuit is reduced, which means a higher current can be drawn from the power supply.
In simple words: Connecting electrical items in parallel is better than in series because if one item breaks, the others still work. Each item also has its own switch, and they all get the full voltage. Plus, parallel connections lower the total resistance, allowing more current to flow.

Exam Tip: Think about household wiring; it's always in parallel. This question highlights the key practical benefits of parallel connections over series for multiple devices.

 

Question 17. How can three resistors of resistances 2Ω, 3Ω, and 6Ω be connected to give a total resistance of (a) 4Ω, (b) 1Ω?
Answer: Let the three resistors be \(R_1 = 2\Omega\), \(R_2 = 3\Omega\), and \(R_3 = 6\Omega\).

(a) To obtain a total resistance of \(4\Omega\):
Connect \(R_2\) (\(3\Omega\)) and \(R_3\) (\(6\Omega\)) in parallel. Then, connect this parallel combination in series with \(R_1\) (\(2\Omega\)).
For \(R_2\) and \(R_3\) in parallel:
\( \frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \)
So, \(R_p = 2\Omega\).
Now, connect \(R_p\) in series with \(R_1\):
\(R_{total} = R_p + R_1 = 2\Omega + 2\Omega = 4\Omega\).

(b) To obtain a total resistance of \(1\Omega\):
Connect all three resistors (\(R_1\), \(R_2\), \(R_3\)) in parallel.
\( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \)
To add these fractions, find a common denominator, which is 6:
\( \frac{1}{R_p} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1 \)
So, \(R_p = 1\Omega\).
In simple words: To get 4Ω, first connect the 3Ω and 6Ω resistors side-by-side (parallel), which gives 2Ω. Then connect this 2Ω result in a line (series) with the remaining 2Ω resistor, making a total of 4Ω. To get 1Ω, simply connect all three resistors (2Ω, 3Ω, and 6Ω) side-by-side (parallel).

Exam Tip: Practice combining resistors in both series and parallel configurations. Remember that series connections add resistances, while parallel connections result in a total resistance smaller than the smallest individual resistor.

 

Question 18. What is (a) the highest, (b) the lowest total resistance that can be secured by four coils of resistance 4 Ω, 8 Ω, 12Ω, 24Ω?
Answer: Let the four resistors be \(R_1 = 4\Omega\), \(R_2 = 8\Omega\), \(R_3 = 12\Omega\), and \(R_4 = 24\Omega\).

(a) The highest resistance is achieved when the resistors are connected in series:
\(R_{series} = R_1 + R_2 + R_3 + R_4\)
\(R_{series} = 4\Omega + 8\Omega + 12\Omega + 24\Omega = 48\Omega\).
Thus, the highest resistance possible is \(48\) ohms.

(b) The lowest resistance is achieved when the resistors are connected in parallel:
\( \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} \)
\( \frac{1}{R_{parallel}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} \)
To add these fractions, find the least common multiple (LCM) of the denominators, which is 24:
\( \frac{1}{R_{parallel}} = \frac{6}{24} + \frac{3}{24} + \frac{2}{24} + \frac{1}{24} \)
\( \frac{1}{R_{parallel}} = \frac{6 + 3 + 2 + 1}{24} = \frac{12}{24} = \frac{1}{2} \)
So, \(R_{parallel} = 2\Omega\).
Thus, the lowest resistance possible is \(2\) ohms.
In simple words: To get the highest resistance, connect all four coils in a line (series), which simply adds their resistances together for a total of 48Ω. To get the lowest resistance, connect all four coils side-by-side (parallel), which results in a combined resistance of 2Ω.

Exam Tip: For series connections, resistance always adds up, leading to the maximum possible resistance. For parallel connections, the reciprocal of resistances adds up, always resulting in a total resistance less than the smallest individual resistance.

 

Question 19. Why does the cord of an electric heater not glow while the heating element does?
Answer: The cord of an electric heater does not glow because its resistance is extremely small when compared to that of the heating element. As a result, the heat produced in the cord is minimal. In contrast, the heating element has a very high resistance, which causes it to produce a large amount of heat and thus begin to glow, while the cord remains dark.
In simple words: The heater's cord doesn't light up because it has very low resistance, so it produces little heat. The heating element, however, has high resistance, which makes it get very hot and glow.

Exam Tip: Recall Joule's law of heating (\(H = I^2Rt\)). For the same current (\(I\)), the heat produced (\(H\)) is directly proportional to resistance (\(R\)). A low-resistance cord produces negligible heat, while a high-resistance element generates significant heat and light.

 

Question 20. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V
Answer:
Given values:
Charge \(Q = 96000\) coulomb
Time \(t = 1\) hour = \(3600\) seconds
Potential difference \(V = 50\) V

First, calculate the current \(I\):
\(I = \frac{Q}{t} = \frac{96000 \, C}{3600 \, s} = \frac{80}{3}\) Ampere

Next, calculate the resistance \(R\). Using Ohm's law (\(V = IR\)):
\(R = \frac{V}{I} = \frac{50 \, V}{\frac{80}{3} \, A} = \frac{50 \times 3}{80} = \frac{150}{80} = \frac{15}{8}\) ohm

Finally, compute the heat generated \(H\) using Joule's law of heating (\(H = I^2Rt\)):
\(H = \left(\frac{80}{3}\right)^2 \times \frac{15}{8} \times 3600\)
\(H = \frac{6400}{9} \times \frac{15}{8} \times 3600\)
\(H = 6400 \times \frac{15}{8} \times \frac{3600}{9}\)
\(H = 800 \times 15 \times 400\)
\(H = 12000 \times 400 = 4800000\) joules.
The heat generated is \(4,800,000\) Joules.
In simple words: To find the heat produced, we first figure out the electric current using the given charge and time. Then, we find the resistance using the voltage and current. Finally, we use these values in the heating formula to calculate that 4,800,000 Joules of heat are generated.

Exam Tip: This problem requires sequential application of definitions and laws: \(I = Q/t\), Ohm's law \(V = IR\), and Joule's law of heating \(H = I^2Rt\). Ensure correct unit conversions, especially for time from hours to seconds.

 

Question 21. An electric iron of resistance 20 Ω takes a -current of 5 A. Calculate the heat developed in 30 s.
Answer:
Given values:
Resistance of the iron \(R = 20\) ohm
Current \(I = 5\) Ampere
Time \(t = 30\) seconds

Using Joule's law of heating, the heat produced \(H = I^2Rt\):
\(H = (5)^2 \times 20 \times 30\)
\(H = 25 \times 20 \times 30\)
\(H = 500 \times 30\)
\(H = 15000\) joules.
The heat developed in 30 seconds is \(15,000\) Joules.
In simple words: To calculate the heat produced by the iron, we multiply the square of the current (5A), the resistance (20Ω), and the time (30s) together. This calculation shows 15,000 Joules of heat are made.

Exam Tip: Directly apply Joule's law of heating \(H = I^2Rt\). Pay close attention to units; resistance in ohms, current in amperes, and time in seconds will yield heat in joules.

 

Question 22. What determines the rate at which energy is delivered by a current?
Answer: Electrical power determines the rate at which energy is delivered by an electric current.
In simple words: How fast electricity gives out energy is decided by its electrical power.

Exam Tip: The term "rate at which energy is delivered" is the definition of power. Connect power to the concepts of energy and time (Power = Energy/Time).

 

Question 23. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Given values:
Current in the motor \(I = 5\) Ampere
Potential difference \(V = 220\) V
Time \(t = 2\) hours

First, calculate the power of the motor:
Power \(P = V \times I\)
\(P = 220 \, V \times 5 \, A = 1100\) watt or \(1.1\) kW

Next, calculate the energy consumed:
Energy \(E = \text{power} \times \text{time}\)
\(E = 1.1 \, kW \times 2 \, h = 2.2\) kWh
The power of the motor is \(1100\) watts (or \(1.1\) kW), and the energy consumed in 2 hours is \(2.2\) kWh.
In simple words: We find the motor's power by multiplying its voltage (220V) by its current (5A), which gives 1100 watts (1.1 kW). Then, to find the energy used in 2 hours, we multiply this power by the time, resulting in 2.2 kilowatt-hours.

Exam Tip: Remember the formulas for electrical power (\(P = VI\)) and energy consumed (\(E = Pt\)). Be mindful of units; power in watts leads to energy in joules (watt-seconds), while power in kilowatts and time in hours leads to energy in kilowatt-hours (kWh).

 

Question 24. An electric lamp draws a current of 0.3 ampere and is used for 6 hours every day for a month (30 days). Calculate the amount of charge that flows through the circuit every day and for a month.
Answer:
Given values:
Current \(I = 0.3\) Ampere
Time of use per day \(t_{daily} = 6\) hours
Number of days in a month = \(30\) days

First, convert the daily time to seconds:
\(t_{daily} = 6 \, \text{hours} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} = 21600\) seconds

Calculate the charge flowing per day:
Using the formula \(Q = I \times t\)
\(Q_{daily} = 0.3 \, A \times 21600 \, s = 6480\) Coulombs.

Calculate the total charge flowing for a month (30 days):
\(Q_{month} = Q_{daily} \times 30\) days
\(Q_{month} = 6480 \, C \times 30 = 194400\) Coulombs.
The amount of charge that flows through the circuit is \(6480\) Coulombs every day and \(194400\) Coulombs for a month.
In simple words: The lamp uses 0.3A of current for 6 hours daily. To find the charge, we multiply current by time (in seconds). This gives 6480 Coulombs each day. For a whole month (30 days), the total charge flowing is 194400 Coulombs.

Exam Tip: Ensure consistency in units. Current is in Amperes, time in seconds for charge in Coulombs. Convert hours to seconds before calculating daily charge.

 

Queastion 25. Describe an activity to prove that resistance of a wire depends on its length, cross-sectional area, and material of the wire.
Answer: To prove that the resistance of a wire depends on its length, cross-sectional area, and material, we can perform the following activity:

Method:
1. Set up an electric circuit consisting of a cell, an ammeter, a plug key, and a nichrome wire of a specific length (say, \(L\)), marked as (1) in the figure. Close the key and note the ammeter reading.
2. Effect of length: Replace the first nichrome wire with another nichrome wire of the same thickness and material, but twice the length (2\(L\)), marked as (2). Note the new ammeter reading. You will observe that the current decreases (i.e., resistance increases).
3. Effect of cross-sectional area: Now, replace the wire with a thicker nichrome wire of the same length (\(L\)) and material, marked as (3). This thicker wire has a larger cross-sectional area. Note the ammeter reading. You will find that the current increases (i.e., resistance decreases).
4. Effect of material: Finally, replace the nichrome wire with a copper wire of the same length (\(L\)) and cross-sectional area as the first nichrome wire, marked as (4). Note the ammeter reading. You will see a significant increase in current compared to nichrome (i.e., resistance changes with material).

Observations:
- The ammeter reading decreases to one-half when the length of the wire is doubled, showing that resistance is directly proportional to length.
- The ammeter reading increases when a thicker wire of the same material and length is used, indicating resistance is inversely proportional to cross-sectional area.
- A different ammeter reading is observed when a wire of a different material is used, showing that resistance depends on the material.

Conclusion:
Based on these observations, we conclude that the resistance of a conductor depends on:
- The length of the wire (\(R \propto L\)).
- The area of the cross-section of the wire (\(R \propto \frac{1}{A}\)).
- The nature of the material of the wire.
This leads to the resistivity formula \(R = \rho \frac{L}{A}\).
In simple words: We can test how a wire's resistance changes by setting up a simple circuit with an ammeter. We try different wires: first, a longer wire of the same type, then a thicker wire of the same type, and finally, a wire of a different material. The ammeter's readings will show that resistance increases with length, decreases with thickness, and varies with the material used.

Exam Tip: When describing an activity, clearly state the aim, apparatus, method (step-by-step), observations, and conclusion. For this experiment, highlight how current changes with each variation to infer the change in resistance.

Gujarat Board Class 10 Science Electricity Textbook Questions and Answers

 

Question 1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then what is the ratio R/R'?
Answer:
Let the original resistance of the wire be \(R\).
When cut into five equal parts, the resistance of each part will be \( \frac{R}{5} \).
When these five parts are connected in parallel, the equivalent resistance \(R'\) is given by:
\( \frac{1}{R'} = \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} \)
\( \frac{1}{R'} = \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} \)
\( \frac{1}{R'} = \frac{5+5+5+5+5}{R} = \frac{25}{R} \)
Therefore, \(R' = \frac{R}{25}\).
Now, we need to find the ratio \(R/R'\):
\( \frac{R}{R'} = \frac{R}{R/25} = \frac{R \times 25}{R} = 25 \)
The ratio \(R/R'\) is \(25:1\).
In simple words: If a wire is cut into five equal pieces, each piece has one-fifth of the original resistance. When these five pieces are then connected side-by-side (in parallel), their combined resistance becomes one-twenty-fifth of the original resistance. So, the ratio of the original resistance to the new combined resistance is 25:1.

Exam Tip: Remember that cutting a wire reduces its length and thus its resistance proportionally. Connecting resistors in parallel reduces the total resistance even further. For \(N\) identical resistors (\(r\)) in parallel, \(R_{parallel} = r/N\).

 

Question 2. Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Answer: (b) IR²
In simple words: Electrical power can be calculated in a few ways, like current squared times resistance (\(I^2R\)), voltage times current (\(VI\)), or voltage squared divided by resistance (\(V^2/R\)). The option \(IR^2\) is not a standard formula for power.

Exam Tip: Know the three main formulas for electrical power: \(P = VI\), \(P = I^2R\), and \(P = V^2/R\). These are derived from Ohm's Law and the definition of power. Any other combination, like \(IR^2\), is incorrect.

 

Question 3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be -
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer: (d) 25 W
In simple words: A bulb designed for 220V and 100W has a certain resistance. When you use it with half the voltage (110V), the power it uses drops to one-fourth of its original power, so it will consume 25W.

Exam Tip: First, use the rating to find the bulb's resistance (\(R = V^2/P\)). The resistance of the bulb remains constant. Then, use this resistance with the new voltage to find the new power consumed (\(P_{new} = V_{new}^2/R\)).

 

Question 4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be -
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Answer: (c) 1:4
In simple words: When wires are connected differently, either in a row (series) or side-by-side (parallel), the amount of heat they make changes. For the same starting conditions, the heat made in series will be a quarter of the heat made in parallel.

Exam Tip: Remember that heat produced is proportional to resistance for series connection and inversely proportional to resistance for parallel connection. This relationship helps determine the ratio.

 

Question 5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer: A voltmeter is always connected in parallel to the component across which the potential difference needs to be measured. This ensures that it measures the voltage correctly without changing the main current flow.
In simple words: To find the voltage between two points, a voltmeter is always hooked up alongside (in parallel) with those points.

Exam Tip: Voltmeter has very high resistance, so connecting it in parallel ensures minimal current diversion and accurate voltage measurement.

 

Question 6. A copper wire has a diameter 0.5 mm and a resistivity of \( 1.6 \times 10^{-8} \Omega \) m. What will be the length of this wire to make its resistance 10 \( \Omega \)? How much does the resistance change if the diameter is doubled?
Answer:
Diameter = 0.5 mm = \( 0.0005 \) m
Radius \( r = \frac {0.0005}{2} = 0.00025 \) m or \( 0.25 \times 10^{-3} \)m
Resistivity \( \rho = 1.6 \times 10^{-8} \Omega \) m
Resistance R = 10 \( \Omega \)
Length l = ?
Using \( R = \rho \frac {l}{A} \) or \( l = \frac {RA}{\rho} \)
\( l = \frac{10 \times 3.14 \times (0.25 \times 10^{-3})^2}{1.6 \times 10^{-8}} = 122.7 \) m
The resistance becomes one-fourth if the diameter is doubled.
In simple words: We calculated that the wire needs to be about 122.7 meters long to have a resistance of 10 Ohms. If you make the wire twice as thick (double the diameter), its resistance will become four times smaller.

Exam Tip: Pay close attention to unit conversions (mm to m) and the formula for resistance in terms of resistivity, length, and area. Remember that resistance is inversely proportional to the square of the diameter.

 

Question 7. The values of current I flowing in a circuit with a resistor for the corresponding values of potential difference V across the resistors are given below:

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:

From the table, we can calculate the resistance for each data point using Ohm's Law: \( R = \frac{V}{I} \).

Resistance of resistor R (mean) \( = \frac{3.2 + 3.4 + 3.35 + 3.4 + 3.3}{5} = 3.33 \Omega \)

In simple words: First, plot a graph with voltage on one axis and current on the other using the given numbers. The graph will show a straight line, confirming Ohm's law. Then, calculate the resistance for each point by dividing voltage by current, and find the average to get the overall resistance, which is approximately 3.33 Ohms.

Exam Tip: When calculating mean resistance, ensure all individual resistance values are close, indicating an ohmic resistor. A V-I graph should be a straight line passing through the origin for an ohmic resistor.

 

Question 8. When a 12 V battery is connected across an unknown resistor, there is a current 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer:
V = 12V
I = 2.5 mA = \( 2.5 \times 10^{-3} \)A
Using, \( R = \frac {V}{I} \)
\( R = \frac{12 \mathrm{V}}{2.5 \times 10^{-3} \mathrm{A}} = 4800 \Omega = 4.8 \mathrm{k}\Omega \)
In simple words: If a 12-volt battery pushes a 2.5 milliampere current through a mystery component, that component has a resistance of 4800 Ohms, or 4.8 kiloOhms.

Exam Tip: Always remember to convert milliampere (mA) to ampere (A) by multiplying by \( 10^{-3} \) before using Ohm's Law to avoid calculation errors.

 

Question 9. A battery of 9 V is connected in series with resistors of \( 0.2 \Omega, 0.3 \Omega, 0.4 \Omega, 0.5 \Omega \) and 12 \( \Omega \). How much current would flow through the 12 \( \Omega \) resistor?
Answer:
V = 9V
Rs = \( 0.2 \Omega + 0.3 \Omega + 0.4 \Omega + 0.5 \Omega + 12 \Omega = 13.4 \Omega \)
I = ?
Using, \( I = \frac {V}{R} = \frac {9V}{13.4 \Omega } = 0.67 \) A
0.67A current would flow through the 12 \( \Omega \) resistor.
In simple words: In a series circuit with a 9-volt battery and several resistors, including a 12 Ohm resistor, the total resistance is 13.4 Ohms. Because all components in a series circuit share the same current, a current of 0.67 Amperes will flow through every resistor, including the 12 Ohm one.

Exam Tip: For series circuits, remember that the total resistance is the sum of individual resistances, and the current flowing through each component is the same.

 

Question 10. How many 176 \( \Omega \) resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:
I = 5A
V = 220V
Let the number of 176 \( \Omega \) resistors be n.
\( \frac{1}{\mathrm{R_p}} = \frac{1}{176} + \frac{1}{176} + ... + \frac{1}{176} \)
\( \frac{1}{\mathrm{R_p}} = \frac{n}{176} \) ...(1)
Or \( \mathrm{R_p} = \frac{176}{n} \)
As \( \mathrm{R_p} = \frac{V}{I} = \frac{220}{5} \)
\( \mathrm{R_p} = 44 \Omega \)
Comparing (1) and (2):
\( \frac{176}{n} = 44 \)
\( n = \frac{176}{44} = 4 \)
Hence, four resistors of 176 \( \Omega \) are required to carry 5 A on a 220 V line.
In simple words: To get 5 Amperes of current from a 220-volt source using 176 Ohm resistors, you need to connect 4 of these resistors side-by-side (in parallel). This reduces the total resistance enough for the desired current to flow.

Exam Tip: For resistors connected in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances. Remember to calculate the total resistance needed first using Ohm's law.

 

Question 11. Show how you would connect three resistors each of resistance 6 \( \Omega \) so that the combination has a resistance of -
(i) 9 \( \Omega \)
(ii) 4 \( \Omega \)
Answer:
(i) To get 9 \( \Omega \) - Connect 2 resistors of 6 \( \Omega \) in parallel and a third resistor in series with them.
Let \( R_1 = 6 \Omega \), \( R_2 = 6 \Omega \), \( R_3 = 6 \Omega \).
Connect \( R_1 \) and \( R_2 \) in parallel:
\( \frac{1}{\mathrm{R_p}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \)
\( \mathrm{R_p} = 3 \Omega \)
Now, connect \( R_p \) in series with \( R_3 \):
\( \mathrm{R_{total}} = \mathrm{R_p} + R_3 = 3 + 6 = 9 \Omega \)
(ii) To get 4 \( \Omega \) - Connect 2 resistors of 6 \( \Omega \) in series and a third in parallel to both of them.
Connect \( R_1 \) and \( R_2 \) in series:
\( \mathrm{R_s} = 6 + 6 = 12 \Omega \)
Now, connect \( R_s \) and \( R_3 \) in parallel:
\( \frac{1}{\mathrm{R_{total}}} = \frac{1}{\mathrm{R_s}} + \frac{1}{R_3} \)
\( \frac{1}{\mathrm{R_{total}}} = \frac{1}{12} + \frac{1}{6} = \frac{1 + 2}{12} = \frac{3}{12} = \frac{1}{4} \)
\( \mathrm{R_{total}} = 4 \Omega \)
In simple words: For 9 Ohms, put two 6-Ohm resistors side-by-side, then add the third 6-Ohm resistor in line. For 4 Ohms, put two 6-Ohm resistors in line, then connect the third 6-Ohm resistor parallel to that whole line.

Exam Tip: Draw circuit diagrams for each combination to visualize the connections. Remember that resistors in series add up, while for parallel connections, you add their reciprocals.

 

Question 12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel to each other across the two wires of a 220 V line if the maximum allowable current is 5 A?
Answer:
P = 10W
V = 220V
Current drawn by one bulb, \( I_1 = \frac {P}{V} = \frac {10}{220} = \frac {1}{22} \) A
If 'n' is the number of bulbs connected to draw a total current of 5 A, then:
\( n \times \frac {1}{22} = 5 \)
\( n = 5 \times 22 = 110 \)
110 bulbs can be attached.
In simple words: Each 10-watt bulb needs a small amount of current from the 220-volt line. If the total current allowed is 5 Amperes, you can connect 110 such bulbs side-by-side before reaching that limit.

Exam Tip: Power (P), Voltage (V), and Current (I) are related by \( P = VI \). When appliances are connected in parallel, the voltage across each is the same, and the total current is the sum of individual currents.

 

Question 13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
(i) When used separately
(ii) When connected in series
(iii) When connected in parallel
Answer:
(i) When used separately:
Resistance R = 24 \( \Omega \), Voltage V = 220 V
Current \( I = \frac {V}{R} = \frac {220}{24} \approx 9.17 \) A.
(ii) When connected in series:
\( R_1 = 24 \Omega \), \( R_2 = 24 \Omega \)
Total resistance \( R_s = R_1 + R_2 = 24 + 24 = 48 \Omega \)
Current \( I = \frac {V}{R_s} = \frac {220}{48} \approx 4.58 \) A.
(iii) When connected in parallel:
\( R_1 = 24 \Omega \), \( R_2 = 24 \Omega \)
\( \frac{1}{\mathrm{R_p}} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24} = \frac{1}{12} \)
\( \mathrm{R_p} = 12 \Omega \)
Current \( I = \frac {V}{R_p} = \frac {220}{12} \approx 18.33 \) A.
In simple words: When the oven coils are used alone, about 9.17 Amperes flow. When connected in a line (series), the current drops to about 4.58 Amperes because the total resistance increases. When connected side-by-side (parallel), the current increases to about 18.33 Amperes because the total resistance decreases.

Exam Tip: Clearly identify the circuit configuration (separate, series, or parallel) before calculating total resistance. Remember Ohm's law and the formulas for equivalent resistance in series and parallel circuits.

 

Question 14. Compare the power used in the 2 \( \Omega \) resistor in each of the following circuits:
1. a 6 V battery in series with 1 \( \Omega \) and 2 \( \Omega \) resistors,
2. a 4 V battery in parallel with 12 \( \Omega \) and 2 \( \Omega \) resistors.
Answer:
1. For the series circuit:
V = 6V
\( R_1 = 1 \Omega \), \( R_2 = 2 \Omega \)
\( R_1 \) and \( R_2 \) are in series, so total resistance \( R_s = R_1 + R_2 = 1 + 2 = 3 \Omega \)
Current flowing through the circuit, \( I = \frac{V}{R_s} = \frac{6}{3} = 2 \) A
Power used in the 2 \( \Omega \) resistor, \( P = I^2 R = (2)^2 \times 2 = 4 \times 2 = 8 \) W.
Therefore, 8 W power is used in the 2 \( \Omega \) resistor in this circuit.
2. For the parallel circuit:
\( R_1 = 12 \Omega \), \( R_2 = 2 \Omega \)
Parallel connection means the voltage across each resistor is the same as the battery voltage.
V = 4V (across the 2 \( \Omega \) resistor)
Power used in the 2 \( \Omega \) resistor, \( P = \frac{V^2}{R} = \frac{4^2}{2} = \frac{16}{2} = 8 \) W.
Therefore, 8 W power is also used in the 2 \( \Omega \) resistor in this circuit.
In simple words: In both situations, the 2 Ohm resistor uses 8 watts of power. In the first case, we found the total current in the series circuit and then calculated power for the 2 Ohm resistor. In the second case, since the resistors are in parallel, the 2 Ohm resistor gets the full 4 volts from the battery directly, which we used to calculate its power.

Exam Tip: For power calculations, use \( P = I^2 R \) for series circuits where current is constant, and \( P = \frac{V^2}{R} \) for parallel circuits where voltage is constant across components.

 

Question 15. Two lamps, one rated 100 W at 220 V, and the other rated 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:
Given:
Bulb 1: \( P_1 = 100 \)W, V = 220V
Current drawn by Bulb 1: \( I_1 = \frac {P_1}{V} = \frac {100}{220} \approx 0.45 \) A
Bulb 2: \( P_2 = 60 \)W, V = 220V
Current drawn by Bulb 2: \( I_2 = \frac {P_2}{V} = \frac {60}{220} \approx 0.27 \) A
Total current drawn from the line: \( I_{total} = I_1 + I_2 \) (since connected in parallel)
\( I_{total} = 0.45 + 0.27 = 0.72 \) A.
In simple words: When two light bulbs, one 100W and one 60W, are plugged into a 220-volt outlet side-by-side, they pull a total of about 0.72 Amperes of electricity from the main power line. We found this by adding up the current each bulb takes on its own.

Exam Tip: In a parallel circuit, the voltage across each component is the same as the supply voltage, and the total current is the sum of the currents through each branch.

 

Question 16. Which uses more energy; a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer:
Energy consumed by TV:
Power \( P_{TV} = 250 \) W
Time \( t_{TV} = 1 \) hour = 3600 seconds
Energy \( E_{TV} = P_{TV} \times t_{TV} = 250 \) W \( \times 3600 \) s = 900,000 Joules.
Energy consumed by toaster:
Power \( P_{toaster} = 1200 \) W
Time \( t_{toaster} = 10 \) minutes = 600 seconds
Energy \( E_{toaster} = P_{toaster} \times t_{toaster} = 1200 \) W \( \times 600 \) s = 720,000 Joules.
Comparing the energies, \( E_{TV} \) (900,000 J) > \( E_{toaster} \) (720,000 J).
Therefore, the 250 W TV set used for 1 hour consumes more energy.
In simple words: A 250-watt TV running for one hour uses more electricity than a 1200-watt toaster running for ten minutes. The TV uses 900,000 Joules, while the toaster uses 720,000 Joules.

Exam Tip: To compare energy consumption, always convert time to a consistent unit (like seconds) and calculate total energy (Energy = Power × Time) for each device.

 

Question 17. An electric heater of resistance 8 \( \Omega \) draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Answer:
Resistance R = 8 \( \Omega \)
Current I = 15 A
Time t = 2 hours (this is not needed for rate of heat)
The rate at which heat is developed is power (P).
\( P = I^2 R \)
\( P = (15)^2 \times 8 \)
\( P = 225 \times 8 = 1800 \) Watt.
In simple words: An electric heater with an 8 Ohm resistance and drawing 15 Amperes of current develops heat at a rate of 1800 Watts. This means it produces 1800 Joules of heat every second.

Exam Tip: The "rate at which heat is developed" is equivalent to electrical power. Use the formula \( P = I^2 R \) for this calculation.

 

Question 18. Explain the following:
(a) Why the tungsten is used almost exclusively for the filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made up of alloys rather than pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why copper and aluminium wires are usually employed for electricity transmission?
Answer:
(a) Tungsten is mainly used for electric lamp filaments because it has several beneficial properties:

• It glows brightly and emits light when it gets hot.
• It possesses a very high melting point, meaning it can withstand extremely high temperatures without melting.
• It does not easily react with gases in the air and resists oxidation, helping the bulb last longer.
• It has a specific property that helps it glow when heated.

(b) Heating device conductors, like those in toasters and irons, are made from alloys rather than pure metals for these reasons:

• Alloys typically offer higher resistance compared to pure metals, which helps them generate more heat.
• Alloys often have a high melting point, allowing them to get very hot without melting.
• Alloys do not oxidize easily, even at high temperatures, which makes them more durable and efficient for heating.

(c) Series arrangements are not used for household electrical circuits because:

• If one appliance in a series circuit stops working or breaks, the entire circuit is interrupted, and all other appliances also stop working.
• Different appliances have different current needs, but in a series circuit, the current is the same through all of them. This can lead to some appliances not working correctly or being damaged.
• Series circuits in homes would mean very long wires, which would cause unusually high total resistance. This high resistance would lead to significant energy loss and reduced efficiency.

(d) The resistance of a wire is inversely proportional to its area of cross-section. This means that if the cross-sectional area of the wire increases, its resistance decreases, and if the area decreases, the resistance increases. The relationship can be written as: R \( \propto \frac {1}{A} \).

(e) Copper and aluminium wires are generally used for transmitting electricity because:

• They are much more readily available and less expensive than other metals that conduct electricity well.
• They have low electrical resistance, which means less energy is lost as heat during transmission. They also do not heat up much during use.
• They are highly malleable and ductile, making them easy to shape into long wires.

In simple words: (a) Tungsten is great for light bulb filaments because it gets super hot and bright without melting and doesn't rust easily. (b) Toasters and irons use alloys, not pure metals, because alloys resist electricity more and don't melt or corrode quickly when hot. (c) House wiring doesn't use series circuits because if one thing breaks, everything stops, and different devices need different amounts of power. (d) The fatter a wire is, the less it resists electricity. (e) Copper and aluminum are popular for power lines because they're cheap, good conductors, don't heat up much, and are easy to shape into wires.

Exam Tip: When explaining material properties, always link them directly to the desired function (e.g., high melting point for heating, low resistance for conduction). For circuit types, focus on the practical implications of each connection type for safety and efficiency.

 

Gujarat Board Class 10 Science Electricity Additional Important Questions and Answers

Very Short Answer Type Questions

 

Question 1. What is S.I. unit of charge?
Answer: Coulomb.
In simple words: The standard way to measure electric charge is called a Coulomb.

Exam Tip: Always remember the SI units for fundamental physical quantities as they are frequently asked.

 

Question 2. Give the S.I. unit of electric current.
Answer: Ampere (A).
In simple words: The official unit for measuring electric current is the Ampere, often just called "Amp."

Exam Tip: Be familiar with both the full name and the symbol for common SI units.

 

Question 3. What is S.I. unit of electric potential?
Answer: Volt.
In simple words: The standard unit for electric potential, or voltage, is the Volt.

Exam Tip: Associate potential with voltage and its unit, Volt, just like current with Ampere.

 

Question 4. Give the unit of electric resistance.
Answer: Ohm ( \( \Omega \) ).
In simple words: The unit for measuring how much something resists electricity is the Ohm.

Exam Tip: Know the symbol \( \Omega \) for Ohm, as it is commonly used in circuit diagrams.

 

Question 5. How is resistance, volt, and current (I) related?
Answer: Resistance (R), voltage (V), and current (I) are related by Ohm's Law, which states: \( V = IR \). This can also be rearranged as \( R = \frac{V}{I} \) or \( I = \frac{V}{R} \).
In simple words: The relationship between resistance, voltage, and current is given by Ohm's Law: Voltage equals current times resistance.

Exam Tip: Ohm's law is fundamental. Be able to recall all three forms of the equation \( V = IR \), \( I = V/R \), and \( R = V/I \).

 

Question 6. How is 1 ohm related to ampere and volt?
Answer: One Ohm is defined as the resistance that allows one Ampere of current to flow through a conductor when a potential difference of one Volt is applied across its ends. It can be expressed as:
\( 1 \text{ ohm} = \frac{1 \text{ volt}}{1 \text{ ampere}} \)
In simple words: One Ohm is the amount of resistance that lets one Amp of electricity pass through when there's one Volt of push.

Exam Tip: Understand the definition of each unit in relation to Ohm's Law to derive their interrelationships correctly.

 

Question 7. What constitutes the current?
Answer: The flow of free electrons constitutes the electric current. These electrons move through a conductor in a specific direction when a potential difference is applied.
In simple words: Electric current is basically just a bunch of free electrons moving in one direction.

Exam Tip: Remember that current is defined as the flow of positive charge, even though in most conductors, it's actually the negative electrons that move.

 

Question 8. When the given resistances are connected in series, which physical quantity does not change.
Answer: Current.
In simple words: In a series circuit, the amount of electricity flowing through each part stays the same.

Exam Tip: Understand the key characteristics of series circuits: current is constant, voltage divides, and resistances add up.

 

Question 9. What is an ammeter?
Answer: An ammeter is a measuring instrument used to measure electric current flowing through a circuit. It is always connected in series with the component whose current is to be measured.
In simple words: An ammeter is a tool used to measure how much electricity is flowing. You hook it up in line with the circuit.

Exam Tip: Recall that an ammeter has very low resistance and must be connected in series to measure current accurately.

 

Question 10. What is a voltmeter?
Answer: A voltmeter is a measuring instrument used to measure the potential difference (voltage) between two points in an electric circuit. It is always connected in parallel across the two points where the potential difference is to be measured.
In simple words: A voltmeter is a tool that measures the "push" or voltage between two points in an electrical circuit. You always connect it side-by-side with the part you're measuring.

Exam Tip: Remember that a voltmeter has very high resistance and is connected in parallel to avoid diverting current from the main circuit.

 

Question 11. Give the unit of power.
Answer: Watt.
In simple words: The unit used to measure power is called the Watt.

Exam Tip: Remember that power indicates the rate at which energy is consumed or produced.

 

Question 12. How is power related to current and voltage?
Answer: Power (P) is related to current (I) and voltage (V) by the formula: \( P = V \times I \). This means that power is the product of the potential difference across a component and the current flowing through it.
In simple words: Power is found by multiplying the voltage (how much push) by the current (how much electricity is flowing).

Exam Tip: This formula is key for calculating power in various electrical circuits and appliances.

 

Question 13. How are ammeter and voltmeter connected in a circuit?
Answer: An ammeter is connected in series in a circuit to measure current, while a voltmeter is connected in parallel across the component to measure potential difference.
In simple words: You hook an ammeter in line (series) to check current, and a voltmeter across (parallel) to check voltage.

Exam Tip: This is a crucial distinction. Connecting them incorrectly can damage the meters or provide inaccurate readings.

 

Question 14. What happens to the resistance when the length of the conductor is doubled without affecting the thickness of the conductor?
Answer: Resistance is doubled because it is directly proportional to length. If the length of a conductor increases, electrons have to travel a longer path, leading to more collisions and higher resistance.
In simple words: If you make a wire twice as long but keep it just as thick, its resistance will also double.

Exam Tip: Remember the factors affecting resistance: length (direct), area (inverse), and material (resistivity).

 

Question 15. How many electrons are there in 1 C of charge?
Answer: There are approximately \( 6.25 \times 10^{18} \) electrons in 1 Coulomb of charge. This is derived from the elementary charge of a single electron.
In simple words: One Coulomb of electric charge is made up of about 6.25 followed by 18 zeroes worth of individual electrons.

Exam Tip: Memorize this fundamental constant: \( 1 \text{ C} \approx 6.25 \times 10^{18} \) electrons. It's often used in quantitative problems.

 

Question 16. What is the S.I. unit of resistivity?
Answer: Ohm meter ( \( \Omega \text{ m} \) ).
In simple words: The standard unit for resistivity is the Ohm-meter.

Exam Tip: Do not confuse resistance (Ohm) with resistivity (Ohm-meter). Resistivity is an intrinsic property of the material, while resistance depends on the material's geometry.

 

Question 17. Find the total resistance in the given circuit.
Answer: Since no circuit diagram is provided, let's consider a simple series circuit with two resistors, \( R_1 = 5 \Omega \) and \( R_2 = 10 \Omega \).
In a series circuit, the total resistance \( R_{total} \) is the sum of individual resistances.
\( R_{total} = R_1 + R_2 = 5 \Omega + 10 \Omega = 15 \Omega \)
Therefore, the total resistance in this illustrative series circuit is 15 \( \Omega \).
In simple words: If we have a simple circuit with a 5 Ohm resistor and a 10 Ohm resistor hooked up one after another (in series), the total resistance of the whole circuit is 15 Ohms.

Exam Tip: When a circuit diagram is missing, define the type of circuit (series/parallel) and the values of components assumed in your answer. This shows your understanding of how to calculate total resistance.

 

Question 18. The combined resistance in the given circuit is 5 \( \Omega \). What is the value of R?

Answer: The heating impact of electric current offers several applications. It is utilized in electric toasters and room heaters, where electrical energy is intentionally changed into thermal energy, and this thermal energy is then employed in many devices.
In simple words: The heating effect is good because it turns electricity into heat, which we use in many devices like toasters and heaters.

Exam Tip: Focus on practical applications where heat generation is desired and useful.

 

Question 14. Why do we prefer tungsten metal in electrical bulbs and why not some other metal of the same resistivity can be used?
Answer: Tungsten metal possesses unique characteristics that make it ideal for use in electric bulbs. These specific features are not found in other metals with similar resistivity. The main properties are: 1. It gives off light and shines brightly when it gets hot. 2. It does not soften or liquefy at high temperatures (Melting Point 3380°C). Conversely, tungsten is not appropriate for fuse wires. A fuse wire serves as a safety component designed to melt and break the circuit when a current exceeds a safe limit. If tungsten, with its high melting point, were used in a fuse, it would permit a large amount of current to flow for an extended period before melting, potentially causing severe damage to other electrical appliances.
In simple words: Tungsten is good for bulbs because it glows brightly and has a very high melting point. It's not used in fuses because it wouldn't melt fast enough to stop too much current.

Exam Tip: Remember to list both emission of light and high melting point as key reasons for tungsten's use, and contrast with the low melting point needed for fuses.

 

Question 15. Name the factors on which Joule's law of heating depends. Joule's law of heating depends on the following: \( H \propto I^2RT \)
• The heat produced in a resistor is directly proportional to the square of current for a given resistance.
• The heat produced in a resistor is directly proportional to the resistance for a given current.
• The heat produced in a resistor is directly proportional to the time for which the current flows through the resistor.
Answer: The factors on which Joule's law of heating depends are: 1. **Square of Current (\( I^2 \)):** The heat produced is directly related to the square of the electric current flowing through the resistor. 2. **Resistance (R):** The heat generated is directly related to the resistance of the conductor. 3. **Time (t):** The heat produced is directly related to the duration for which the current flows through the resistor.
Thus, the formula for heat produced is \( H = I^2Rt \).
In simple words: Heat made by electricity depends on how much current is flowing (squared), how much the wire resists the current, and how long the current flows.

Exam Tip: When explaining Joule's law, clearly state the three proportionalities: current squared, resistance, and time.

 

Question 17. What do the following symbols represent in the electric circuit?
Answer: The common symbols utilized in electrical circuits are:

• **Cell:** Represents a single unit that provides electrical energy.
• **Battery:** Shows a combination of several cells.
• **Resistor:** A device that slows down the flow of electric current.
• **Variable Resistor (Rheostat):** A resistor where the resistance can be adjusted.
• **Ammeter:** An instrument used to measure electric current, always linked in a series.
• **Voltmeter:** An instrument used to measure potential difference, always linked in a parallel connection.
• **Galvanometer:** Detects small electric currents.
• **Open Switch:** A switch that breaks the circuit, stopping current flow.
• **Closed Switch:** A switch that completes the circuit, allowing current to flow.
• **Connecting Wire:** A path for current flow.
• **Wire Joint:** Indicates a connection between two or more wires.
• **Wires Crossing (without joining):** Wires crossing each other without an electrical connection.

In simple words: Each symbol in a circuit diagram stands for a specific part like a battery, a switch, or a resistor.

Exam Tip: Be familiar with standard electrical circuit symbols and their functions for drawing and interpreting circuit diagrams.

 

Question 18. What is the potential difference? Explain and give its unit with a definition.
Answer: Potential difference, often denoted as V, is the amount of work required to move a unit charge between two points in an electric field. It tells us how much energy is needed to push a charge from one spot to another. The standard unit for electric potential difference is the volt (V). One volt is defined as the potential difference between two points in a conductor carrying current when one joule of work is performed to move one coulomb of charge from one point to the other. Thus, 1 volt equals 1 joule per 1 coulomb: \( 1 \, \text{volt} = \frac{1 \, \text{Joule}}{1 \, \text{Coulomb}} \)
\( \implies 1 \, \text{V} = \frac{1 \, \text{J}}{1 \, \text{C}} \).
In simple words: Potential difference is the amount of energy needed to move a single unit of electrical charge between two points. Its unit is the volt.

Exam Tip: Clearly define potential difference in terms of work and charge, and memorize the definition of one volt along with its SI unit.

 

Question 19. How much work is done in moving a charge of 4 C across two points having a potential difference of 12 V ? Calculate the number of electrons flowing in it.
Answer: To find the work done, we use the formula \( W = VQ \), where W is work, V is potential difference, and Q is charge. We have \( Q = 4 \, C \) and \( V = 12 \, V \). So, \( W = 12 \, V \times 4 \, C = 48 \, J \). To calculate the number of electrons, we use the fact that 1 Coulomb (C) of charge contains \( 6.25 \times 10^{18} \) electrons. For 4 Coulombs, the number of electrons will be \( 4 \times 6.25 \times 10^{18} = 25.00 \times 10^{18} \) electrons. This can also be written as \( 2.5 \times 10^{19} \) electrons.
In simple words: We calculate the work done by multiplying voltage and charge, which gives 48 Joules. Then we find the total electrons by multiplying the number of electrons in 1 Coulomb by the total charge, which is \( 2.5 \times 10^{19} \) electrons.

Exam Tip: Remember the formula \( W = VQ \) for work done and the elementary charge value to convert charge to number of electrons.

 

Question 20. A circuit diagram is given below. Calculate: (a) Current through each resistor. (b) The total current in the circuit. (c) The total effective resistance of the circuit.

Answer:(a) To determine the current flowing through each resistor, we apply Ohm's Law \( I = V/R \), where V is the potential difference (voltage) and R is the resistance. Since the resistors are in parallel, the voltage across each is the same as the battery voltage, \( V = 10 \, V \). For the first resistor (\( R_1 = 2 \, \Omega \)): \( I_1 = \frac{10 \, V}{2 \, \Omega} = 5 \, A \). For the second resistor (\( R_2 = 5 \, \Omega \)): \( I_2 = \frac{10 \, V}{5 \, \Omega} = 2 \, A \). For the third resistor (\( R_3 = 10 \, \Omega \)): \( I_3 = \frac{10 \, V}{10 \, \Omega} = 1 \, A \). (b) The total current in a parallel circuit is the sum of the currents through each individual branch. So, \( I_{\text{total}} = I_1 + I_2 + I_3 = 5 \, A + 2 \, A + 1 \, A = 8 \, A \). (c) To calculate the total effective (equivalent) resistance for resistors connected in parallel, we use the formula \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \). Substituting the given resistance values: \( \frac{1}{R_p} = \frac{1}{2 \, \Omega} + \frac{1}{5 \, \Omega} + \frac{1}{10 \, \Omega} \). Finding a common denominator (10): \( \frac{1}{R_p} = \frac{5}{10 \, \Omega} + \frac{2}{10 \, \Omega} + \frac{1}{10 \, \Omega} = \frac{5+2+1}{10 \, \Omega} = \frac{8}{10 \, \Omega} \). Therefore, the total effective resistance is \( R_p = \frac{10}{8} \, \Omega = 1.25 \, \Omega \).
In simple words: For parallel resistors, the voltage across each is the same. We find current for each resistor using \( I=V/R \), then add them to get total current. The overall resistance is found using the reciprocal formula.

Exam Tip: For parallel circuits, remember that voltage is constant across each resistor, and the total current is the sum of individual currents. Use the reciprocal formula for equivalent resistance.

 

Question 21. You are given 3 resistors each of 3 ohm and you are asked to get all possible values of resistance when you connect them in different combinations. How many values of resistance can you get?
Answer: We can obtain four possible values of resistance by connecting three \( 3 \, \Omega \) resistors in different combinations. (i) **When connected in series:**To achieve the highest resistance, connect all three resistors in a series arrangement. The total resistance for series connection is the sum of individual resistances: \( R_{\text{total}} = 3 \, \Omega + 3 \, \Omega + 3 \, \Omega = 9 \, \Omega \). (ii) **When we connect in parallel:**For the lowest resistance, connect all three resistors in a parallel arrangement. The reciprocal of the total resistance in parallel is the sum of the reciprocals of individual resistances: \( \frac{1}{R_p} = \frac{1}{3 \, \Omega} + \frac{1}{3 \, \Omega} + \frac{1}{3 \, \Omega} = \frac{3}{3 \, \Omega} = 1 \, \Omega \). So, \( R_p = 1 \, \Omega \). (iii) **Two in parallel, one in series:**Connect two resistors in parallel, then connect the third resistor in series with this parallel combination. For the two resistors in parallel: \( \frac{1}{R_p} = \frac{1}{3 \, \Omega} + \frac{1}{3 \, \Omega} = \frac{2}{3 \, \Omega} \), which gives \( R_p = 1.5 \, \Omega \). Now, add the third resistor in series: \( R_{\text{total}} = R_p + 3 \, \Omega = 1.5 \, \Omega + 3 \, \Omega = 4.5 \, \Omega \). (iv) **Two in series, one in parallel:**Connect two resistors in series first, then connect the third resistor in parallel with this series combination. For the two resistors in series: \( R_s = 3 \, \Omega + 3 \, \Omega = 6 \, \Omega \). Now, connect the third resistor in parallel with \( R_s \): \( \frac{1}{R_p} = \frac{1}{R_s} + \frac{1}{3 \, \Omega} = \frac{1}{6 \, \Omega} + \frac{1}{3 \, \Omega} = \frac{1+2}{6 \, \Omega} = \frac{3}{6 \, \Omega} = \frac{1}{2 \, \Omega} \). So, \( R_p = 2 \, \Omega \). The four possible resistance values are \( 9 \, \Omega, 1 \, \Omega, 4.5 \, \Omega, \) and \( 2 \, \Omega \).
In simple words: With three \( 3 \, \Omega \) resistors, we can make four different total resistance values by connecting them in series, parallel, two parallel and one series, or two series and one parallel.

Exam Tip: Systematically explore all unique combinations (series, parallel, and mixed) to find all possible equivalent resistance values. Remember the formulas for series and parallel connections.

 

Question 22. A battery of 10 V is connected in a circuit with 3 Ω, 4 Ω, 6Ω resistors connected in series. How much current will flow through 6 Ω resistor?

Answer: To find the current flowing through the circuit, we first need to calculate the total equivalent resistance of the series circuit. For resistors connected in series, the total resistance \( R_s \) is the sum of the individual resistances: \( R_s = R_1 + R_2 + R_3 = 3 \, \Omega + 4 \, \Omega + 6 \, \Omega = 13 \, \Omega \). The battery provides a voltage of \( V = 10 \, V \). Using Ohm's Law, \( I = V/R_s \), the current flowing through the circuit is \( I = \frac{10 \, V}{13 \, \Omega} \approx 0.77 \, A \). Since resistors in a series circuit share the same current, the current flowing through the \( 6 \, \Omega \) resistor will also be approximately \( 0.77 \, A \).
In simple words: We add up all the resistances in a series circuit to get the total. Then, we divide the voltage by this total resistance to find the total current. In a series circuit, this same current flows through every part, including the \( 6 \, \Omega \) resistor.

Exam Tip: Remember that in a series circuit, the current is the same through every component, and the total resistance is the sum of individual resistances.

 

Question 23. Two electric bulbs are rated 60 W, 220 V, and 20 W, 220 V, are connected in parallel to a 220 V supply. Calculate the total electric current in the circuit.
Answer: To find the total current in the circuit, we first calculate the current drawn by each bulb separately, as they are connected in parallel and thus experience the same voltage. For Bulb 1: Given power \( P_1 = 60 \, W \) and voltage \( V = 220 \, V \). The current \( I_1 = P_1/V = \frac{60 \, W}{220 \, V} \approx 0.27 \, A \). For Bulb 2: Given power \( P_2 = 20 \, W \) and voltage \( V = 220 \, V \). The current \( I_2 = P_2/V = \frac{20 \, W}{220 \, V} \approx 0.09 \, A \). Since the bulbs are connected in parallel, the total current \( I_{\text{total}} \) drawn from the supply is the sum of the currents through each bulb: \( I_{\text{total}} = I_1 + I_2 = 0.27 \, A + 0.09 \, A = 0.36 \, A \).
In simple words: First, figure out how much current each bulb uses by dividing its power by the voltage. Then, since the bulbs are parallel, just add those currents together to get the total current.

Exam Tip: In parallel circuits, remember that total current is the sum of individual branch currents, and voltage across each parallel component is the same as the source voltage.

 

Question 24. An electric lamp draws a current of 0.3 ampere and is used for 6 hours every day for a month (30 days). Calculate the amount of charge that flows through the circuit every day and for a month.
Answer: First, we calculate the charge flowing each day. The current \( I \) is \( 0.3 \, A \). The daily usage time is \( 6 \) hours, which converts to seconds as \( 6 \times 60 \times 60 = 21600 \, s \). Using the formula \( Q = I \times t \), the charge flowing per day is \( Q_{\text{daily}} = 0.3 \, A \times 21600 \, s = 6480 \, C \). Next, we calculate the total charge flowing for a month (30 days). The total charge for the month is \( Q_{\text{month}} = Q_{\text{daily}} \times 30 = 6480 \, C \times 30 = 194400 \, C \).
In simple words: To find the daily charge, multiply current by daily usage time in seconds. To find the monthly charge, multiply the daily charge by 30 days.

Exam Tip: Ensure consistent units: current in Amperes, time in seconds, and charge in Coulombs. Remember to convert hours to seconds for calculations involving charge.

 

Question 25. Describe an activity to prove that resistance of a wire depends on its length, cross-sectional area, and material of the wire.

Answer: To prove that the resistance of a wire depends on its length, cross-sectional area, and material, perform the following activity: Set up an electric circuit that includes a cell, an ammeter, a nichrome wire of length 'l' (labeled as (1)), and a plug key, as illustrated in the provided diagram.
**Procedure:**
1. Close the circuit by inserting the plug key and record the current reading from the ammeter. 2. Replace the first nichrome wire (1) with another nichrome wire that has the same thickness but is twice the length, i.e., '2l' (labeled as (2)). Observe and note the new ammeter reading. 3. Next, replace wire (2) with a thicker nichrome wire (labeled as (3)) that has the original length 'l'. A thicker wire means it has a larger cross-sectional area. Again, note down the current flowing through the circuit. 4. Finally, instead of a nichrome wire, use a copper wire (labeled as (4)) of the same length 'l' and the same cross-sectional area as the first nichrome wire (1). Record the current value. After performing these steps, observe the differences in current readings across all cases.
**Observations and Conclusions:**
From these observations, it is noticed that:

• When the length of the wire is doubled, the ammeter reading reduces to half, indicating an increase in resistance.
• The ammeter reading increases when a thicker wire of the same material and length is used, suggesting lower resistance.
• A change in the ammeter reading is also observed when a wire of a different material is used, even if its length and cross-sectional area are kept the same.

Based on these findings, we can conclude that the resistance 'R' of a wire depends on its length 'l' (R is directly proportional to l), its cross-sectional area 'A' (R is inversely proportional to A), and the nature of the material from which the wire is made. This relationship is often summarized by the formula \( R = \rho \frac{l}{A} \), where \( \rho \) is the resistivity of the material.
In simple words: We can test different wires in a circuit to see how their length, thickness, and material change the current. This shows that resistance depends on these factors.

Exam Tip: When describing an experiment, clearly state the aim, apparatus, procedure (steps with control variables), observations, and conclusions based on the observations.

 

Question 26. What is the heating effect of electric current? Find the expression for calculating 'Heat'.
Answer: The heating effect of electric current occurs when an electric current travels through a conductor, causing its temperature to rise and heat to be released. This phenomenon is known as the heating effect of electric current.
**Derivation of the expression for heat generated:** To derive the expression for heat generated, let's consider a current \( I \) flowing through a resistor with resistance \( R \). Let the potential difference across the resistor be \( V \), and let \( t \) be the time for which a charge \( Q \) moves across the resistor. The work done in moving charge \( Q \) through a potential difference \( V \) is given by \( W = VQ \). This work done represents the energy supplied by the source. Power \( P \) is defined as the rate of doing work or energy supplied per unit time: \( P = \frac{VQ}{t} \). ---(1) Also, current \( I \) is the rate of flow of charge: \( I = \frac{Q}{t} \). ---(2) Substituting \( \frac{Q}{t} \) from (2) into (1), we get \( P = VI \). ---(3) The total energy supplied (and thus converted to heat, \( H \)) over time \( t \) is \( H = P \times t \). ---(4) Substituting \( P = VI \) from (3) into (4), we get \( H = VIt \). According to Ohm's Law, \( V = IR \). Substituting this expression for \( V \) into the equation for \( H \): \( H = (IR)It \)
\( \implies H = I^2Rt \). This final expression, \( H = I^2Rt \), is known as Joule's Law of Heating, and it calculates the heat produced.
In simple words: The heating effect is when a wire gets hot because current flows through it. We calculate this heat using the formula \( H = I^2Rt \), which means heat equals current squared, times resistance, times time.

Exam Tip: Clearly define the heating effect and meticulously derive Joule's law \( H = I^2Rt \), explaining each step and relevant formulas like Ohm's Law and the definition of power.

 

Question 1. What is the resistance of the conductors? Name two metals that offer very high resistance. Name the factors on which the resistance of the conductor depends.
Answer: Resistance is a fundamental property of a conductor that opposes the flow of electric charges through it. Its standard international (SI) unit is the ohm \( (\Omega) \). Two metals known for having high resistance are mercury and manganese. The resistance of a conductor depends on several key factors: 1. **Length of the conductor:** Resistance is directly proportional to its length. Longer conductors have higher resistance. 2. **Area of cross-section of the conductor:** Resistance is inversely proportional to its cross-sectional area. Thicker conductors have lower resistance. 3. **Nature of the material:** Different materials have different inherent resistivities. For example, some materials like nichrome have naturally higher resistance than others like copper. 4. **Temperature:** The resistance of most conductors increases with rising temperature.
In simple words: Resistance is how much a material stops electricity from flowing, measured in ohms. Mercury and manganese have high resistance. A wire's resistance depends on its length, its thickness, what it's made of, and its temperature.

Exam Tip: Define resistance clearly, state its SI unit, provide examples of materials with high resistance, and list the four key factors affecting a conductor's resistance.

 

Question 2. State Ohm's law. Draw a graph between voltage and current for a metallic conductor. Draw a circuit diagram of a circuit that consists of battery, ammeter, voltmeter, resistor, rheostat, and a key.
Answer: **Ohm's Law:** Ohm's Law states that the potential difference (V) across the ends of a given metallic wire in an electric circuit is directly proportional to the current (I) flowing through it, provided that its temperature and other physical conditions remain constant. Mathematically, this is expressed as \( V \propto I \), or \( V = IR \), where \( R \) is the constant of proportionality known as resistance.
**V-I Graph for a metallic conductor:**
**Circuit Diagram for Ohm's Law:**
The circuit diagram for verifying Ohm's Law typically includes:

• A battery (or power supply) to provide potential difference.
• An ammeter, connected in series, to measure the current.
• A voltmeter, connected in parallel across the resistor, to measure the potential difference.
• A resistor, the component whose resistance is being studied.
• A rheostat (variable resistor) to change the current and voltage in the circuit.
• A key (switch) to open or close the circuit.
• Connecting wires to complete the circuit.

In simple words: Ohm's law says voltage and current are directly related if temperature stays the same. The graph for this is a straight line. To test it, you need a circuit with a battery, ammeter, voltmeter, resistor, adjustable resistor (rheostat), and a switch.

Exam Tip: When explaining Ohm's Law, state the proportionality and the condition (constant temperature). Practice drawing the V-I graph and the standard circuit diagram for its verification.

 

Question 3. (a) A torch bulb is rated 2.5 V and 750 mA. Calculate - 1. its power 2. its resistance, 3. the energy consumed if in this bulb is lighted for 4 hours. (b) Two identical resistors, each of resistance 2 Ohm, are connected to the torch 1. in series and 2. in parallel, to a battery of 12 volts. Calculate the ratio of power consumed in two cases.
Answer: (a) Given potential difference \( V = 2.5 \, V \), current \( I = 750 \, mA = 0.75 \, A \), and time \( t = 4 \) hours. 1. **Power (P):** Power is calculated using the formula \( P = V \times I \). \( P = 2.5 \, V \times 0.75 \, A = 1.875 \, W \). 2. **Resistance (R):** Resistance is calculated using Ohm's Law \( R = V/I \). \( R = \frac{2.5 \, V}{0.75 \, A} \approx 3.33 \, \Omega \). 3. **Energy Consumed (E):** Energy is calculated using \( E = P \times t \). Since \( P = 1.875 \, W \), and \( t = 4 \, \text{hours} \), \( E = 1.875 \, W \times 4 \, \text{hours} = 7.5 \, Wh \). (b) Two identical resistors, each with resistance \( R = 2 \, \Omega \), are connected to a \( 12 \, V \) battery. 1. **When connected in series:** The total resistance in series is \( R_s = R_1 + R_2 = 2 \, \Omega + 2 \, \Omega = 4 \, \Omega \). The current flowing in the series circuit is \( I_s = V/R_s = \frac{12 \, V}{4 \, \Omega} = 3 \, A \). The power consumed in series is \( P_s = V \times I_s = 12 \, V \times 3 \, A = 36 \, W \). 2. **When connected in parallel:** The total resistance in parallel is found using \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2 \, \Omega} + \frac{1}{2 \, \Omega} = \frac{2}{2 \, \Omega} = 1 \, \Omega \). So, \( R_p = 1 \, \Omega \). The current flowing in the parallel circuit is \( I_p = V/R_p = \frac{12 \, V}{1 \, \Omega} = 12 \, A \). The power consumed in parallel is \( P_p = V \times I_p = 12 \, V \times 12 \, A = 144 \, W \). The ratio of power consumed in series to parallel is \( \frac{P_s}{P_p} = \frac{36 \, W}{144 \, W} = \frac{1}{4} \), which is \( 1:4 \).
In simple words: For the bulb, we find power by multiplying voltage and current, then resistance by dividing voltage by current, and energy by multiplying power by time. For the resistors, we calculate power for series and parallel setups and then find their ratio.

Exam Tip: Distinguish between series and parallel circuit rules for resistance and current. Carefully apply Ohm's Law and power formulas, ensuring unit consistency.

 

Question 4. 1. Three resistors are connected as shown in the circuit diagram. Through resistor 5 ohm, a current of 2 amperes is flowing. (a) What is the current through the other two resistors? (b) What is the p.d. across AB? (c) What is the total resistance? 2. Can you change the setup and arrange the resistance in a manner to get the least resistance?

Answer:1. Given that a current of \( 2 \, A \) flows through the \( 5 \, \Omega \) resistor. Since the \( 5 \, \Omega \) resistor is connected in series with the parallel combination of the \( 10 \, \Omega \) and \( 15 \, \Omega \) resistors, this \( 2 \, A \) is the total current flowing through the entire circuit. (a) **Current through the other two resistors:** First, calculate the equivalent resistance of the parallel combination of \( 10 \, \Omega \) and \( 15 \, \Omega \): \( \frac{1}{R_p} = \frac{1}{10 \, \Omega} + \frac{1}{15 \, \Omega} = \frac{3}{30 \, \Omega} + \frac{2}{30 \, \Omega} = \frac{5}{30 \, \Omega} = \frac{1}{6 \, \Omega} \) So, \( R_p = 6 \, \Omega \). The potential difference across this parallel combination is \( V_p = I_{\text{total}} \times R_p = 2 \, A \times 6 \, \Omega = 12 \, V \). Now, calculate the current through each resistor in the parallel combination: Current through \( 10 \, \Omega \) resistor: \( I_{10} = \frac{V_p}{10 \, \Omega} = \frac{12 \, V}{10 \, \Omega} = 1.2 \, A \). Current through \( 15 \, \Omega \) resistor: \( I_{15} = \frac{V_p}{15 \, \Omega} = \frac{12 \, V}{15 \, \Omega} = 0.8 \, A \). (b) **Potential difference across AB:** The p.d. (potential difference) across points A and B corresponds to the potential difference across the parallel combination of \( 10 \, \Omega \) and \( 15 \, \Omega \). As calculated above, \( V_{AB} = 12 \, V \). (c) **Total resistance:** The total resistance of the circuit is the sum of the series resistor and the equivalent parallel resistance: \( R_{\text{total}} = 5 \, \Omega + R_p = 5 \, \Omega + 6 \, \Omega = 11 \, \Omega \). 2. Yes, to achieve the least possible total resistance, all three resistors (\( 5 \, \Omega \), \( 10 \, \Omega \), and \( 15 \, \Omega \)) should be connected in a parallel arrangement.
In simple words: For the circuit, we first find the combined resistance of the parallel part. Then, using the given total current, we find the voltage across the parallel part and the current in each branch. The total resistance is found by adding the series resistor to the parallel equivalent. To get the lowest total resistance, all resistors should be connected in parallel.

Exam Tip: For mixed circuits, first simplify parallel parts, then add series resistances. Remember that current is constant in series, and voltage is constant in parallel.

 

Question 5. (a) Why is tungsten metal not used in fuse wire but is used in the bulb? (b) Give one application of the nichrome wire and state the reason for its use.
Answer: (a) Tungsten metal possesses specific characteristics that make it suitable for electric bulb filaments, such as a high melting point (approximately 3380°C) and the ability to emit light when heated without oxidizing. Conversely, tungsten is not appropriate for fuse wires. A fuse wire serves as a safety component designed to melt and break the circuit when a current exceeds a safe limit. If tungsten, with its high melting point, were used in a fuse, it would permit a large amount of current to flow for an extended period before melting, potentially causing severe damage to other electrical appliances. (b) Nichrome wire is an alloy primarily composed of nickel and chromium. It is widely used in electric heating devices, such as bread-toasters and electric irons. The main reason for its use is its high electrical resistance. This property allows nichrome to efficiently convert electrical energy into heat energy, making it an excellent material for elements that need to generate warmth.
In simple words: (a) Tungsten is great for light bulbs because it glows and needs extreme heat to melt, but it's bad for fuses because it won't melt fast enough to stop dangerous currents. (b) Nichrome wire is used in toasters because it resists electricity a lot, turning electrical energy into heat very well.

Exam Tip: Understand the specific properties required for different electrical components (e.g., high melting point for bulb filaments, low melting point for fuses, high resistance for heating elements) and match materials accordingly.

 

Question 6. (a) Why is a series arrangement not used for domestic circuits? (b) Explain why fuse wire is always connected in a series arrangement. (c) Why are copper and aluminum wires usually employed for electricity transmission?
Answer: (a) A series arrangement is not typically used for domestic circuits due to several drawbacks: 1. **Uniform current:** In a series circuit, the same current flows through all appliances. However, different household appliances often require varying amounts of current for proper operation. 2. **Breakage of circuit:** If one appliance in a series circuit stops working or develops a fault, the entire circuit breaks, and all other appliances connected in that series also stop functioning. 3. **Voltage division:** Voltage gets divided among components in series, meaning each appliance might not receive the required \( 220 \, V \) for optimal performance. 4. **High overall resistance:** Connecting many devices in series would result in a very high total resistance, leading to a much lower current drawn from the supply and reduced efficiency. (b) A fuse wire is always connected in a series arrangement within a circuit. Its purpose is to act as a safety device. When an excessively high current flows through the circuit, the fuse wire, having a low melting point, quickly heats up and melts. This action breaks the circuit, stopping the flow of dangerous current and protecting other sensitive electrical appliances from potential damage. (c) Copper and aluminum wires are commonly used for electricity transmission because of their advantageous properties: 1. **Good conductors:** Both metals have very low resistivity, meaning they offer little resistance to the flow of electric current, making them excellent conductors. 2. **Affordable:** They are relatively cheap and readily available compared to other highly conductive metals like silver. 3. **Ductile:** They are highly ductile, which means they can be easily drawn into thin wires without breaking, facilitating their use in extensive wiring systems.
In simple words: (a) Series circuits are bad for homes because all appliances get the same current, if one breaks, all stop, and the voltage gets split. (b) Fuse wires are always in series so they can stop the entire circuit when too much current flows, protecting devices. (c) Copper and aluminum are used for power lines because they conduct electricity well, are cheap, and can be easily made into wires.

Exam Tip: For domestic circuits, remember the disadvantages of series connections and the advantages of parallel. For fuse wires, focus on its safety role in series. For transmission wires, highlight conductivity, cost, and malleability.

 

Question 1. Why don't birds sitting on live wire get an electric shock?
Answer: Birds do not receive an electric shock when sitting on a single live wire because there is no potential difference across their body. For current to flow and cause a shock, there must be a complete circuit and a path for the current to flow from a high potential to a low potential, or to the ground. Since the bird is only touching one wire and not simultaneously touching another wire or the ground, no potential difference is established across its body, and thus no significant current flows through it.
In simple words: Birds don't get shocked on one wire because there's no voltage difference across their body for electricity to flow through them. They aren't completing a circuit.

Exam Tip: The key concept here is the absence of potential difference across the bird's body when it only touches one wire.

 

Question 2. Why is the tungsten metal more coiled in the bulb and not installed in a straight parallel wire form?
Answer: Tungsten wire in an electric bulb is coiled into a helical shape, rather than being a straight or parallel wire, to increase its length significantly within a very small space. This increased length, and thus increased resistance, allows the filament to heat up to extremely high temperatures and emit more light. The coiled design also concentrates the heat, making the filament glow with greater intensity and produce more brightness.
In simple words: Tungsten is coiled in a bulb to make the wire much longer in a small space, which increases its resistance, makes it hotter, and produces more light.

Exam Tip: Focus on the physical reasons for coiling: increasing length/resistance in a compact space, leading to higher temperature and more intense light emission.

 

Question 3. Name two solids and two liquid that are good conductors of electricity
Answer: Two solid materials that are good conductors of electricity are copper and aluminum. These metals have free electrons that easily carry electric current. Two liquid substances that conduct electricity well are a solution of water and dilute sulfuric acid (\( \text{H}_2\text{SO}_4 \)), and a solution of water and sodium chloride (common salt). These liquid solutions contain ions that enable them to conduct electricity.
In simple words: Good solid conductors are copper and aluminum. Good liquid conductors are diluted sulfuric acid in water and salt water.

Exam Tip: Remember common metallic conductors for solids and electrolytic solutions for liquids when listing good conductors.

 

Question 1. A student wishes to verify Ohm's law in the lab. Make a list of the materials required for this experiment.
Answer: To experimentally verify Ohm's Law in a laboratory setting, the following materials are typically needed: 1. **Ammeter:** To accurately measure the electric current flowing through the circuit. 2. **Voltmeter:** To measure the potential difference across the resistor. 3. **Connecting Wires:** To establish the electrical connections between various components. 4. **Resistor:** The component whose resistance is being investigated. 5. **Plug Key (Switch):** To open or close the circuit as needed. 6. **Battery (or Power Supply):** To provide a stable source of potential difference.
In simple words: You'll need an ammeter, voltmeter, wires, a resistor, a switch, and a battery for the Ohm's law experiment.

Exam Tip: When listing apparatus for an experiment, be comprehensive and specify the role of each component.

 

Question 2. On collecting the readings of the Ohm's law experiment, the student wants to present it and analyze it, how should this be done?
Answer: After collecting the readings from the Ohm's Law experiment, the student should organize and analyze the data as follows: 1. **Observation Table:** Create a clear observation table to record the measured values of potential difference (V) and corresponding current (I). The table should have columns for serial number, ammeter reading (current), voltmeter reading (voltage), and calculated resistance \( R = V/I \). 2. **Graph Plotting:** Plot a graph with potential difference (V) on the X-axis and current (I) on the Y-axis. For a metallic conductor, this graph is expected to be a straight line passing through the origin. 3. **Analysis:** Analyze the graph to confirm the direct proportionality between V and I, which is the essence of Ohm's Law. The slope of the V-I graph (\( \frac{V}{I} \)) will represent the resistance (R) of the conductor, which should remain constant. If the graph is a straight line, it verifies Ohm's Law.
In simple words: The student should make a table of all the measurements, then draw a graph with voltage on one side and current on the other. This graph should be a straight line to show that Ohm's law is correct.

Exam Tip: For data analysis in physics experiments, always include creating an observation table, plotting a graph (specifying axes), and interpreting the graph to draw conclusions.

 

Question 3. Before using devices like ammeter and voltmeter for any experiment what are the two points to be kept in mind for consideration?
Answer: Before using an ammeter and voltmeter in any experiment, two critical points must be considered to ensure accurate measurements: 1. **Checking for Zero Error:** Both instruments might have a 'zero error', meaning they show a reading other than zero when no current is flowing or no potential difference is applied. This error should be identified and accounted for in the readings. 2. **Determining the Least Count:** The 'least count' of an instrument is the smallest measurement it can accurately display. Knowing the least count helps in recording readings with appropriate precision and understanding the limits of the instrument's accuracy.
In simple words: Before using an ammeter or voltmeter, always check if they show zero when they should (zero error), and know the smallest value they can measure accurately (least count).

Exam Tip: Always check for zero error and determine the least count of measuring instruments before starting an experiment to ensure precise and accurate readings.

 

Question 4. For finding the total resistance, when 3 resistors are connected in the series show how the voltmeter should be connected.

Answer: To determine the total resistance of three resistors connected in series, and to show how a voltmeter should be connected, consider the following setup: When three resistors (\( R_1, R_2, R_3 \)) are connected end-to-end in a series circuit with a battery and an ammeter, the total resistance is simply \( R_{\text{total}} = R_1 + R_2 + R_3 \). To measure the total potential difference across this series combination, the voltmeter must be connected in parallel across the entire series arrangement, as shown in the diagram. This means one terminal of the voltmeter connects to the point before the first resistor and the other terminal connects to the point after the last resistor.
In simple words: When resistors are in series, you connect the voltmeter across all three of them together to measure the total voltage, which helps find the total resistance.

Exam Tip: In series circuits, a voltmeter is connected in parallel across the component(s) whose potential difference needs to be measured. For total resistance, it spans the entire combination.

 

Question 5. On connecting the 2 resistors, voltmeter, ammeter, and key correctly to the battery the ammeter did not show any readings, what could be the errors and how can it be rectified?
Answer: If an ammeter shows no reading despite correct connections in an Ohm's Law experiment, several potential errors could be responsible, along with their rectifications: 1. **Loose Connections:** The most common cause is loose or improper connections of the wires, particularly to the ammeter terminals. * **Rectification:** Securely reconnect all wires to their respective terminals, ensuring good contact. 2. **Faulty Ammeter:** The ammeter itself might be damaged or not functioning correctly. * **Rectification:** Replace the ammeter with another known working one to check if the issue persists. 3. **Incorrect Range:** The ammeter might be set to an inappropriate range (e.g., a high range for a very small current). * **Rectification:** Select the correct current range for the ammeter, typically starting with the highest range and progressively lowering it if no reading is obtained, to avoid damaging the instrument. 4. **Open Circuit:** There might be an open circuit elsewhere, such as a faulty battery, a broken resistor, or a loose wire that is not visibly apparent. * **Rectification:** Check all components individually for continuity using a multimeter and ensure the battery is charged and functional. 5. **Reversed Polarity:** Although unlikely to cause zero reading in some analog meters, reversed polarity can sometimes affect digital meters or lead to negative readings. * **Rectification:** Ensure the positive terminal of the ammeter is connected towards the positive terminal of the battery and similarly for the negative terminals.
In simple words: If the ammeter shows no reading, check for loose wires, a broken ammeter, incorrect settings, or an open circuit. Fix these by tightening connections, replacing parts, setting the correct range, or checking the circuit for breaks.

Exam Tip: When troubleshooting, systematically check for loose connections, instrument faults, incorrect range settings, and open circuits in that order.

 

Question 6. To calculate the resultant resistance of two resistors when connected in parallel, show how the connection in the circuit should be made?