CBSE Class 10 Science Electricity VBQs Set 03

Question. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer: A voltmeter is connected in parallel to a circuit with its +ve terminal to the point at higher potential and -ve terminal to the point at lower potential.

Question. A copper wire has diameter 0.5 mm and resistivity of \( 1.6 \times 10^{-8} \text{ } \Omega\text{m} \). What will be the length of this wire to make its resistance \( 10 \text{ } \Omega \)? How much does the resistance change if the diameter is doubled?
Answer: \( d = 0.5 \text{ mm}, r = \frac{0.5}{2} \times 10^{-3} \text{ m} \), \( \rho = 1.6 \times 10^{-8} \text{ } \Omega\text{m}, R = 10 \text{ } \Omega \)
Using \( R = \frac{\rho l}{A} \), we get \( l = \frac{R\pi r^2}{\rho} = \frac{10 \times 3.14 \times (0.25 \times 10^{-3})^2}{1.6 \times 10^{-8}} = 122.6 \text{ m} \approx 123 \text{ m} \).
If the diameter is doubled, the area becomes 4 times. Since \( R \propto 1/A \), the resistance becomes one-fourth of the original one.

Question. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer: \( V = IR \)
\( \implies R = \frac{V}{I} \)
\( \implies R = \frac{12}{2.5 \times 10^{-3}} = 4800 \text{ } \Omega = 4.8 \text{ k}\Omega \).

Question. A battery of 9 V is connected in series with resistors of \( 0.2 \text{ } \Omega, 0.3 \text{ } \Omega, 0.4 \text{ } \Omega, 0.5 \text{ } \Omega \) and \( 12 \text{ } \Omega \), respectively. How much current would flow through the \( 12 \text{ } \Omega \) resistor?
Answer: \( R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 \text{ } \Omega \), \( V = 9 \text{ V} \),
Current drawn, \( I = \frac{V}{R} = \frac{9}{13.4} \text{ A} = 0.67 \text{ A} \).
Since all the resistors are in series, the same current, i.e. 0.67 A flows through the \( 12 \text{ } \Omega \) resistor.

Question. How many \( 176 \text{ } \Omega \) resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer: When \( N \) resistors each \( R \text{ } \Omega \) are in parallel, \( R_P = \frac{R}{N} \).
Current drawn from cell, \( I = \frac{V}{R_P} = \frac{VN}{R} \)
\( \therefore N = \frac{IR}{V} = \frac{5 \times 176}{220} = 4 \).

Question. 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 with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer: \( N \) bulbs of power \( P \) each connected in parallel will make the total power of \( NP \), therefore, using \( I = \frac{NP}{V} \)
\( I = 5 = \frac{N \times 10}{220} \)
\( \therefore N = \frac{5 \times 220}{10} = 110 \)

Question. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of \( 24 \text{ } \Omega \) resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer: When used individually, \( I = \frac{220}{24} = 9.16 \text{ A} \) in both of them.
When used in series, \( R_s = 24 + 24 = 48 \text{ } \Omega \),
\( \implies I_s = \frac{220}{48} \text{ A} = 4.58 \text{ A} \)
When used in parallel, \( R_P = \frac{24 \times 24}{48} = 12 \text{ } \Omega \)
\( \implies I_P = \frac{220}{12} \text{ A} = 18.3 \text{ A} \).

Question. Compare the power used in the \( 2 \text{ } \Omega \) resistor in each of the following circuits.
(i) a 6 V battery in series with \( 1 \text{ } \Omega \) and \( 2 \text{ } \Omega \) resistors, and
(ii) a 4 V battery in parallel with \( 12 \text{ } \Omega \) and \( 2 \text{ } \Omega \) resistors.
Answer: (i) \( I = \frac{6}{1 + 2} = 2 \text{ A} \). Since current flowing is same in both resistors, power used in \( 2 \text{ } \Omega \) resistor \( P_1 = I^2R = (2)^2 \times 2 = 8 \text{ W} \).
(ii) Since both \( 12 \text{ } \Omega \) and \( 2 \text{ } \Omega \) are in parallel to the 4 V source, Power used in \( 2 \text{ } \Omega \) resistor \( P_2 = \frac{V^2}{R} = \frac{4^2}{2} = \frac{16}{2} = 8 \text{ W} \).
Comparison between the power used in both cases = \( \frac{P_1}{P_2} = \frac{8 \text{ W}}{8 \text{ W}} = 1 \).

Question. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer: Current drawn by 100 W bulb = \( \frac{100}{220} \text{ A} = 0.45 \text{ A} \).
Current drawn by 60 W bulb = \( \frac{60}{220} \text{ A} = 0.27 \text{ A} \).
Total current drawn from the line = \( 0.45 \text{ A} + 0.27 \text{ A} = 0.72 \text{ A} \).

Question. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer: Energy consumed by 250 W TV set in 1 h = \( 250 \times 1 = 250 \text{ Wh} \).
Energy consumed by 1200 W toaster in 10 min = \( 1200 \times \frac{10}{60} = 200 \text{ Wh} \).
\( \therefore \) Energy consumed by TV set is more than the energy consumed by toaster in the given timings.

Question. An electric heater of resistance \( 8 \text{ } \Omega \) draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Answer: \( R = 8 \text{ } \Omega, I = 15 \text{ A}, t = 2 \text{h} \)
Rate of heat developed = \( \frac{H}{t} = \frac{I^2Rt}{t} = 15^2 \times 8 = 225 \times 8 = 1800 \text{ Js}^{-1} \).

Question. Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a 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 are copper and aluminium wires usually employed for electricity transmission?
Answer: (a) It has high melting point and emits light at a high temperature.
(b) It has more resistivity and less temperature coefficient of resistance.
(c) (i) All appliances do not get same potential in series arrangement. (ii) All appliances cannot be individually operated.
(d) \( R \propto \frac{1}{\text{Area of cross - section}} \)
(e) They are very good conductors of electricity.

Question. A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly
(a) \( 10^{20} \)
(b) \( 10^{16} \)
(c) \( 10^{18} \)
(d) \( 10^{23} \)
Answer: (a) \( 10^{20} \)

Question. A cylindrical conductor of length l and uniform area of cross-section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross-section
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A
Answer: (c) 2A

Question. If the current I through a resistor is increased by 100 % (assume that temperature remains unchanged), the increase in power dissipated will be
(a) 100 %
(b) 200 %
(c) 300 %
(d) 400 %
Answer: (c) 300 %

Question. The resistivity does not change if
(a) the material is changed
(b) the temperature is changed
(c) the shape of the resistor is changed
(d) both material and temperature are changed
Answer: (c) the shape of the resistor is changed

Question. In an electrical circuit two resistors of \( 2 \text{ } \Omega \) and \( 4 \text{ } \Omega \) respectively are connected in series to a 6 V battery. The heat dissipated by the \( 4 \text{ } \Omega \) resistor in 5 s will be
(a) 5 J
(b) 10 J
(c) 20 J
(d) 30 J
Answer: (c) 20 J

Question. Electrical resistivity of a given metallic wire depends upon
(a) its length
(b) its thickness
(c) its shape
(d) nature of the material
Answer: (d) nature of the material

Question. What is the minimum resistance which can be made using five resistors each of 1/5 \( \Omega \)?
(a) 1/5 \( \Omega \)
(b) 1/25 \( \Omega \)
(c) 1/10 \( \Omega \)
(d) 25 \( \Omega \)
Answer: (b) 1/25 \( \Omega \)

Question. In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?
(a) Brightness of all the bulbs will be the same
(b) Brightness of bulb A will the maximum
(c) Brightness of bulb B will be more than that of A
(d) Brightness of bulb C will be less than that of B
Answer: (c) Brightness of bulb B will be more than that of A

Question. Should the resistance of an ammeter be low or high? Give reason.
Answer: The resistance of an ammeter should be low so that it will not disturb the magnitude of current of the circuit when connected in series in a circuit.

Question. How does use of a fuse wire protect electrical appliances?
Answer: The fuse wire is always connected in series with the live wire or electrical devices. If the flow of current exceeds the specified preset value due to some reason, the heat produced melts it and disconnects the circuit or the device from the mains.

Question. What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?
Answer: The resistance offered by a metallic wire of unit length and unit cross-sectional area is called electrical resistivity. We know that \( R = \frac{\rho l}{A} \) and \( V = IR \). So, \( R \propto l \) and \( I \propto 1/R \) (V is constant). Hence, when the length of wire is doubled the resistance becomes double and current decreases to half.

Question. Electric current originates from which part of a conductor?
(a) Nucleus
(b) Positively charged protons
(c) Negatively charged electrons
(d) All atoms of a conductor
Answer: (c) Negatively charged electrons

Question. Ohm’s law states which relationship between electrical quantities?
(a) Potential difference = current times resistance
(b) Potential difference = current divides resistance
(c) Potential difference = charge divides time
(d) Potential different = charge times current
Answer: (a) Potential difference = current times resistance

Question. The resistance of wire varies inversely as
(a) length
(b) resistivity
(c) temperature
(d) area of cross-section
Answer: (d) area of cross-section

Question. How much energy is consumed when a current of 5 A flows through the filament of a room heater having resistance of \( 10 \text{ } \Omega \) for 2 hours. Express it in joules.
(a) \( 18 \times 10^5 \text{ J} \)
(b) \( 1.8 \times 10^6 \text{ J} \)
(c) \( 1.8 \times 10^7 \text{ J} \)
(d) \( 3.6 \times 10^6 \text{ J} \)
Answer: (c) \( 1.8 \times 10^7 \text{ J} \)

Question. Assertion: The Ohm’s law equation does not apply to non-Ohmic conducting devices.
Reason: For Ohmic conducting devices, the value of R is independent of the value of ‘V’.
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.

Question. Assertion: When a wire is stretched such that its area of cross-section is halved, its resistance would become 16 times.
Reason: Resistance is inversely proportional to the fourth power of area of cross-section of the wire.
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (d) The statement of the Assertion is false but the Reason is true.

Question. How do we connect voltmeter and ammeter in an electric circuit? What is likely to happen if the position of these instruments are interchanged?
Answer: An ammeter is always connected in series in a circuit, while a voltmeter is connected in parallel across the points where the potential difference is to be measured. If their positions are interchanged, the ammeter (which has very low resistance) being in parallel will draw a very high current and may get damaged. The voltmeter (which has very high resistance) being in series will significantly increase the total circuit resistance, reducing the current to almost zero.

Question. A bulb cannot be used in place of resistor to verify Ohm’s law. Justify this statement with reasons.
Answer: A bulb filament is made of tungsten, whose resistance increases significantly as its temperature rises when current flows through it. Since Ohm's law is only valid when physical conditions like temperature remain constant, a bulb (a non-ohmic conductor) cannot be used to verify the linear relationship between \( V \) and \( I \).

Question. Define resistance. Write the SI unit of resistance and define it. Match the correct range of resistivity with the materials given.
(a) Conductors (i) \( 10^{-6} \text{ } \Omega\text{m} \)
(b) Alloys (ii) \( 10^{12} \) to \( 10^{17} \text{ } \Omega\text{m} \)
(c) Insulators (iii) \( 10^{-6} \) to \( 10^{-8} \text{ } \Omega\text{m} \)
Answer: Resistance is the property of a conductor to resist the flow of charges through it. Its SI unit is Ohm (\( \Omega \)). One Ohm is the resistance of a conductor such that when a potential difference of 1 V is applied across its ends, a current of 1 A flows through it.
Matching:
(a) — (iii)
(b) — (i)
(c) — (ii)

Question. Nichrome wire of length ‘l’ and radius ‘r’ has resistance of \( 10 \text{ } \Omega \). How would the resistance of wire change when
(i) only the diameter is doubled?
(ii) only length of wire is doubled?
Answer: Resistance \( R = \frac{\rho l}{A} = \frac{\rho l}{\pi r^2} \).
(i) If diameter is doubled, radius \( r \) is doubled. New resistance \( R' = \frac{\rho l}{\pi (2r)^2} = \frac{1}{4} R = \frac{10}{4} = 2.5 \text{ } \Omega \).
(ii) If length is doubled, \( R'' = \frac{\rho (2l)}{\pi r^2} = 2R = 2 \times 10 = 20 \text{ } \Omega \).

Question. (a) Consider a conductor of resistance ‘R’, length ‘l’, thickness ‘d’ and resistivity \( \rho \). Now this conductor is cut into four equal parts. What will be the new resistivity of each of these parts? Why?
(b) Find the resistance if all of these parts are connected in (i) parallel (ii) series
(c) Out of the combinations of resistors mentioned above in the previous part, for a given voltage which combination will consume more power and why?
(d) An electric iron consumes energy at the rate of 420 W when heating is at maximum rate and 180 W when heating is at minimum. The voltage is 220 V. What is the current and resistance in each case?
Answer: (a) The resistivity remains the same (\( \rho \)) because resistivity is a characteristic property of the material and does not depend on the dimensions of the conductor.
(b) Each part has resistance \( r = R/4 \).
(i) Parallel: \( \frac{1}{R_p} = \frac{1}{R/4} \times 4 = \frac{16}{R} \implies R_p = \frac{R}{16} \).
(ii) Series: \( R_s = \frac{R}{4} \times 4 = R \).
(c) Parallel combination will consume more power because \( P = \frac{V^2}{R_{eq}} \). Since the equivalent resistance is lower in parallel, the power consumption is higher.
(d) At Maximum: \( I = \frac{P}{V} = \frac{420}{220} \approx 1.91 \text{ A} \); \( R = \frac{V}{I} = \frac{220}{1.91} \approx 115.18 \text{ } \Omega \).
At Minimum: \( I = \frac{P}{V} = \frac{180}{220} \approx 0.82 \text{ A} \); \( R = \frac{V}{I} = \frac{220}{0.82} \approx 268.29 \text{ } \Omega \).

Electric Potential and Potential Difference

Question. Some work is done to move a charge Q from infinity to a point A in space. The potential of the point A is given as V. What is the work done to move this charge from infinity in terms of Q and V?

Answer: The work done (\( W \)) to move a charge \( Q \) from infinity to a point where the potential is \( V \) is given by the product of the charge and the potential.
\( \implies W = Q \times V \)

Factors on which the Resistance of a Conductor Depends

Question. A complete circuit is left on for several minutes, causing the connecting copper wire to become hot. As the temperature of the wire increases, the electrical resistance of the wire
(a) decreases
(b) remains the same
(c) increases
(d) increases for some time and then decreases.

Answer: (c) increases

Resistance of a System of Resistors

Question. Two unequal resistances are connected in parallel. If you are not provided with any other parameters (e.g., numerical values of I and R), what can be said about the voltage drop across the two resistors?

Answer: When resistors are connected in parallel, the potential difference (voltage drop) across each resistor is the same, regardless of their individual resistance values.

Electric Power

Question. (a) It would cost a man Rs. 3.50 to buy 1.0 kW h of electrical energy from the main electricity board. His generator has a maximum power of 2.0 kW. The generator produces energy at this maximum power for 3 hours. Calculate how much it would cost to buy the same amount of energy from the main electricity board.
(b) A student boils water in an electric kettle for 20 minutes. Using the same mains supply he wants to reduce the boiling time of water. To do so should he increase or decrease the length of the heating element? Justify your answer.

Answer: (a) Energy produced by the generator \( E = \text{Power} \times \text{time} = 2.0\text{ kW} \times 3\text{ h} = 6.0\text{ kWh} \).
Cost to buy 1 kWh = Rs. 3.50.
Total cost = \( 6.0 \times 3.50 = \text{Rs. } 21.00 \).
(b) The heat required to boil the water (\( H \)) is constant. The mains supply voltage (\( V \)) is also constant. The heat produced is given by \( H = \frac{V^2 t}{R} \).
To reduce the boiling time \( t \), the resistance \( R \) must be decreased (since \( t \propto R \)).
As resistance \( R = \rho \frac{l}{A} \), decreasing the length (\( l \)) of the heating element will decrease its resistance. Thus, the student should decrease the length of the heating element.