Electricity — Class 10 Science PYQs

22 previous-year board questions (2026) with marking-scheme solutions, grouped by topic and marks.

Try each question first, then press (or tap Show Solution) to reveal the answer. Press again for the next question.

Marks:

Electric current and potential difference

1 Mark Questions
11 Mark · March 2026 · Standardopen ↗
For Questions number 8 and 9, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Assertion (A): There are generally a greater number of individuals at the lower trophic levels of an ecosystem.
Reason (R) : The greatest number is of the producers.
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(B) / Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
21 Mark · March 2026 · Standardopen ↗
The correct way to connect an ammeter and a voltmeter in an electric circuit is :
  • (a)Ammeter in parallel and voltmeter in series
  • (b)Ammeter and voltmeter both in parallel
  • (c)Ammeter in series and voltmeter in parallel
  • (d)Ammeter and voltmeter both in series
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(C) /Ammeter in series and voltmeter in parallel.
31 Mark · March 2026 · Standardopen ↗
Work done in displacing an electron between two points having potential difference of $\text{1 V}$, is :
  • (a)$\text{– 1.6 \times 10^{-19} C}$
  • (b)$\text{– 1.6 \times 10^{-19} J}$
  • (c)$\text{+ 1.6 \times 10^{-19} J}$
  • (d)$\text{– 1.6 \times 10^{-19} V}$
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(B) / $-1.6 \times 10^{-19}$ J (1 Mark)
2 Marks Questions
42 Marks · March 2026 · Standardopen ↗
(a) Define volt, the unit of potential difference.
(b) Calculate the work done required to move an electron between two points A and B located at a potential difference of $100\,\text{V}$ in an electric field.
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1 Volt is the potential difference between two points in a current carrying conductor when one Joule work is done to move a charge of 1Coulomb from one point to the other. / $1V = \frac{1J}{1C}$
$V = \frac{W}{Q}$
$W=QxV$
$W = 1.6 \times 10^{-19} \times 100$
$W = 1.6 \times 10^{-17} J$

Ohm's law and the V-I relationship

1 Mark Questions
51 Mark · March 2026 · Standardopen ↗
In an electric circuit carrying current $I$, a $$\begin{aligned}& 2 \\ & \Omega\end{aligned}$$ resistor is replaced by an $$\begin{aligned}& 8 \\ & \Omega\end{aligned}$$ resistor without changing the connected (ideal) battery. Then :
  • (a)current becomes $\frac{I}{4}$ and potential difference $\frac{V}{4}$
  • (b)current becomes $\frac{I}{4}$ and potential difference remains $V$
  • (c)current becomes $4I$ and potential difference remains $V$
  • (d)current becomes $4I$ and potential difference $\frac{V}{4}$
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(B)/ current becomes $\frac{1}{4}$ and potential difference remains V.

Resistance, resistivity, factors affecting resistance

2 Marks Questions
62 Marks · March 2026 · Standardopen ↗
(a) The resistance of a wire of $$\begin{aligned}& 0.01 \\ & \text{cm}\end{aligned}$$ radius and $$\begin{aligned}& 1.0 \\ & \text{cm}\end{aligned}$$ length is $$\begin{aligned}& 7 \\ & \Omega\end{aligned}$$. Calculate its resistivity.
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(a) $r = 0.01 \text{ cm}=1 \times 10^{-4} \text{ m}$
$l = 1 \text{ cm} = 0.01 \text{ m}$
$R=\rho \frac{l}{A}$
$\rho = \frac{RA}{l} = \frac{R \times \pi r^2}{l}$
$\rho = \frac{7 \times 22 \times 10^{-8}}{7 \times 0.01}$
$\rho = 22 \times 10^{-8} \times 10^2$
$\rho = 22 \times 10^{-6} \Omega m =2.2 \times 10^{-5} \Omega m$ (1)
(b) Resistance of electric heater $R = \frac{V}{I}$
$R = \frac{220}{11}$
$R = 20 \Omega$ (½)
$P = \frac{V^2}{R}$
$P = \frac{200 \times 200}{20}$
$P = 2000\text{W}/2 \text{ kW}$ (½)
4 Marks Questions
74 Marks · March 2026 · Standardopen ↗
Three students Shweta, Ayesha and Samridhi were performing an experiment to understand the factors on which the resistance of a conductor depends. Each one of them completed electric circuit with the help of a cell, an ammeter, a plug key and wire.
Shweta put nichrome wire of length ‘l’ in the circuit and after plugging the key, noted current in the ammeter.
Ayesha put nichrome wire of same thickness but twice the length i.e. ‘2l’ in the circuit and after plugging the key, noted current in the ammeter.
Samridhi took copper wire of length ‘l’ and same thickness in the circuit and after plugging the key, noted current in the ammeter.
(a) If the ammeter reading is X ampere with nichrome wire of length ‘l’, then what will be the ammeter reading if the length of nichrome wire is doubled with same area of cross-section ?
(b) What happens to the ammeter reading if the area of cross-section of nichrome wire is doubled, keeping the length of wire ‘l’ the same ?
(c) Define ‘resistivity’. Write its SI unit. Compare the resistivity of an alloy with its constituents metals.
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(a) Ammeter reading becomes $\frac{X}{2}$ / halved (1 Mark)
(b) Ammeter reading becomes $2X$ / doubled (1 Mark)
(c) • Resistivity is equal to electrical resistance of a conductor of unit length and unit area of cross section. (1 Mark)
• SI unit = $\Omega\text{ m}$ / ohm metre (½ Mark)
• Resistivity of an alloy is higher than its constituent metals. (½ Mark)
OR
(c) (i) It has high melting point. (1 Mark)
(ii) The resistivity of an alloy is generally higher than that of its constituent metals. / Alloys do not oxidise (burn) readily at high temperatures. (1 Mark)
5 Marks Questions
85 Marks · March 2026 · Standardopen ↗
(a) (i) Due to change in length and area of cross-section of a conductor, resistance of conductor changes while resistivity does not change. Why?
(ii) Conductors of electric toasters and electric iron are made of an alloy rather than a pure metal. Why ?
(iii) Define the S.I. unit of electric current.
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(a)
(i) As Resistance, $R = \rho \frac{l}{A}$, it changes with change in length and
area of cross section of conductor.
But resistivity of conductor is the characteristic property of
material and hence it does not change. (1)
(ii) The resistivity of an alloy is generally higher than that of its
constituent metals. /
Alloys do not oxidise (burn) readily at high temperatures. (1)
(iii) 1 Ampere is constituted by the flow of 1 Coulomb of
charge per second. /
$1\text{A}=1\text{C}/1\text{s}$ (1)
95 Marks · March 2026 · Standardopen ↗
(B) (i) Define electrical resistivity of a material.
(ii) How will the resistivity and resistance of a conducting wire change if the length and area of cross-section of wire both are doubled? Justify your answer.
(iii) Calculate the electric current flowing through a $$\begin{aligned}& 4 \\ & \Omega\end{aligned}$$ resistor that produces $$\begin{aligned}& 100 \\ & \text{J}\end{aligned}$$ of heat every second.
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(i) Resistivity is the resistance offered by a wire of unit length and unit area of cross section in an electric circuit. (1 Mark)
(ii) • Resistivity of conducting wire will remain same. (½ Mark)
• Resistivity is the characteristic property of the material. (½ Mark)
• Resistance will remain same (½ Mark)
• $R = \rho \frac{l}{A}$
If $l' = 2l$, $A' = 2A$
$R' = \rho \frac{l'}{A'}$
$R' = \rho \frac{2l}{2A}$
$R' = R$ (½ Mark)
(iii) $H = I^2 R t$ (1 Mark)
$I = \sqrt{\frac{H}{Rt}}$
$I = \sqrt{\frac{100}{4 \times 1}}$ (1 Mark)
$I = 5\text{ A}$
105 Marks · March 2026 · Standardopen ↗
Attempt either subpart (a) or (b) :
OR
(b) (i) State the relationship of the resistance (R) of a wire with its area of cross-section (A) and its length (l). With the help of this relation, define resistivity of the material of the wire.
(ii) In an electric circuit, $\text{2 A}$ electric current flows through a $\text{40 \Omega}$ resistor for $\text{50}$ seconds. (I) Calculate the heat produced in the wire. (II) If another $\text{40 \Omega}$ resistor is connected in parallel to the given $\text{40 \Omega}$ resistor, calculate the heat produced in the circuit in $\text{50}$ seconds.
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(b) (i) • $R = \rho \frac{l}{A}$ (1 Mark)
• Resistivity of a conductor is the resistance offered by a conductor of length 1 m and area of cross section 1 m² (1 Mark)
(ii) H = $I^2Rt$ (½ Mark)
H = $(2)^2 \times 40 \times 50$ (½ Mark)
H = $8000$ J (½ Mark)
Voltage applied across $40\Omega$, $V=I \times R$ (½ Mark)
= $2 \times 40 = 80$ V (½ Mark)
Equivalent resistance $\frac{1}{R} = \frac{1}{40} + \frac{1}{40}$ (½ Mark)
$R = 20\Omega$ (½ Mark)
Current flowing through circuit
$I = \frac{V}{R}$ (½ Mark)
$I = \frac{80}{20}$ (½ Mark)
$I = 4$ A (½ Mark)
H = $I^2Rt$ (½ Mark)
H = $(4)^2 \times 20 \times 50$ (½ Mark)
H = $16000$ J (½ Mark)

Series and parallel combinations of resistors

5 Marks Questions
115 Marks · March 2026 · Standardopen ↗
(b) (i) How many bulbs of resistance $$\begin{aligned}& 8 \\ & \Omega\end{aligned}$$ each should be connected in parallel combination to draw a current of $$\begin{aligned}& 2 \\ & \text{A}\end{aligned}$$ from a battery of $$\begin{aligned}& 4 \\ & \text{V}\end{aligned}$$?
(ii) Name the device used for measuring electric current. How is it connected in a circuit ?
(iii) State Joule's law of heating.
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(b)
(i) $V=4\text{V}, I= 2\text{A}$
Resistance of circuit $R = \frac{V}{I}$
$R=\frac{4}{2}$
$R = 2\Omega$ (½)
Let 'n' be the number of bulbs
$\frac{1}{R} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + ---n$
$\frac{1}{R} = \frac{n}{8}$
$\frac{1}{2} = \frac{n}{8}$
$n=4$
Therefore, 4 bulbs of resistance $8 \Omega$ should be connected in parallel. (1)
(ii)
• Ammeter (1)
• In series (1)
(iii) Heat generated through a current carrying conductor is directly
proportional to square of current, resistance of conductor and time
for which current flows in conductor. (1)
/
$H=I^2Rt$ (1)

Heating effect of electric current

1 Mark Questions
121 Mark · March 2026 · Standardopen ↗
Read the following passage and answer the questions that follow :
Swati, a class 10 student, observes that when she passes close to the refrigerator in her kitchen, she feels the heat, although the things kept inside the refrigerator are cool.
Describe the cause of heating in the above-mentioned case.
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(a) A part of current is consumed into useful work and rest is expended in heat to raise the temperature of gadget.
(any other suitable explanation)
(b)
$W=V \times Q= VIt = IR \times It$
$H = I^2Rt$ / $H = \frac{V^2}{R}t$
131 Mark · March 2026 · Standardopen ↗
A current $\text{I}$ flows through a resistor of resistance $\text{R}$ when the potential difference across it is $\text{V}$. Applying Ohm's law, write the formula for amount of heat produced by the resistor in time $\text{t}$.
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(a) A part of current is consumed into useful work and rest is expended in heat to raise the temperature of gadget.
(any other suitable explanation)
(b)
$W=V \times Q= VIt = IR \times It$
$H = I^2Rt$ / $H = \frac{V^2}{R}t$
141 Mark · March 2026 · Standardopen ↗
Read the following passage and answer the questions that follow :
Swati, a class 10 student, observes that when she passes close to the refrigerator in her kitchen, she feels the heat, although the things kept inside the refrigerator are cool.
(a) Describe the cause of heating in the above-mentioned case.
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(a) A part of current is consumed into useful work and rest is expended in heat to raise the temperature of gadget. (1 Mark) (any other suitable explanation)
(b) $W=V \times Q= VIt = IR \times It$
$H = I^2Rt$ / $\frac{V^2}{R}t$ (1 Mark)
(c) (i) Electric heater, Oven, Electric iron (Any two, Any other) (1+1 Marks)
OR
(c) (ii) When 1 kilowatt of power is used for 1 hour then energy consumed is 1 kWh (1 Mark)
1 kWh = $3.6 \times 10^6\text{J}$ (1 Mark)
2 Marks Questions
152 Marks · March 2026 · Standardopen ↗
Write any two practical applications of heating effect of electric current.
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(a) A part of current is consumed into useful work and rest is expended in heat to raise the temperature of gadget.
(any other suitable explanation)
(b)
$W=V \times Q= VIt = IR \times It$
$H = I^2Rt$ / $H = \frac{V^2}{R}t$
3 Marks Questions
163 Marks · March 2026 · Standardopen ↗
(a) Name a device which is used to :
(i) Maintain a constant potential difference in a circuit.
(ii) Change the electric current in an electric circuit.
(b) When the potential difference between the terminals of an electric heater is $$\begin{aligned}& 110 \\ & \text{V}\end{aligned}$$, a current of $$\begin{aligned}& 5 \\ & \text{A}\end{aligned}$$ flows through it. What will be the value of current flowing through it when the potential difference is increased to $$\begin{aligned}& 220 \\ & \text{V}\end{aligned}$$ ?
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(a) (i) Battery / Electric cell
(ii) Rheostat / Variable resistance
(b) Resistance of the heater, $R = \frac{V}{I} = \frac{110}{5} = 22 \Omega$
Current through the heater, $I' = \frac{V'}{R} = \frac{220}{22} = 10 A$
Alternate answer for (b):
• According to Ohm's law, potential difference is directly proportional to current.
• When the potential difference is doubled, current will also be doubled. So, value of current will be 10A.
4 Marks Questions
174 Marks · March 2026 · Standardopen ↗
Read the following passage and answer the questions that follow :
Swati, a class 10 student, observes that when she passes close to the refrigerator in her kitchen, she feels the heat, although the things kept inside the refrigerator are cool.
(a) Describe the cause of heating in the above-mentioned case.
(b) A current $\text{I}$ flows through a resistor of resistance $\text{R}$ when the potential difference across it is $\text{V}$. Applying Ohm's law, write the formula for amount of heat produced by the resistor in time $\text{t}$.
(c) (i) Write any two practical applications of heating effect of electric current.
OR
(c) (ii) Define the commercial unit of electric energy and express it in Joules (J).
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(c) (i) Electric heater, Oven, Electric iron (Any two, Any other)

Electric power and energy

2 Marks Questions
182 Marks · March 2026 · Standardopen ↗
(b) An electric heater is rated $$\begin{aligned}& 220 \\ & \text{V}\end{aligned}$$; $$\begin{aligned}& 11 \\ & \text{A}\end{aligned}$$. Calculate the power consumed if the heater is operated at $$\begin{aligned}& 200 \\ & \text{V}\end{aligned}$$.
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(a) $r = 0.01 \text{ cm}=1 \times 10^{-4} \text{ m}$
$l = 1 \text{ cm} = 0.01 \text{ m}$
$R=\rho \frac{l}{A}$
$\rho = \frac{RA}{l} = \frac{R \times \pi r^2}{l}$
$\rho = \frac{7 \times 22 \times 10^{-8}}{7 \times 0.01}$
$\rho = 22 \times 10^{-8} \times 10^2$
$\rho = 22 \times 10^{-6} \Omega m =2.2 \times 10^{-5} \Omega m$ (1)
(b) Resistance of electric heater $R = \frac{V}{I}$
$R = \frac{220}{11}$
$R = 20 \Omega$ (½)
$P = \frac{V^2}{R}$
$P = \frac{200 \times 200}{20}$
$P = 2000\text{W}/2 \text{ kW}$ (½)
192 Marks · March 2026 · Standardopen ↗
Define the commercial unit of electric energy and express it in Joules (J).
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Solution not available.
5 Marks Questions
205 Marks · March 2026 · Standardopen ↗
(i) How much electric current will an electric iron draw from $220\,\text{V}$ source if the resistance of its heating element when hot, is $55\,\Omega$ ? Calculate the power consumed by the electric iron when it is operated at $220\,\text{V}$.
(ii) In a house, 3 bulbs of $100\,\text{watt}$ each, are lit for $5\,\text{hours}$ daily and an electric heater of $1.0\,\text{kW}$ is used for half an hour daily. Calculate the total energy consumed in a month of $30\,\text{days}$ and its cost at the rate of ₹ $3.60$ per $\text{kWh}$.
(iii) With reason explain, why are alloys commonly used to make elements of electrical heating devices.
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$I = \frac{V}{R}$
$I = \frac{220}{55}$
$I = 4 A$
Power of electric iron $P = VI$
$P = 220 \times 4$
$P = 880 W$
$E=Pxt$
Energy (3 bulbs) $= 3 \times 100 \times 5$
$= 1500$
$= 1.5 \text{ kWh}$
Energy (electric heater) $= 1.0 \times 0.5$
$= 0.5 \text{ kWh}$
Total energy consumed (1 day) $= 1.5 + 0.5 = 2 \text{ kWh}$
Total energy consumed (30 days) $= 30 \times 2$
$= 60 \text{ kWh}$
$= 60 \text{ units}$
Total cost = Units $\times$ Rate
$= 60 \times 3.60$
$= 216$
The resistivity of an alloy is generally higher than that of its constituent metals. / Alloys do not oxidise (burn) readily at high temperatures.
215 Marks · March 2026 · Standardopen ↗
(i) Define the SI unit of power. Express the electric power of an electric appliance in terms of potential difference and current flowing through it.
(ii) In a house, 3 bulbs of $100\text{ W}$ each, are lit for $5\text{ hours}$ daily and an electric heater of $1.0\text{ kW}$ is used for half an hour daily. Calculate the total energy consumed in a month of 30 days and its cost at the rate of ₹ $3.60$ per $\text{kWh}$.
(iii) Convert the commercial unit of electric energy into Joule (J).
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(i) • Power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V / If one joule energy is consumed in one second then power of instrument is said to be 1 watt / 1 W = 1 volt $\times$ 1 ampere (1 Mark)
• $P = VI$ (1 Mark)
(ii) $E = P \times t$ (½ Mark)
Energy (3 bulbs) $= 3 \times 100 \times 5 = 1500\text{ Wh} = 1.5\text{ kWh}$
Energy (electric heater) $= 1.0 \times 0.5 = 0.5\text{ kWh}$
Total energy consumed (1 day) $= 1.5 + 0.5 = 2\text{ kWh}$ (½ Mark)
Total energy consumed (30 days) $= 30 \times 2 = 60\text{ kWh} = 60\text{ units}$
Total cost = Units $\times$ Rate $= 60 \times 3.60 = \text{\text{Rs} } 216$ (1 Mark)
(ii) $1\text{ kW h} = 1000\text{ watt} \times 3600\text{ second} = 3.6 \times 10^6\text{ watt second} = 3.6 \times 10^6\text{ joule (J)}$ (1 Mark)
225 Marks · March 2026 · Standardopen ↗
Attempt either subpart (a) or (b) :
(a) (i) In a house, $\text{5}$ fans, rated $\text{100 W}$; $\text{220 V}$ each are used for $\text{4}$ hours daily and an electric heater rated $\text{1000 W}$; $\text{220 V}$ is used for $\text{5}$ hours daily. Calculate the cost of electrical energy consumed by the fans and the electric heater in a month of $\text{30}$ days if the cost of $\text{1}$ unit of electrical energy is $\text{\text{Rs}5}$.
(ii) You have two fuse wires X and Y with electric current rating $\text{5 A}$ and $\text{10 A}$ respectively. Which of the two would you use with a heater rated $\text{2000 W}$; $\text{220 V}$ ? Justify your answer.
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(a) (i) E=Pxt (½ Mark)
E (fan) = $5 \times 100 \times 4$ (½ Mark)
= $2000$ Wh (½ Mark)
= $2$ kWh (½ Mark)
E(heater) = $1000 \times 5$ (½ Mark)
= $5000$ Wh (½ Mark)
= $5$ kWh (½ Mark)
Total energy, E = E(fan) + E(heater) (½ Mark)
= $2 + 5 = 7$ kWh (½ Mark)
Cost = $7 \times 30 \times 5$= Rs $1050$ (1 Mark)
(ii) P=VI (½ Mark)
Current required for heater
$I = \frac{P}{V}$ (½ Mark)
$I = \frac{2000}{220}$ (½ Mark)
$I = 9.09$ A (½ Mark)
Current rating of Y wire is slightly greater than required value. Hence fuse wire Y will be used. (1 Mark)