Class 10 Science · Chapter 11

Electricity

363 Class 10 Science questions on Electricity from the NCERT textbook, NCERT Exemplar and CBSE board papers 2018–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 8 of 19

1. Q141 CBSE 2025
   Prove that ​if ​the ​current through a resistor is increased ​by $100\%$, then ​the ​increase in ​power dissipated through the resistor will ​be $300\%$.
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   A 100% increase doubles the current, so the power becomes four times; the increase in power is 300%.

   The power dissipated in a resistor is $P=I^{2}R$. Increasing the current by $100\%$ makes it $I^{\prime }=2I$; the resistance is unchanged, so $P^{\prime }=(2I{)}^{2}R=4I^{2}R=4P$. The increase in power is $P^{\prime }-P=3P$, and $\frac{3P}{P}\times 100=300\%$. Hence the power dissipated increases by $300\%$.

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2. Q142 CBSE 2025
   Calculate potential ​difference ​across a $4\Omega$ resistor ​that ​produces ​100 J of heat ​every second.
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   The potential difference across the resistor is 20 V.

   The heat produced each second is the power: $P=\frac{H}{t}=\frac{100\text{J}}{1\text{s}}=100$ W. From $P=\frac{V^2}{R}$, $V^2=PR=100\times 4=400$, so the potential difference across the resistor is $V=20$ V (the current through it is $5$ A).

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3. Q143 CBSE 2025
   State ​Ohm’s ​law. Give reason ​why the shape of voltage versus ​current $(V-I)$ graph for a conductor should ​be a ​straight line passing through the origin.
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   At constant temperature $V\propto I$ (Ohm's law); the graph is a straight line through the origin because $R$ is constant and $V=0$ when $I=0$.

   Ohm’s law: at constant temperature, the potential difference $V$ across the ends of a metallic conductor is directly proportional to the current $I$ flowing through it, so $V\propto I$ and hence $V=IR$, where the constant $R$ is the resistance of the conductor. The graph is a straight line because $R$ is constant, so $V=IR$ has the form $y=mx$: equal increases in current give equal increases in potential difference and every reading has the same ratio $V/I$. It passes through the origin because putting $I=0$ gives $V=0$ — with no current there is no potential difference across the conductor — so $(0,0)$ is itself a reading. The slope of the line is the resistance $R$.

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4. Q144 CBSE 2025
   A wire ​of resistance R is cut into three equal parts. If ​these three parts are then joined in parallel, ​calculate the total ​resistance ​of ​the combination ​so formed.
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   Each part has resistance $\frac{R}{3}$; the three in parallel give a total resistance of $\frac{R}{9}$.

   Each part has resistance $\frac{R}{3}$, because cutting changes only the length and $R\propto L$ at constant material and cross-section. In parallel, $\frac{1}{R_p}=\frac{3}{R}+\frac{3}{R}+\frac{3}{R}=\frac{9}{R}$, so $R_p=\frac{R}{9}$. The total resistance of the combination is $\frac{R}{9}$, that is one-ninth of the resistance of the original wire.

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5. Q145 CBSE 2025
   (A) State ​two applications ​of Joule's heating in domestic electric ​circuit. (B) Establish the relationship between ​the commercial unit of electric energy and the ​SI ​unit of ​electric energy.
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   Joule heating is used in heating appliances and the electric fuse; $1\text{kW h}=3.6\times 10^6\text{J}$.

   (A) Two applications of Joule's heating in a domestic circuit: (i) heating appliances such as the electric iron, room heater, toaster or geyser, in which a coil of high-resistance wire converts electrical energy into heat; and (ii) the electric fuse, a wire of low melting point that melts and breaks the circuit when the current becomes excessive. (The filament of an incandescent lamp is another example.) (B) The commercial unit is the kilowatt hour and the SI unit is the joule. $1\text{kW h}=1\text{kW}\times 1\text{h}=1000\text{W}\times 3600\text{s}=3.6\times 10^6\text{J}$.

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6. Q146 CBSE 2025
   Define electric ​power. ​When do we say ​that ​the power ​consumed in ​an electric circuit is 1 watt?
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   Electric power is the rate of consumption of electric energy; it is 1 watt when 1 ampere flows under 1 volt.

   Electric power is the rate at which electric energy is consumed or dissipated in an electric circuit, that is the electric work done per unit time: $P=\frac{W}{t}=VI=I^{2}R=\frac{V^2}{R}$. Its SI unit is the watt (W). The power consumed in a circuit is said to be 1 watt when a current of 1 ampere flows through it under a potential difference of 1 volt: $1\text{W}=1\text{V}\times 1\text{A}$. Equivalently, the power is 1 watt when 1 joule of electrical energy is consumed every second, $1\text{W}=1{\text{J s}}^{-1}$.

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7. Q147 CBSE 2025
   (A) State ​two applications of Joule's heating ​in domestic electric circuit. (B) ​Determine the total ​resistance of ​the parallel ​combination of three ​resistances ​of $2\Omega$, $4\Omega$ and $6\Omega$.
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   Joule heating is used in heating appliances and in the fuse; the parallel combination has $R_p=1.09\Omega$.

   (A) Two applications of Joule's heating in a domestic circuit: (i) heating appliances such as the electric iron, room heater, toaster or geyser, whose element is a coil of high-resistance wire; and (ii) the electric fuse, a wire of low melting point that melts and breaks the circuit when the current becomes excessive. (The filament of an incandescent lamp is another example.) (B) $\frac{1}{R_p}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{6+3+2}{12}=\frac{11}{12}$, so $R_p=\frac{12}{11}\Omega =1.09\Omega$ (to two decimal places).

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8. Q148 CBSE 2025
   An electric kettle is rated $750$ ​W; $220$ V. ​Can this ​kettle be used in a ​circuit which ​has a fuse of current rating $3$ ​A ? Give reason for ​your ​answer.
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   No; the kettle draws about 3.41 A, more than the 3 A rating, so the fuse would melt.

   No, it cannot. The current drawn by the kettle is $I=P/V=750/220\approx 3.41$ A, which is greater than the $3$ A rating of the fuse, so the fuse wire would melt and break the circuit. A fuse of higher rating, such as $5$ A, would be needed.

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9. Q149 CBSE 2025
   An ​electric kettle ​is rated ​230 V; ​1000 W. Calculate ​the resistance of its ​heating element when in ​operation.
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   The resistance of the heating element is $52.9\Omega$.

   $R=\frac{V^2}{P}=\frac{(230{)}^2}{1000}=52.9\Omega$

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10. Q150 CBSE 2025
    An electric ​fuse is rated $8$ A. ​Can ​it be used ​with an electrical appliance ​of rating $5$ ​kW, $200$ V ​? Justify your answer.
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    No: the appliance draws 25 A, more than the 8 A rating, so the fuse would melt.

    No, the fuse cannot be used. The appliance draws $I=P/V=5000/200=25$ A, which is more than the $8$ A rating of the fuse, so the fuse wire would melt and break the circuit as soon as the appliance is switched on. A fuse rated a little above $25$ A is needed.

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11. Q151 CBSE 2025
    A voltage source sends a current of 2 A to ​a resistor of 40 Ω connected ​across ​it for 5 ​minutes. Calculate the electrical ​energy ​supplied by ​the ​source.
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    The voltage source supplies $4.8\times 10^4$ J (48 kJ) of electrical energy.

    $W=I^{2}Rt=(2{)}^2\times 40\times 300=48000$ J, that is $4.8\times 10^4$ J (48 kJ).

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12. Q152 CBSE 2025
    Four resistors, each of resistance $2.0\Omega$, are joined end to end to form a ​square ABCD as shown. ​Using appropriate ​formula, determine ​the equivalent resistance ​of ​the combination between its two ends ​A and B.

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    The equivalent resistance between A and B is $1.5\Omega$.

    The resistor AB ($2.0\Omega$) is in parallel with the series path A–D–C–B ($2.0+2.0+2.0=6.0\Omega$), so $\frac{1}{R_{AB}}=\frac{1}{2.0}+\frac{1}{6.0}=\frac{4}{6.0}$ and $R_{AB}=1.5\Omega$.

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13. Q153 CBSE 2025
    1. (a)Define ​one ampere.
    2. (b)The resistance of a wire of $0.01$ cm ​radius is $14\Omega$. If the ​resistivity of the material of the wire ​is $44\times 10^{-8}\Omega \text{m}$, ​find the length of ​the wire. ​(Given $\pi =\frac{22}{7}$ )
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    One ampere is one coulomb of charge passing a cross-section per second; the wire is $1$ m long.

    (a) One ampere is the current that flows through a conductor when one coulomb of charge passes through any cross-section of it in one second: $1\text{A}=\frac{1\text{C}}{1\text{s}}$. (b) With $r=0.01$ cm $=1\times 10^{-4}$ m, $A=\pi r^2=\frac{22}{7}\times 10^{-8}$ m${}^2$, so $L=\frac{RA}{\rho }=\frac{14\times \frac{22}{7}\times 10^{-8}}{44\times 10^{-8}}=1$ m. The length of the wire is $1$ m.

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14. Q154 CBSE 2025
    1. (a)Define one volt potential difference between ​two points in an electric ​field.
    2. (b)​Draw a schematic ​diagram of an electric circuit of a cell of $1.5$ V, $5\Omega$ and $10\Omega$ ​resistor and a plug key, all connected in series. Calculate the current drawn ​from the cell when ​the ​key is closed.
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    One volt is 1 joule of work per coulomb; the current drawn from the cell is $0.1$ A.

    (a) One volt is the potential difference between two points in an electric field when $1$ joule of work is done to move a charge of $1$ coulomb from one point to the other: $1\text{V}=\frac{1\text{J}}{1\text{C}}$. (b) The cell, the $5\Omega$ resistor, the $10\Omega$ resistor and the plug key are joined one after another in a single loop, as in the circuit diagram. The total resistance is $R_s=5+10=15\Omega$, so the current drawn from the cell when the key is closed is $I=\frac{V}{R_s}=\frac{1.5}{15}=0.1$ A.

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15. Q155 CBSE 2025
    Show the ​arrangement of three resistors of $2\Omega$, $3\Omega$ and $6\Omega$ ​which are connected in ​such a ​way that the total/equivalent resistance ​of the combination is ​

    1. (a)$1\Omega$, ​and
    2. (b)$4\Omega$. ​Justify your answer in each case.
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    All three in parallel give $1\Omega$; the $3\Omega$ and $6\Omega$ in parallel, in series with the $2\Omega$, give $4\Omega$.

    (a) All three resistors are connected in parallel. Then $\frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1$, so $R=1\Omega$. This is possible only in parallel, because a parallel combination is always smaller than the smallest resistor and $1\Omega$ is smaller than $2\Omega$. (b) The $3\Omega$ and $6\Omega$ resistors are connected in parallel with each other, and this combination is joined in series with the $2\Omega$ resistor. For the pair, $\frac{1}{R^{\prime }}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$, so $R^{\prime }=2\Omega$, and in series $R=2+2=4\Omega$.

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16. Q156 CBSE 2025
    1. (a)​Write mathematical ​expression ​for ​Joule’s law of heating.
    2. (b)​Calculate the amount ​of heat ​generated while transferring $72000$ coulomb of charge in two hours ​through a potential difference of $220$ V.
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    Joule's law is $H=I^{2}Rt$, and the heat generated is $1.584\times 10^7$ J.

    (a) Joule's law of heating: $H=I^{2}Rt$, the heat produced in a resistor of resistance $R$ that carries a current $I$ for a time $t$. (b) $t=2$ h $=7200$ s, so $I=\frac{Q}{t}=\frac{72000}{7200}=10$ A and $H=VIt=220\times 10\times 7200=1.584\times 10^7$ J (equally, $H=VQ=220\times 72000$ J).

    Step-by-step solution →
17. Q157 CBSE 2025
    Consider the following electric circuit ​: Calculate the values of the following :

    1. (a)The total resistance ​of the circuit
    2. (b)The total current drawn from ​the ​source
    3. (c)Potential difference across ​the ​parallel combination ​of $10\Omega$ and $15\Omega$ resistors

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    Total resistance is $30\Omega$, the current is $0.5$ A, and the $10\Omega$ and $15\Omega$ parallel pair has $3$ V across it.

    (a) Total resistance $R=6\Omega +24\Omega =30\Omega$; (b) total current $I=\frac{15}{30}=0.5$ A; (c) potential difference across the parallel combination of the $10\Omega$ and $15\Omega$ resistors $=0.5\times 6=3$ V.

    Step-by-step solution →
18. Q158 CBSE 2025
    Three resistors of $2\Omega$, $3\Omega$ ​and $6\Omega$ are connected in

    1. (i)​series, and ​
    2. (ii)parallel. Draw the arrangements of ​the resistors ​and find the equivalent resistance of each ​arrangement.
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    In series the equivalent resistance is $11\Omega$; in parallel it is $1\Omega$.

    In series the three resistors lie end to end in one path and the equivalent resistance is $R_s=2+3+6=11\Omega$. In parallel all three are joined between the same two points and $\frac{1}{R_p}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$, so $R_p=1\Omega$.

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19. Q159 CBSE 2025
    1. (a)What is an ​electric fuse ? Explain its role in a domestic ​circuit.
    2. (b)An electric heater of rating ​2 kW; 220 V ​is to be operated in a room having fuse rating of 5 A. What is ​likely to happen ​when the heater ​is switched on ? Justify your answer with ​necessary calculations.
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    A fuse melts and breaks the circuit on excess current; the heater draws 9.09 A, more than 5 A, so the fuse will blow.

    (a) An electric fuse is a safety device made of a short piece of thin wire of a metal or alloy of high resistance and low melting point (such as a tin-lead alloy), connected in series in the **live** wire of a circuit. Its role in a domestic circuit is to protect the circuit and the appliances from damage due to a short circuit or overloading: so long as the current is below the rated value the fuse simply carries it, but if the current exceeds that value the heat produced ($H=I^{2}Rt$) melts the fuse wire, the circuit is broken and the current stops, preventing damage to the appliances and the wiring and preventing fire. (b) Current drawn by the heater $I=\frac{P}{V}=\frac{2000}{220}=9.09$ A. This is much greater than the $5$ A rating of the fuse (the excess is about $4.09$ A). Therefore the fuse wire will melt — the fuse will blow — as soon as the heater is switched on; the circuit will break and the heater will not run. Equivalently, a $5$ A fuse at $220$ V can pass at most $220\times 5=1100$ W $=1.1$ kW, whereas the heater needs $2$ kW. A fuse of at least $10$ A would be required.

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20. Q160 CBSE 2025
    The ​electrical resistivity ​of three materials A, B and C at 20°C is ​given below :

    1. (i)Classify these materials as conductor, alloy and insulator.
    2. (ii)Give one example of each of these materials and state ​one use of each material in ​the design ​of ​an electrical appliance say an electric stove or an electric iron.

    Electrical resistivity of three materials at 20°C

    Material	Resistivity ($\Omega$ m)
    A	$10^{17}$
    B	$44\times 10^{-6}$
    C	$1.62\times 10^{-8}$

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    A is an insulator (ebonite), B an alloy (manganin), C a conductor (copper); the alloy is used for the heating element.

    (i) A is an insulator, B is an alloy and C is a conductor. (ii) C is copper ($1.62\times 10^{-8}$ $\Omega$ m); it is used for the connecting wires and the lead that carry the current to the stove or the iron, because its very low resistivity wastes almost no energy on the way. B is an alloy such as manganin ($44\times 10^{-6}$ $\Omega$ m); alloy wire is used for the heating element or coil of the stove or iron, because its high resistivity produces a large amount of heat and it does not oxidise readily at a high temperature. A is an insulator such as ebonite or hard rubber ($10^{17}$ $\Omega$ m); it is used for the handle of the iron and for the covering of the wires, so that the current cannot pass to the user.

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