Class 10 Science · Chapter 11

Electricity

218 Class 10 Science questions on Electricity from the NCERT textbook, NCERT Exemplar and CBSE board papers 2023–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 5 of 11

1. Q81 CBSE 2026
   (A) In the ​given electric circuit of Ohm's law ​experiment, ​identify P and ​X. (B) Write the ​uses ​of Q ​and X in the ​given circuit.

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   P is a voltmeter and X is a rheostat; Q (ammeter) measures the current, and X varies it.

   (A) P is a voltmeter - it is connected in parallel across the resistor R, so it reads the potential difference $V$ across R. X is a rheostat, that is, a variable resistance, shown by the slanting arrow drawn across the zig-zag. (B) Q is the ammeter: it is connected in series, so the whole circuit current passes through it and it measures the current $I$ flowing in the circuit, through R. X, the rheostat, is used to change (regulate) the current in the circuit without changing the voltage source, so that several pairs of readings of $V$ and $I$ can be taken with the same resistor and $V/I$ shown to be constant; it also keeps the current within safe limits.

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2. Q82 CBSE 2026
   In the ​given electric ​circuit, ​what ​will be the reading ​in ​instrument P and Q ​?

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   P is a voltmeter reading 12 V and Q is an ammeter reading 1 A.

   P is connected across the 12 $\Omega$ resistor, so it is a voltmeter, and it reads the full potential difference of the battery, $12$ V. Q is connected in series in the loop, so it is an ammeter, and it reads the current $I=\frac{V}{R}=\frac{12\text{V}}{12\Omega }=1$ A.

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3. Q83 CBSE 2026
   A fuse ​in ​electric circuit ​is ​rated 4 A. Can it be used with an electric heater of ​rating ​2 kW, ​200 ​V ? Explain your answer.
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   No; the heater draws 10 A, more than the 4 A rating, so the fuse would melt.

   No, a $4$ A fuse cannot be used. The heater draws $I=\frac{P}{V}=\frac{2000\text{W}}{200\text{V}}=10$ A in normal working. The current rating of a fuse is the largest current its wire can carry without melting, and $10$ A is two and a half times the fuse's $4$ A rating, so the fuse wire would melt and break the circuit the moment the heater is switched on. A fuse rated just above $10$ A must be used instead.

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4. Q84 CBSE 2026
   Calculate the ​total resistance of ​the given circuit and ​find the ​total ​current flowing in it.

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   The total resistance is $6\Omega$ and the total current is 2 A.

   Total resistance of the circuit $R=6\Omega$, and the total current flowing in it $I=2$ A.

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5. Q85 CBSE 2026
   How ​would ​you join ​three resistors each of resistance $6\Omega$, so ​that the ​equivalent resistance of the ​combination ​is

   1. (i)$9\Omega$, and
   2. (ii)$4\Omega$ ?
   Show answer

   For $9\Omega$, join two in parallel and the third in series with them; for $4\Omega$, join two in series and the third across them.

   (i) $9\Omega$: join two of the $6\Omega$ resistors in parallel and connect the third in series with that pair — $\frac{6\times 6}{6+6}+6=3+6=9\Omega$. (ii) $4\Omega$: join two of the $6\Omega$ resistors in series and connect the third in parallel across that pair — $\frac{(6+6)\times 6}{(6+6)+6}=\frac{72}{18}=4\Omega$.

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6. Q86 CBSE 2026
   1. (a)Define volt, the unit of potential difference.
   2. (b)​Calculate the work done required ​to move an electron between two points A ​and ​B located at a ​potential ​difference of 100 V in an ​electric field.
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   One volt is one joule of work per coulomb moved; the work done in moving the electron is $1.6\times 10^{-17}$ J.

   (a) One volt is the potential difference between two points in a current-carrying conductor when one joule of work is done to move a charge of one coulomb from one point to the other; that is, $1\text{V}=\frac{1\text{J}}{1\text{C}}$. (b) $V=\frac{W}{Q}$, so $W=QV=(1.6\times 10^{-19}\text{C})\times (100\text{V})=1.6\times 10^{-17}$ J.

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7. Q87 CBSE 2026
   What do the symbols ​given below ​represent in an electric circuit ? Write one ​function of each.

   1. (i)[symbol ​printed - ​see figure]
   2. (ii)[symbol printed - ​see figure]

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   Symbol (i) is a voltmeter, which measures potential difference; symbol (ii) is a variable resistance (rheostat), which changes the current.

   (i) The symbol represents a voltmeter. Function: it measures the potential difference, in volts, between two points of a circuit - in practice across a component. It is connected in parallel with that component, with its $+$ terminal towards the positive terminal of the cell, and its very high resistance means it draws almost no current from the circuit. (ii) The symbol represents a variable resistance, that is, a rheostat (the arrow is its sliding contact). Function: connected in series in the circuit, it lets the resistance be changed while the circuit is running, and so varies the current in the circuit without changing the cell - by Ohm's law $I=V/R$, more resistance means less current.

   Step-by-step solution →
8. Q88 CBSE 2026
   1. (a)​State ​the law which gives the relation ​between the electric current ​flowing through a ​given conductor and the potential difference applied across it.
   2. (b)Draw ​V-I graph for a given ​conductor as per the law stated in ​above question (a).
   Show answer

   Ohm's law: at constant temperature $V$ is directly proportional to $I$; the $V$–$I$ graph is a straight line through the origin.

   (a) The law is Ohm's law. It states that the potential difference across the ends of a given metallic conductor is directly proportional to the electric current flowing through it, provided its temperature and other physical conditions remain unchanged. In symbols, $V\propto I$, so $V=IR$, where the constant of proportionality $R$ is the resistance of the conductor. (b) The graph of $V$ against $I$ - potential difference on the vertical axis, current on the horizontal axis - is a straight line passing through the origin, drawn in the figure in this solution. Its slope is constant and equals $V/I=R$, the resistance of the conductor; for the line drawn there, $2.0\text{V}÷0.4\text{A}=5\Omega$.

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9. Q89 CBSE 2026
   A bulb consumed 24 ​J ​of energy per second, ​when ​connected to a 12 V ​supply. ​Calculate the power consumed when ​it is connected to a 6 V supply.
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   $P_2=6$ W: the bulb consumes 6 W on the 6 V supply.

   $P_2=6$ W. The bulb consumes $24$ J per second at $12$ V, so $P_1=24$ W and its resistance is $R=V_1^2/P_1=144/24=6\Omega$. The same resistance on a $6$ V supply gives $P_2=V_2^2/R=36/6=6$ W.

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10. Q90 CBSE 2026
    The resistance of a ​wire of ​0.01 ​cm radius ​and 1.0 ​cm ​length is ​7 Ω. Calculate ​its resistivity.
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    The resistivity of the wire's material is $\rho =2.2\times 10^{-5}\Omega \text{m}$.

    The resistivity of the material of the wire is $\rho =2.2\times 10^{-5}\Omega$ m (that is, $2.2\times 10^{-3}\Omega$ cm).

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11. Q91 CBSE 2026
    An ​electric heater is rated 220 ​V; 11 ​A. Calculate the ​power consumed if ​the ​heater is operated at ​200 V.
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    $P=2000$ W: the heater consumes 2 kW when operated at 200 V.

    The heater consumes $2000$ W ($2$ kW) at $200$ V. Its resistance is $R=\frac{220}{11}=20\Omega$, so $P=\frac{V^2}{R}=\frac{(200{)}^2}{20}=2000$ W.

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12. Q92 CBSE 2026
    (a) Name a device which is used to ​:

    1. (i)Maintain a constant potential difference in a ​circuit.
    2. (ii)Change the ​electric current in an electric circuit. (b) When ​the potential ​difference between the terminals of an electric heater is 110 V, a current of 5 A flows ​through it. What will be ​the value of current flowing through it when the potential difference is increased to ​220 V ?
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    A cell or battery keeps the potential difference constant, a rheostat changes the current, and the heater draws $10$ A at 220 V.

    (a) (i) A cell or a battery maintains a constant potential difference; (ii) A rheostat (variable resistance) changes the current. (b) The resistance of the heater is $R=\frac{110\text{V}}{5\text{A}}=22\Omega$, so at $220$ V the current is $I=\frac{220\text{V}}{22\Omega }=10$ A.

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13. Q93 CBSE 2026
    The values ​of current $I$ flowing in a given ​resistor for ​the ​corresponding values ​of potential difference $V$ applied across the ends of resistor ​are given below in the table : Plot ​a graph between $V$ and $I$ and calculate the resistance of that resistor.

    $I$ (Amperes)	0.4	0.8	1.2	1.6	2.0
    $V$ (Volts)	1.2	2.4	3.6	4.8	6.0

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    The graph is a straight line through the origin; its slope gives the resistance $R=3\Omega$.

    Plotting $V$ (vertical axis) against $I$ (horizontal axis) gives a straight line through the origin, so the resistor obeys Ohm's law. Its resistance is the slope of the line: $R=\frac{\Delta V}{\Delta I}=\frac{6.0-1.2}{2.0-0.4}=\frac{4.8}{1.6}=3\Omega$. The resistance of the resistor is $3\Omega$.

    Step-by-step solution →
14. Q94 CBSE 2026
    Consider ​the given electric circuit : Calculate the following ​:

    1. (a)Total ​resistance of ​the circuit
    2. (b)​The electric current drawn from the battery
    3. (c)​Potential ​difference between points P and Q

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    Total resistance is $5\Omega$, the battery supplies $2$ A, and the potential difference between P and Q is $5$ V.

    (a) Total resistance $=5\Omega$: between S and R, $(1+4)\Omega =5\Omega$ in parallel with $5\Omega$ gives $2.5\Omega$; between Q and P, $(3+2)\Omega =5\Omega$ in parallel with $5\Omega$ gives $2.5\Omega$; in series, $2.5+2.5=5\Omega$. (b) Current drawn from the battery $I=\frac{10\text{V}}{5\Omega }=2$ A. (c) Potential difference between P and Q $=2\text{A}\times 2.5\Omega =5$ V.

    Step-by-step solution →
15. Q95 CBSE 2026
    1. (a)State Ohm's law and justify that V ​– I graph for Ohm's law will always be a straight line passing through the origin.
    2. (b)The ​figure ​shows V – I graphs for two identical wires A and ​B ​having ​the same length and area of ​cross-section, but made of different materials. Which of these wires will be more suitable for making the heating element of an electric heater and why ?

    Show answer

    At constant temperature $V\propto I$, so the graph is a straight line through the origin; wire B is better for heating.

    (a) Ohm's law: the potential difference $V$ across the ends of a given metallic wire is directly proportional to the current $I$ flowing through it, provided its temperature remains the same; that is, $V=IR$, where the resistance $R$ is a constant. Justification: with $R$ constant, $V=RI$ has the form $y=mx$, which is a straight line of constant slope $R$, and $V=0$ when $I=0$, so the line passes through the origin. (b) Wire B. Its $V$-$I$ graph is the steeper one, so for the same current it needs a larger potential difference and its resistance $R=\frac{V}{I}$ is larger. Since both wires have the same length and the same area of cross-section, $R=\rho \frac{l}{A}$ shows that B's material has the higher resistivity; a heating element is made of a high-resistivity material, which produces more heat ($H=I^{2}Rt$) for the current it carries.

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16. Q96 CBSE 2026
    1. (a)Why ​does ​an electric bulb become dim when an electric heater in parallel circuit ​is switched ON ?
    2. (b)How ​to connect three resistors ​each of resistance $8\Omega$, so that the ​equivalent ​resistance of the combination is $12\Omega$ ​? Draw diagram of the combination and justify your answer.
    Show answer

    The heater's large current lowers the voltage across the bulb; two $8\Omega$ resistors in parallel plus one in series give $12\Omega$.

    (a) A bulb and a heater in a house are joined in parallel across the supply. The heater has a low resistance, so when it is switched on it draws a large current and the total current drawn from the mains increases greatly. This larger current flows through the supply wires, which have some resistance $r$, so a larger potential difference $Ir$ is used up in them and the potential difference left across the bulb decreases. As a result the current through the bulb decreases, its power $P=\frac{V^2}{R}$ falls, and the bulb glows dim. (b) Connect two of the $8\Omega$ resistors in parallel and join this combination in series with the third $8\Omega$ resistor. Justification: for the parallel pair $\frac{1}{R_p}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}$, so $R_p=4\Omega$; in series with the third resistor, $R_{\text{eq}}=4+8=12\Omega$.

    Step-by-step solution →
17. Q97 CBSE 2026
    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.

    1. (a)Describe the cause of heating in the above-mentioned case.[1]
    2. (b)A current $I$ flows through a resistor of resistance $R$ when the potential difference across it is $V$. Applying Ohm's law, write the formula for amount of heat produced by the resistor in time $t$.[1]
    3. (c) (i)Write any two practical applications of heating effect of electric current.[2]
    4. (c) (ii)OR : Define the commercial unit of electric energy and express it in Joules (J).[2]
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    The fridge warms the kitchen air by the heating effect of current, $H=I^{2}Rt$; irons and bulbs use it; or $1\text{kWh}=3.6\times 10^6$ J.

    (a) The heating is caused by the heating effect of electric current. The refrigerator runs on electricity, and its current-carrying parts - the windings of the compressor motor and the connecting wires - have resistance, so part of the electrical energy they draw is continuously converted into heat, which is given out into the kitchen air near the refrigerator. (b) The heat produced is $H=VIt$. By Ohm's law $V=IR$, so $H=I^{2}Rt$ (equivalently $H=\frac{V^{2}t}{R}$). (c) (i) Any two of these: the electric iron, toaster, room heater, kettle and geyser, whose resistive heating elements get hot; the electric bulb, whose filament becomes so hot that it glows; the electric fuse, whose wire melts and breaks the circuit when too large a current flows. OR (c) (ii) The commercial unit of electric energy is the kilowatt hour (kWh), the energy used by an appliance of power $1$ kW in $1$ hour: $1\text{kWh}=1000\text{W}\times 3600\text{s}=3.6\times 10^6$ J.

    Step-by-step solution →
18. Q98 CBSE 2026
    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.

    1. (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 ?
    2. (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 ?
    3. (c)Define ‘resistivity’. Write its SI unit. Compare the resistivity of an alloy with its constituents metals.
    4. (c)OR : Give reason : (i) Tungsten is used almost exclusively for making the filament of electric lamps. (ii) Conductors of bread-toasters are made of an alloy rather than a pure metal.
    Show answer

    Current halves to $\frac{X}{2}$ for double length and doubles to $2X$ for double area; resistivity is $\rho =\frac{RA}{l}$, in ohm metre.

    (a) Doubling the length doubles the resistance ($R=\rho l/A$), so with the same cell the current is halved: the ammeter reads $\frac{X}{2}$ ampere. (b) Doubling the area of cross-section halves the resistance, so the ammeter reading is doubled, to $2X$ ampere. (c) Resistivity is the resistance of a conductor of the material having unit length and unit area of cross-section, $\rho =\frac{RA}{l}$; its SI unit is the ohm metre ($\Omega$ m). The resistivity of an alloy is generally much higher than that of its constituent metals - nichrome, for instance, has about $100\times 10^{-6}\Omega$ m against nickel's $6.84\times 10^{-8}\Omega$ m - and alloys do not oxidise readily at high temperatures. OR (c) (i) Tungsten has a very high melting point (about $3380^{\circ }$C) and a high resistivity, and it can be drawn into a fine, strong wire, so its filament can be kept white hot and glow without melting. (ii) An alloy has a much higher resistivity than a pure metal, so for the same current ($H=I^{2}Rt$) the element gives out far more heat than the connecting wires, and an alloy does not oxidise (burn) readily at high temperature, so the heating conductor lasts.

    Step-by-step solution →
19. Q99 CBSE 2026
    Four cells, a thin wire AB, a ​voltmeter, an ​ammeter, a plug key and some thick connecting wires are ​taken. The circuit is set ​up as shown below. The key ​is inserted into the plug and readings of ammeter (I) and voltmeter (V) are noted. Then the plug key is removed. Now two cells are connected in the circuit in series. Plugging in ​the key the ​new values of ‘V’ and ‘I’ are noted. The experiment is ​repeated by connecting three cells in series and then four cells. In each case ‘V’ and ‘I’ are noted.

    1. (A)Why was the plug key removed after each measurement of ‘V’ and ‘I’ and inserted again for the next step ?
    2. (B)What change takes place in the reading of the voltmeter when a new cell is added in series in the circuit ? Give reason.
    3. (C)In which case will the readings of voltmeter and ammeter be maximum in the above experiment ? What conclusion can be drawn from the straight line graph between ‘V’ and ‘I’ ?
    4. (C)OR : Suppose this experiment is carried out using only one cell but wires of same length & thickness but made of different materials like nichrome wire, fuse wire, copper wire and iron wire which are replaced one by one. Out of V and I which will change and why ?
    Show answer

    Key removed so wire AB stays cool; each added cell raises the voltmeter reading; four cells give maximum readings; the line shows Ohm's law.

    (A) So that current flows through the circuit only while a reading is being taken. A current left flowing through the thin wire AB would heat it ($H=I^{2}Rt$); the resistance of a metal rises with temperature, so $V/I$ would drift and the V-I graph would not come out straight, since Ohm's law holds only at constant temperature. Removing the key lets AB cool to room temperature between readings and also saves the cells. (B) The voltmeter reading increases. Cells joined in series have their emfs added, so each extra cell applies a larger potential difference across AB; as the wire is unchanged its resistance is unchanged, so the current increases in the same proportion and $V/I$ stays constant. (C) Both readings are maximum when all four cells are connected in series, since that gives the greatest total emf. The straight line through the origin shows that $V\propto I$, i.e. $V/I$ is constant for the conductor at constant temperature - this is Ohm's law - and that constant, the slope of the line, is the resistance $R$ of the wire AB. (C) OR: $I$ will change and $V$ will remain the same. The same single cell applies the same potential difference in every trial, while the wires have the same length $l$ and the same area of cross-section $A$, so by $R=\rho l/A$ their resistances differ only through the resistivity $\rho$ of the material; hence $I=V/R$ is different for each wire - largest for copper, which has the lowest resistivity of the four, and smallest for nichrome, which has the highest. This shows that resistance depends on the nature of the material.

    Step-by-step solution →
20. Q100 CBSE 2026
    1. (i)​The ​given ​electric circuit is a ​part of an electrical device. Use the information given in the electric circuit diagram ​to calculate : (I) Potential ​difference across the ends of resistor ${\mathrm{R}}_2$. (II) Value of resistor ${\mathrm{R}}_2$. (III) Value of resistor ${\mathrm{R}}_1$.
    2. (ii)Write ​the ​factors on which resistance of a conductor depends and derive the formula for resistance of a given conductor.

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    $V_{R_2}=6.0$ V, $R_2=12\Omega$, $R_1=1.0\Omega$; resistance depends on length, area, material and temperature, and $R=\rho \frac{l}{A}$.

    (i) (I) The potential difference across ${\mathrm{R}}_2$ is $6.0$ V, the same as across the $4.0\Omega$ branch beside it, since $1.5\times 4.0=6.0$ V. (II) ${\mathrm{R}}_2=6.0/0.5=12\Omega$, the current through it being $2.0-1.5=0.5$ A. (III) ${\mathrm{R}}_1=1.0\Omega$, because the $2.0\Omega$ resistor takes $4.0$ V and the parallel section $6.0$ V, leaving $2.0$ V across ${\mathrm{R}}_1$ at $2.0$ A. (ii) The resistance of a conductor depends on its length $l$ (directly), on its area of cross-section $A$ (inversely), on the nature of its material and on its temperature. Since $R\propto l$ and $R\propto 1/A$, we get $R\propto l/A$, that is $R=\rho l/A$, where the constant $\rho$ is the resistivity of the material, measured in $\Omega$ m.

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