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 4 of 11

1. Q61 Exemplar · Short Answer Questions
   Three 2 ​Ω resistors, ​A, B and C, are connected as ​shown ​in ​Figure 12.7. Each ​of them dissipates energy and can withstand a maximum ​power of 18W without melting. Find the maximum current that can flow through the three resistors?

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   The maximum current is 3 A, limited by resistor A, which carries the whole current and dissipates its rated 18 W.

   The maximum current that can flow is $I=3$ A. At that current resistor A carries the whole $3$ A and dissipates its full rated $18$ W, while B and C each carry $1.5$ A and dissipate only $4.5$ W, so A is the resistor that sets the limit.

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2. Q62 Exemplar · Short Answer Questions
   Should ​the resistance ​of an ammeter be ​low or ​high? ​Give ​reason.
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   An ammeter should have low resistance, ideally zero, so that connecting it in series does not change the current.

   The resistance of an ammeter should be **low**, and ideally zero. An ammeter is connected in series, so its resistance adds to the resistance of the circuit and the current becomes $I=\frac{V}{R+r}$ instead of $\frac{V}{R}$. Only if $r$ is very small compared with $R$ does the meter leave the current essentially unchanged, so that what it reads is the current that was flowing before it was put in.

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3. Q63 Exemplar · Short Answer Questions
   Draw a circuit ​diagram of an electric circuit ​containing a cell, a key, an ammeter, a ​resistor of 2 Ω in series with a combination of two resistors ​(4 Ω each) in parallel and a voltmeter across the ​parallel combination. Will the potential difference across the ​2 Ω resistor be the same as ​that across the parallel combination of 4Ω resistors? Give reason.
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   Yes, the potential differences are equal: the two $4\Omega$ resistors in parallel make $2\Omega$, equal to the series resistor carrying the same current.

   Yes, the two potential differences are the same. The two $4\Omega$ resistors in parallel are equivalent to $R_p=2\Omega$, which is equal to the series resistor. Since the two sections are in series they carry the same current $I$, so by $V=IR$ each has a potential difference of $2I$ — that is, each takes half the potential difference of the cell. With a $6$ V cell, for instance, $I=1.5$ A and each section drops $3$ V.

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4. Q64 Exemplar · Short Answer Questions
   How ​does use ​of ​a fuse ​wire protect ​electrical appliances?
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   A fuse, a high-resistance low-melting-point wire in series, melts when the current is excessive and breaks the circuit, saving the appliance.

   A fuse is a short piece of wire of high resistance and low melting point, connected in series in the live wire, with a current rating slightly above the circuit’s normal working current. It therefore carries the whole current the appliance draws. If a short circuit or an overload makes the current exceed the rating, the heat developed in the fuse, $H=I^{2}Rt$, rises as the square of the current; the fuse wire quickly reaches its melting point and melts, breaking the circuit. The supply is cut off before the large current can overheat the appliance or its wiring, so the fuse is destroyed instead of the appliance.

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5. Q65 Exemplar · Short Answer Questions
   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?
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   Resistivity is the resistance of unit length and unit cross-section; doubling the length doubles $R$, halving the current to 2.5 A.

   Electrical resistivity is the resistance of a conductor of that material of unit length and unit area of cross-section: from $R=\frac{\rho l}{A}$ we get $\rho =\frac{RA}{l}$, and its SI unit is the ohm metre ($\Omega$ m). It is a property of the material and its temperature, not of the size or shape of the conductor. When the length of the wire is doubled, $\rho$ and $A$ stay the same, so $R$ doubles. The same potential difference then drives half the current, $I=\frac{V}{2R}$, and the ammeter reading falls from $5$ A to $2.5$ A.

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6. Q66 Exemplar · Short Answer Questions
   What ​is ​the commercial unit ​of electrical energy? Represent it ​in ​terms of ​joules.
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   The commercial unit is the kilowatt hour, and $1\text{kW h}=3.6\times 10^6$ J.

   The commercial unit of electrical energy is the kilowatt hour (kW h), called one "unit" of electricity on an electricity bill. Since $1$ kW $=1000$ W and $1$ h $=3600$ s, $1$ kW h $=1000\times 3600=3.6\times 10^6$ J.

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7. Q67 Exemplar · Short Answer Questions
   A current of ​1 ampere flows in a series circuit containing an electric ​lamp ​and a ​conductor ​of 5 Ω when connected to a 10 V battery. Calculate ​the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and ​potential difference across the lamp will take place? Give reason.
   Show answer

   The lamp is $5\Omega$; adding the parallel $10\Omega$ resistor changes neither the current through the conductor nor the potential difference across the lamp.

   Resistance of the lamp: the total resistance is $R=V/I=10/1=10\Omega$, so $R_{\text{lamp}}=10-5=5\Omega$. After the $10\Omega$ resistor is joined in parallel with the series combination there is **no** change: the $5\Omega$ conductor still carries $1$ A and the lamp still has $5$ V across it. The reason is that the new resistor is connected across the same two points as the series combination, so the potential difference driving that combination is still $10$ V and its resistance is still $10\Omega$. Only the current drawn from the battery changes, from $1$ A to $2$ A, the extra ampere going through the new branch.

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8. Q68 Exemplar · Short Answer Questions
   Why ​is parallel arrangement ​used in ​domestic wiring?
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   In parallel each appliance gets the full 220 V and works independently, and the supply can deliver the large total current.

   Domestic wiring is in parallel for four reasons. (1) Every appliance is connected directly across the live and neutral wires, so each receives the full supply voltage of $220$ V that it is designed for; in series that voltage would be shared out. (2) Each appliance has its own branch and its own switch, so it can be operated independently and a break in one branch does not stop the others. (3) Each appliance draws the current its own resistance demands, $I=V/R$, so a $100$ W lamp and a $1000$ W heater can work together; in series both would be forced to carry the same current. (4) The equivalent resistance of a parallel combination is smaller than any single branch, so the supply can deliver the large total current the whole household load requires.

   Step-by-step solution →
9. Q69 Exemplar · Short Answer Questions
   $B_1$, $B_2$ and $B_3$ are three identical bulbs ​connected as shown in Figure 12.8. ​When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

   1. (i)What happens to the glow of the other two bulbs when the bulb $B_1$ gets fused?
   2. (ii)What happens to the reading of $A_1$, $A_2$, $A_3$ and A ​when the bulb $B_2$ ​gets fused?
   3. (iii)How much power is dissipated in ​the circuit when all the three bulbs glow ​together?

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   The other bulbs keep glowing; when $B_2$ fuses, $A_2$ reads zero and $A$ reads 2 A; the circuit dissipates 13.5 W.

   (i) The other two bulbs, $B_2$ and $B_3$, go on glowing with exactly the same brightness as before. They are connected in parallel directly across the battery, so each still has the full $4.5$ V across it and still carries $1$ A; only the reading of A falls, from $3$ A to $2$ A. (ii) $A_2$ reads zero, because its branch is broken; $A_1$ and $A_3$ are unaffected and each still reads $1$ A; and A reads $1+1=2$ A instead of $3$ A. (iii) $P=VI=4.5\times 3=13.5$ W.

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10. Q70 Exemplar · Long Answer Questions
    Three incandescent bulbs of ​100 W each are connected in series in an ​electric circuit. In another ​circuit another set of three ​bulbs of the same ​wattage ​are connected in parallel to the same source.

    1. (a)Will the bulb in the two circuits glow with the same brightness? Justify your answer.
    2. (b)Now let ​one bulb in both the circuits get ​fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.
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    No: parallel bulbs are nine times as bright; when one fuses, series bulbs go out but parallel bulbs keep glowing.

    (a) No. Each bulb has resistance $R=V^2/100$. In the parallel circuit every bulb has the full source voltage $V$ across it, so each dissipates its rated $100$ W. In the series circuit the three equal resistances share the voltage, so each bulb has only $V/3$ across it and dissipates $P=\frac{(V/3{)}^2}{R}=\frac{100}{9}\approx 11.1$ W. A parallel bulb is therefore nine times as bright as a series bulb, and the two circuits take $300$ W and $33.3$ W in all. (b) In the series circuit the remaining bulbs do **not** glow: a fused bulb breaks the only path, so no current flows anywhere and all three go out. In the parallel circuit the remaining bulbs **do** go on glowing, with unchanged brightness, because each has its own path across the source and only the fused branch is broken.

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11. Q71 Exemplar · Long Answer Questions
    State ​Ohm’s law? How can it be ​verified experimentally? Does ​it ​hold good ​under ​all ​conditions? ​Comment.
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    Ohm's law is $V=IR$ at constant temperature; a straight $V$–$I$ graph verifies it, but it fails for non-ohmic devices and heated filaments.

    Ohm’s law: the potential difference across the ends of a given metallic conductor is directly proportional to the current through it, provided its temperature and physical state remain unchanged; that is, $V\propto I$, so $V=IR$, where the constant $R$ is the resistance. Verification: connect a nichrome wire in series with a battery, a plug key, a rheostat and an ammeter, and a voltmeter in parallel across the wire. Vary the current with the rheostat and record four or five pairs of readings $(I,V)$. The ratio $V/I$ comes out the same for every pair, and a graph of $V$ against $I$ is a straight line through the origin whose slope is $R$ — which verifies the law. Comment: it does **not** hold under all conditions. It applies only to ohmic conductors whose temperature and physical state are constant. It fails when the conductor heats up (the filament of a lamp, whose V–I graph curves upward) and for non-ohmic devices such as semiconductor diodes, transistors, LEDs, thermistors, electrolytes and gases in a discharge tube, whose V–I graphs are not straight lines.

    Step-by-step solution →
12. Q72 Exemplar · Long Answer Questions
    What is electrical ​resistivity of a ​material? What is its unit? ​Describe ​an experiment to ​study the factors on ​which the resistance of conducting ​wire depends.
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    Resistivity is the resistance of unit length and unit area, in $\Omega$ m; varying wire length, area and material shows $R=\frac{\rho l}{A}$.

    The electrical resistivity of a material is the resistance of a conductor of that material of unit length and unit area of cross-section: from $R=\frac{\rho l}{A}$ we get $\rho =\frac{RA}{l}$. Its SI unit is the ohm metre ($\Omega$ m). It depends only on the material and its temperature, not on the size or shape of the conductor. Experiment: fix a test wire between terminals X and Y, connect X and Y in series with a cell, a plug key and an ammeter, and connect a voltmeter across X and Y; then $R=V/I$. Repeat with different test wires, changing one factor at a time. (i) With the same material and thickness, a wire of double the length gives double the resistance, so $R\propto l$. (ii) With the same material and length, a wire of double the area of cross-section gives half the resistance, so $R\propto 1/A$. (iii) With the same length and thickness but a different metal, the resistance is different, so it depends on the nature of the material. Combining these, $R=\frac{\rho l}{A}$. Heating the wire also raises its resistance, so resistance depends on temperature as well.

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13. Q73 Exemplar · Long Answer Questions
    How will you infer ​with the help of an experiment that ​the same current flows ​through every ​part of the circuit containing three resistances in series ​connected to ​a ​battery?
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    An ammeter gives the same reading at every point of the series circuit, so the same current flows through every part.

    Connect three unequal resistors — say $R_1=1\Omega$, $R_2=2\Omega$ and $R_3=3\Omega$ — in series with a battery and a plug key, leaving four gaps in the loop: P before $R_1$, Q between $R_1$ and $R_2$, S between $R_2$ and $R_3$, and T after $R_3$. Insert the ammeter at P, close the key and note the reading; then open the key, move the same ammeter to Q, then to S, then to T, changing nothing else, and note the reading each time. All four readings are found to be equal — $1$ A each in this example. The inference is that the same current flows through every part of a series circuit. The reason is that there is only one path and no junction at which charge can be added or removed, and charge is conserved, so the same charge passes every cross-section each second. A voltmeter across each resistor, by contrast, reads $1$ V, $2$ V and $3$ V, adding to the $6$ V of the battery: in a series circuit it is the potential difference that is divided, not the current.

    Step-by-step solution →
14. Q74 Exemplar · Long Answer Questions
    How will ​you conclude that the same potential difference (voltage) exists ​across three ​resistors connected in ​a ​parallel arrangement to ​a ​battery?
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    Equal voltmeter readings (6 V) across each resistor and across X and Y show the same potential difference across parallel resistors.

    Join three unequal resistors — say $R_1=2\Omega$, $R_2=3\Omega$ and $R_3=6\Omega$ — between the same two points X and Y, and connect X and Y through a plug key and an ammeter to a battery of, say, $6$ V. Close the key and connect one voltmeter across $R_1$; record the reading. Then move the same voltmeter across $R_2$, then across $R_3$, and finally across X and Y, changing nothing else. All four readings are found to be equal, each $6$ V, and equal to the potential difference of the battery. Hence the same potential difference exists across every resistor of a parallel arrangement. The reason is that each resistor has one end at X and the other at Y, and the potential difference between two given points has one definite value. Measuring the branch currents confirms what does differ: $3$ A, $2$ A and $1$ A, adding to the $6$ A read by the ammeter in the main line — in parallel it is the current that divides, not the voltage.

    Step-by-step solution →
15. Q75 Exemplar · Long Answer Questions
    What is Joule’s heating effect? How can ​it ​be ​demonstrated ​experimentally? List ​its four applications in ​daily ​life.
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    Electrical energy turns into heat in a resistance ($H=I^{2}Rt$), shown by a coil heating water, and used in irons, bulbs, fuses and geysers.

    Joule’s heating effect is the conversion of electrical energy into heat when a current is passed through a conductor having resistance. In a time $t$ a current $I$ carries a charge $Q=It$ through a potential difference $V$, so the work done is $W=VIt$, and in a resistor all of it appears as heat: $H=VIt=I^{2}Rt=\frac{V^{2}t}{R}$. Hence Joule’s law: $H\propto I^2$, $H\propto R$ and $H\propto t$. Demonstration: immerse a nichrome coil in a known mass of water in a calorimeter with a thermometer and a stirrer, and connect it in series with a battery, a key, a rheostat and an ammeter, with a voltmeter across the coil. Pass a steady current for a measured time: the temperature rises, and the heat gained, $mc\Delta \theta$, equals $VIt$. Doubling the current for the same time makes the rise about four times as large ($H\propto I^2$); using a coil of larger resistance at the same current increases the rise in proportion to $R$; doubling the time doubles the rise. Simply put: a thin nichrome wire glows while the thick copper leads carrying the same current stay cool. Four applications: (1) electric heating appliances — electric iron, room heater, toaster and electric kettle, with a nichrome heating element; (2) the filament of an incandescent electric bulb, a tungsten filament heated to about $2500{}^{\circ }$C so that it glows; (3) the electric fuse, which melts and breaks the circuit when the current is excessive; (4) the electric geyser or immersion heater for heating water.

    Step-by-step solution →
16. Q76 Exemplar · Long Answer Questions
    Find out the following in the electric circuit given in Figure 12.9

    1. (a)Effective resistance ​of two 8 Ω resistors in the combination
    2. (b)​Current flowing through ​4 Ω resistor
    3. (c)Potential difference across ​4 Ω resistance
    4. (d)Power ​dissipated in 4 Ω ​resistor
    5. (e)Difference in ammeter readings, if ​any.

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    (a) $4\Omega$; (b) $1$ A; (c) $4$ V; (d) $4$ W; (e) the two ammeters read the same, so the difference is zero.

    (a) The two $8\Omega$ resistors are in parallel between C and D, so $\frac{1}{R_p}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$, giving $R_p=4\Omega$. (b) The total resistance is $4+4=8\Omega$, so the current in the main line, and therefore through the $4\Omega$ resistor, is $I=8/8=1$ A. (c) The potential difference across the $4\Omega$ resistance is $V=IR=1\times 4=4$ V. (d) The power dissipated in it is $P=I^{2}R=(1{)}^2\times 4=4$ W. (e) There is **no** difference: $A_1$ and $A_2$ both lie in the main line outside the parallel section, so each reads the full current of $1$ A and $A_1-A_2=0$.

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17. Q77 CBSE 2026
    The correct ​way to connect an ​ammeter ​and a voltmeter ​in ​an electric ​circuit ​is :

    1. (a)Ammeter in parallel and voltmeter in series
    2. (b)Ammeter and voltmeter both in parallel
    3. (c)Ammeter in series and voltmeter in parallel
    4. (d)Ammeter and voltmeter both in series
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    (c) Ammeter in series and voltmeter in parallel

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18. Q78 CBSE 2026
    Work done in displacing ​an electron between ​two ​points having potential difference of 1 ​V, ​is ​:

    1. (a)$-1.6\times 10^{-19}\text{C}$
    2. (b)$-1.6\times 10^{-19}\text{J}$
    3. (c)$+1.6\times 10^{-19}\text{J}$
    4. (d)$-1.6\times 10^{-19}\text{V}$
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    (b) $-1.6\times 10^{-19}\text{J}$

    Step-by-step solution →
19. Q79 CBSE 2026
    In an electric circuit carrying current $I$, ​a $2\Omega$ resistor is ​replaced ​by ​an $8\Omega$ ​resistor without changing the connected ​(ideal) ​battery. ​Then :

    1. (a)current becomes $\frac{I}{4}$ and potential difference $\frac{V}{4}$
    2. (b)current becomes $\frac{I}{4}$ and potential difference remains $V$
    3. (c)current becomes $4I$ and potential difference remains $V$
    4. (d)current becomes $4I$ and potential difference $\frac{V}{4}$
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    (b) current becomes $\frac{I}{4}$ and potential difference remains $V$

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20. Q80 CBSE 2026
    For Question ​number 32, two statements are given — one labelled as Assertion ​(A) ​and the other labelled ​as Reason (R). Select the correct answer to this question from ​the ​codes (a), (b), (c) and (d) ​as given below.

    Assertion (A): Nichrome is an alloy which is commonly used in electrical heating devices like electric irons.

    Reason (R): The resistivity of nichrome is high and its resistance decreases with an increase in temperature.

    1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
    2. (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
    3. (c)Assertion (A) is true, but Reason (R) is false.
    4. (d)Assertion (A) is false, but Reason (R) is true.
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    (c) Assertion (A) is true, but Reason (R) is false.

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