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 1 of 11

1. Q1 NCERT · Exercises
   A piece of ​wire ​of resistance $R$ is cut into five equal parts. These ​parts are then connected in parallel. If the equivalent ​resistance ​of ​this ​combination is $R^{\prime }$, then the ratio $R/R^{\prime }$ is ​–

   1. (a)$\frac{1}{25}$
   2. (b)$\frac{1}{5}$
   3. (c)$5$
   4. (d)$25$
   Show answer

   (d) $25$

   Step-by-step solution →
2. Q2 NCERT · Intext Questions (Page 172)
   What does ​an ​electric ​circuit mean?
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   An electric circuit is a continuous, closed path along which an electric current flows.

   An electric circuit means a continuous and closed path along which an electric current flows. It is made up of a source of potential difference (a cell or a battery), connecting wires, one or more electrical components such as a bulb or a resistor, and usually a switch or plug key; an ammeter may also be joined in series. When the key is closed the path is complete and a current flows round the loop, conventionally from the positive terminal of the cell through the components to its negative terminal. If the path is broken at any point — for example by opening the key — the current stops flowing everywhere in the circuit.

   Step-by-step solution →
3. Q3 NCERT · Exercises
   Which ​of ​the following terms does not represent electrical ​power ​in ​a ​circuit?

   1. (a)$I^{2}R$
   2. (b)$I{R}^2$
   3. (c)$VI$
   4. (d)$\frac{V^2}{R}$
   Show answer

   (b) $I{R}^2$

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4. Q4 NCERT · Intext Questions (Page 172)
   Define the unit of current.
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   One ampere is the current when one coulomb of charge flows through a cross-section in one second.

   The SI unit of electric current is the ampere (A), named after Andre-Marie Ampere. One ampere is the current constituted by the flow of one coulomb of charge through any cross-section of a conductor in one second: $1\text{A}=1\text{C}/1\text{s}=1{\text{C s}}^{-1}$. Smaller currents are measured in milliampere ($1\text{mA}=10^{-3}$ A) and microampere ($1\mu \text{A}=10^{-6}$ A).

   Step-by-step solution →
5. Q5 NCERT · Exercises
   An electric bulb ​is ​rated $220$ V and $100$ W. When it ​is operated ​on $110$ ​V, ​the ​power consumed will be –

   1. (a)$100$ W
   2. (b)$75$ W
   3. (c)$50$ W
   4. (d)$25$ W
   Show answer

   (d) $25$ W

   Step-by-step solution →
6. Q6 NCERT · Intext Questions (Page 172)
   Calculate ​the ​number of electrons ​constituting ​one ​coulomb ​of charge.
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   $n=6.25\times 10^{18}$: one coulomb of charge is carried by about $6.25\times 10^{18}$ electrons.

   $n=Q/e=1\text{C}÷(1.6\times 10^{-19}\text{C})=6.25\times 10^{18}$ electrons (to the two significant figures of $e$, $6.3\times 10^{18}$; the chapter quotes it as nearly $6\times 10^{18}$ electrons). The number is a pure count and carries no unit.

   Step-by-step solution →
7. Q7 NCERT · Exercises
   Two conducting wires of the ​same ​material and of ​equal lengths and equal diameters are first connected in series and then parallel in a circuit ​across the same potential difference. The ratio of heat produced in series and parallel combinations ​would be ​–

   1. (a)$1:2$
   2. (b)$2:1$
   3. (c)$1:4$
   4. (d)$4:1$
   Show answer

   (c) $1:4$

   Step-by-step solution →
8. Q8 NCERT · Exercises
   How ​is a voltmeter connected ​in ​the circuit ​to measure ​the ​potential difference ​between two ​points?
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   In parallel across the two points, positive terminal to the higher potential; its very high resistance keeps the measured potential difference almost undisturbed.

   A voltmeter is connected in **parallel** — that is, across the two points, so that one terminal is joined to each point and the meter lies in parallel with whatever part of the circuit is between them. Its positive terminal goes to the point at the higher potential. A voltmeter is deliberately built with a very high resistance, so it draws almost no current and therefore hardly disturbs the potential difference it is measuring.

   Step-by-step solution →
9. Q9 NCERT · Exercises
   A copper wire has ​diameter $0.5$ mm and resistivity ​of $1.6\times 10^{-8}\Omega$ m. What will ​be ​the ​length of this wire to make its resistance $10\Omega$? ​How ​much does the resistance change if the diameter ​is doubled?
   Show answer

   $L=122.7$ m; doubling the diameter makes the area four times, so the resistance falls to one-fourth ($2.5\Omega$).

   The wire must be about $122.7$ m long. If the diameter is doubled the cross-section area becomes four times as large, so the resistance becomes $\frac{1}{4}$ times its earlier value — for this same $122.7$ m of copper it would fall from $10\Omega$ to $2.5\Omega$.

   Step-by-step solution →
10. Q10 NCERT · Exercises
    The values of ​current $I$ flowing in a ​given resistor for the corresponding values ​of ​potential difference $V$ across ​the resistor are ​given below – Plot a graph between $V$ and $I$ and calculate the resistance of that ​resistor.

    Current through the resistor and the corresponding potential difference across it

    I (amperes)	V (volts)
    0.5	1.6
    1.0	3.4
    2.0	6.7
    3.0	10.2
    4.0	13.2

    Show answer

    The $V$–$I$ graph is a straight line through the origin, and its slope gives the resistance $R=3.33\Omega$.

    The graph of $V$ against $I$ is a straight line through the origin, which shows that the resistor obeys Ohm's law. Its slope is the resistance: the five readings give $V/I=3.2,3.4,3.35,3.4$ and $3.3\Omega$, whose mean is $16.65/5=3.33\Omega$, and the slope read off the drawn line between $I=1.5$ A and $I=3.5$ A is $6.66/2.0=3.33\Omega$. So $R=3.33\Omega$.

    Step-by-step solution →
11. Q11 NCERT · Exercises
    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.
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    The resistance is $4800\Omega$ ($4.8\text{k}\Omega$).

    The resistance is $4800\Omega$, that is $4.8\text{k}\Omega$. Converting the current to amperes gives $I=2.5\text{mA}=0.0025$ A, and Ohm's law then gives $R=V/I=12/0.0025=4800\Omega$.

    Step-by-step solution →
12. Q12 NCERT · Exercises
    A ​battery of $9$ ​V ​is connected in series with ​resistors of $0.2\Omega$, $0.3\Omega$, $0.4\Omega$, $0.5\Omega$ ​and $12\Omega$, ​respectively. ​How much current would flow through the $12\Omega$ resistor?
    Show answer

    $I=0.67$ A: the same current flows through all five series resistors, including the $12\Omega$ one.

    A current of about $0.67$ A flows through the $12\Omega$ resistor. The five resistors are in series, so their total resistance is $0.2+0.3+0.4+0.5+12=13.4\Omega$ and the circuit current is $I=9/13.4=0.67$ A. Being a series circuit, that same current flows through every one of the five resistors.

    Step-by-step solution →
13. Q13 NCERT · Exercises
    How ​many $176\Omega$ ​resistors (in parallel) ​are required to ​carry $5\text{A}$ on ​a $220\text{V}$ line?
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    Four $176\Omega$ resistors in parallel are required: they give $44\Omega$, so $220$ V drives exactly $5$ A.

    $4$ resistors. Four $176\Omega$ resistors in parallel give $R_p=176/4=44\Omega$, and $220\text{V}$ across $44\Omega$ drives exactly $5\text{A}$.

    Step-by-step solution →
14. Q14 NCERT · Exercises
    Show ​how ​you ​would connect three resistors, each ​of resistance $6\Omega$, ​so that the combination has a resistance ​of ​

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

    (i) Two $6\Omega$ resistors in parallel, in series with the third, give $9\Omega$; (ii) two in series, in parallel with the third, give $4\Omega$.

    (i) $9\Omega$: connect two of the resistors in parallel and the third in series with that pair — the pair is worth $\frac{6\times 6}{6+6}=3\Omega$, and $3+6=9\Omega$. (ii) $4\Omega$: connect two of the resistors in series and the third in parallel across that pair — the pair is worth $6+6=12\Omega$, and $\frac{12\times 6}{12+6}=\frac{72}{18}=4\Omega$.

    Step-by-step solution →
15. Q15 NCERT · Exercises
    Several ​electric bulbs ​designed to be used on a $220\text{V}$ electric supply ​line, are rated $10\text{W}$. How many ​lamps can be connected in parallel with each ​other ​across the ​two wires of $220\text{V}$ ​line if the maximum allowable current is $5\text{A}$?
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    110 lamps can be connected in parallel: each draws $\frac{1}{22}$ A, and $110$ such branch currents add to the allowed $5$ A.

    $110$ lamps. Each $10\text{W}$ lamp draws $\frac{10}{220}=\frac{1}{22}\text{A}$ at $220\text{V}$, and $110$ such branch currents add to exactly $5\text{A}$.

    Step-by-step solution →
16. Q16 NCERT · Exercises
    A hot plate of an electric oven ​connected to a $220\text{V}$ line has ​two resistance coils A and ​B, each of $24\Omega$ resistance, which ​may be used separately, in series, or in parallel. ​What are ​the currents in the three ​cases?
    Show answer

    The currents are about 9.2 A for one coil, 4.6 A for two in series and 18.3 A for two in parallel.

    Coil used separately: $I\approx 9.2\text{A}$ (exactly $\frac{55}{6}\text{A}$). Both coils in series: $I\approx 4.6\text{A}$ (exactly $\frac{55}{12}\text{A}$). Both coils in parallel: $I\approx 18.3\text{A}$ (exactly $\frac{55}{3}\text{A}$). All three are rounded to one decimal place at the final step only.

    Step-by-step solution →
17. Q17 NCERT · Exercises
    Compare the power used in the $2\Omega$ resistor ​in each ​of the ​following circuits:

    1. (i)a $6\text{V}$ battery in series ​with $1\Omega$ ​and $2\Omega$ ​resistors, and
    2. (ii)a $4\text{V}$ ​battery in parallel with $12\Omega$ ​and $2\Omega$ resistors.
    Show answer

    The $2\Omega$ resistor uses $8$ W in both circuits, so the powers are in the ratio $1:1$.

    (i) $P=8\text{W}$; (ii) $P=8\text{W}$. The $2\Omega$ resistor uses the same power in both circuits, so the ratio of the two is $1:1$.

    Step-by-step solution →
18. Q18 NCERT · Exercises
    Two ​lamps, one rated $100\text{W}$ at $220\text{V}$, and ​the ​other $60\text{W}$ at $220\text{V}$, are connected ​in ​parallel to electric mains supply. What current is drawn from the line if the ​supply voltage ​is $220\text{V}$?
    Show answer

    $I=0.73$ A: the two lamps together draw 160 W from the 220 V mains.

    $I\approx 0.73\text{A}$. The two lamps together take $100+60=160\text{W}$ from the $220\text{V}$ mains, so the line current is $\frac{160}{220}=\frac{8}{11}=0.7272\dots \text{A}$, which is $0.73\text{A}$ rounded to two decimal places.

    Step-by-step solution →
19. Q19 NCERT · Exercises
    Which uses ​more ​energy, a $250\text{W}$ ​TV set in $1$ ​hr, ​or ​a $1200\text{W}$ toaster ​in $10$ ​minutes?
    Show answer

    The 250 W TV set in 1 hour uses more energy: 250 Wh against 200 Wh for the toaster.

    The $250\text{W}$ TV set in $1$ hour uses more energy. It consumes $250\times 1=250\text{Wh}$ ($0.25\text{kWh}$, or $9\times 10^5\text{J}$), while the toaster consumes $1200\times \frac{1}{6}=200\text{Wh}$ ($0.20\text{kWh}$, or $7.2\times 10^5\text{J}$) — so the television uses $50\text{Wh}$ more, one and a quarter times as much.

    Step-by-step solution →
20. Q20 NCERT · Exercises
    An electric ​heater ​of ​resistance $44\Omega$ draws $5\text{A}$ from the ​service mains ​for $2$ ​hours. Calculate the rate at which ​heat ​is developed in the heater.
    Show answer

    $P=1100$ W: the heater develops heat at the rate of 1100 W (1.1 kW).

    The rate at which heat is developed is $P=I^{2}R=(5{)}^2\times 44=1100\text{W}$, that is $1.1\text{kW}$ (equivalently $1100\text{J}$ of heat every second). The two-hour running time does not enter the rate; it would only be needed for the total heat, $7.92\times 10^6\text{J}$.

    Step-by-step solution →