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 10 of 19

1. Q181 CBSE 2024
   Study the I-V graph for three resistors ​of ​resistances $R_1$, $R_2$ ​and $R_3$ ​and ​select ​the correct ​statement from the following :

   1. (a)$R_1=R_2=R_3$
   2. (b)$R_1>R_2>R_3$
   3. (c)$R_3>R_2>R_1$
   4. (d)$R_2>R_3>R_1$
   Show answer

   (c) $R_3>R_2>R_1$

   Step-by-step solution →
2. Q182 CBSE 2024
   The minimum resistance ​which ​can be made using five resistors each of ​resistance $10\Omega$ ​is :

   1. (a)$\frac{1}{50}\Omega$
   2. (b)$\frac{1}{5}\Omega$
   3. (c)$2\Omega$
   4. (d)$1\Omega$
   Show answer

   (c) $2\Omega$

   Step-by-step solution →
3. Q183 CBSE 2024
   The resistance of a resistor ​is ​increased four times its initial ​value. If other parameters (current and time) remain unchanged, the ​percentage change in the amount of heat produced ​in ​the resistor would ​be :

   1. (a)100%
   2. (b)200%
   3. (c)300%
   4. (d)400%
   Show answer

   (c) 300%

   Step-by-step solution →
4. Q184 CBSE 2024
   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): Electrons move from lower potential to higher potential in a conductor.

   Reason (R): A dry cell maintains electric potential difference across the ends of a conductor.

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

   (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

   Step-by-step solution →
5. Q185 CBSE 2024
   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): The commercial unit of electric energy is kilowatt hour (kWh).

   Reason (R): $1\text{kWh}=10^6$ joule (J)

   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.
   Show answer

   (c) Assertion (A) is true, but Reason (R) is false.

   Step-by-step solution →
6. Q186 CBSE 2024
   Use Ohm’s law to determine the potential difference ​across ​the $3\Omega$ ​resistor in the circuit shown ​in ​the following diagram when ​key is closed ​:

   Show answer

   The potential difference across the $3\Omega$ resistor is 1 V.

   The resistors are in series, so $R_s=1+2+3=6\Omega$ and the current is $I=\frac{V}{R_s}=\frac{2}{6}=\frac{1}{3}$ A. By Ohm's law, the potential difference across the $3\Omega$ resistor is $V_3=I{R}_3=\frac{1}{3}\times 3=1$ V.

   Step-by-step solution →
7. Q187 CBSE 2024
   Show how ​you would ​connect three ​resistors ​each of resistance $6\Omega$, so ​that the combination ​has a resistance ​of $9\Omega$. Also justify ​your answer.
   Show answer

   Join two $6\Omega$ resistors in parallel and connect the third in series with them to get $9\Omega$.

   Connect two of the $6\Omega$ resistors in parallel, and connect this pair in series with the third $6\Omega$ resistor. Justification: for the parallel pair $\frac{1}{R_p}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$, so $R_p=3\Omega$; in series with the third resistor the total is $R=3+6=9\Omega$.

   Step-by-step solution →
8. Q188 CBSE 2024
   In ​the ​given ​circuit ​calculate ​the power consumed in watts in the resistor of $2\Omega$ ​:

   Show answer

   The power consumed in the $2\Omega$ resistor is 8 W.

   Total resistance $R=1+2=3\Omega$; current $I=\frac{V}{R}=\frac{6}{3}=2$ A; power consumed in the $2\Omega$ resistor $P=I^{2}R=2^2\times 2=8$ W.

   Step-by-step solution →
9. Q189 CBSE 2024
   Two wires A and B of same material, having same lengths ​and diameters $0.2$ mm and $0.3$ mm respectively, ​are ​connected one by one ​in a circuit. ​Which one of these two wires will offer more resistance to ​the flow of current in the circuit ? ​Justify your answer.
   Show answer

   Wire A, the thinner wire, offers more resistance: $R_A$ is 2.25 times $R_B$.

   Wire A (diameter $0.2$ mm) offers more resistance. For wires of the same material and length, $R=\rho \frac{l}{A}$ with $A=\frac{\pi d^2}{4}$, so $R\propto \frac{1}{d^2}$. Hence $\frac{R_A}{R_B}={\left(\frac{d_B}{d_A}\right)}^2={\left(\frac{0.3}{0.2}\right)}^2=\frac{9}{4}=2.25$: the thinner wire A has the smaller area of cross-section and $2.25$ times the resistance of wire B.

   Step-by-step solution →
10. Q190 CBSE 2024
    State Joule’s law of heating. How ​is this ​effect useful ​in ​electric circuits where fuse is used as ​a ​safety device ?
    Show answer

    Heat produced is $H=I^{2}Rt$; in a fuse, excess current heats the wire until it melts and breaks the circuit.

    Joule's law of heating: when a current $I$ flows through a resistor of resistance $R$ for a time $t$, the heat produced is $H=I^{2}Rt$; that is, the heat produced is directly proportional to the square of the current for a given resistance, to the resistance for a given current, and to the time for which the current flows. Use in a fuse: a fuse is a wire of a metal or alloy of suitable melting point connected in series with the appliance, so it carries the same current. If the current becomes larger than the specified value (overloading or short circuit), the heat produced in the fuse wire, which grows as $I^2$, raises its temperature until it melts. This breaks the circuit and stops the current, protecting the appliances and the wiring from damage.

    Step-by-step solution →
11. Q191 CBSE 2024
    Calculate ​the resistance ​of ​a copper ​wire of ​length $1000$ m and area of ​cross-section $2{\text{mm}}^2$. Resistivity of copper is $1.6\times 10^{-8}\Omega \text{m}$.
    Show answer

    The resistance of the copper wire is $8\Omega$.

    The resistance of the copper wire is $R=8\Omega$.

    Step-by-step solution →
12. Q192 CBSE 2024
    An electric ​source can supply a charge of 500 coulomb. If ​the current ​drawn by ​a device ​is 25 ​mA, find the time in ​which the electric source will ​be discharged completely.
    Show answer

    The electric source is discharged completely in $2\times 10^4$ s (about 5.56 hours).

    $t=\frac{Q}{I}=\frac{500\text{C}}{25\times 10^{-3}\text{A}}=2\times 10^4$ s $=20000$ s, which is about $5.56$ h ($5$ h $33$ min $20$ s).

    Step-by-step solution →
13. Q193 CBSE 2024
    An electric source ​can supply a charge of 750 coulomb. ​If ​the current drawn ​by a ​device is ​15 mA, find ​the time in which the ​electric source will be discharged completely.
    Show answer

    The electric source is discharged completely in $5\times 10^4$ s (about 13.9 hours).

    Using $I=\frac{Q}{t}$: $t=\frac{Q}{I}=\frac{750\text{C}}{15\times 10^{-3}\text{A}}=5\times 10^4$ s, that is $50000$ s (about $13.9$ hours).

    Step-by-step solution →
14. Q194 CBSE 2024
    The filament ​of ​an electric lamp draws ​a current of $0.5$ ​A, ​which lights for $2$ hours. Calculate the ​charge that flows through the ​circuit.
    Show answer

    A charge of 3600 C flows through the circuit.

    $Q=It=0.5\text{A}\times 7200\text{s}=3600$ C. A charge of $3600$ C, that is $3.6\times 10^3$ C, flows through the circuit.

    Step-by-step solution →
15. Q195 CBSE 2024
    In the given circuit ​when key ​is closed, determine the ​following :

    1. (i)Total resistance offered by the ​three resistors
    2. (ii)​Reading of ​the ​ammeter
    3. (iii)Reading of the voltmeter

    Show answer

    $R=12\Omega$, the ammeter reads $0.5$ A and the voltmeter across the $4\Omega$ resistor reads $2$ V.

    (i) Total resistance $R=2+4+6=12\Omega$, as the three resistors are in series. (ii) Reading of the ammeter $I=\frac{V}{R}=\frac{6}{12}=0.5$ A. (iii) Reading of the voltmeter, which is across the $4\Omega$ resistor, $=IR=0.5\times 4=2$ V.

    Step-by-step solution →
16. Q196 CBSE 2024
    Draw a schematic diagram of a circuit consisting of ​a battery ​of four dry cells of $1.5$ V ​each, ​a $2\Omega$ resistor, a $6\Omega$ resistor, $16\Omega$ resistor and a plug key all connected in ​series. Put an ammeter to measure the current in the circuit and a voltmeter across ​the $16\Omega$ resistor to measure potential difference across its two ends. Use Ohm’s law to determine

    1. (a)​ammeter reading, and
    2. (b)voltmeter reading when key ​is closed.
    Show answer

    The ammeter reads 0.25 A and the voltmeter across the $16\Omega$ resistor reads 4 V.

    (a) Ammeter reading: $I=\frac{6\text{V}}{24\Omega }=0.25$ A. (b) Voltmeter reading: $V=0.25\text{A}\times 16\Omega =4$ V. In the circuit, the four $1.5$ V cells ($6$ V), the plug key, the $2\Omega$, $6\Omega$ and $16\Omega$ resistors and the ammeter form one series loop, and the voltmeter is connected across the $16\Omega$ resistor.

    Step-by-step solution →
17. Q197 CBSE 2024
    A $2000$ W heater has a resistance of about $25\Omega$, whereas a $100$ W bulb ​has a resistance of $500\Omega$. When $220$ ​V ​is applied on these, then which of the two

    1. (a)can ​carry large currents ?
    2. (b)may be used ​with an electrical circuit having $1.0$ A rating ?
    3. (c)will ​be fitted with a $15$ A electric board and not ​with a $5$ ​A electric board ? Justify your answer in each case.
    Show answer

    Heater carries more current (8.8 A against 0.44 A); bulb suits the 1.0 A circuit; heater needs the 15 A board.

    (a) The heater: $I=\frac{220\text{V}}{25\Omega }=8.8$ A, against $\frac{220\text{V}}{500\Omega }=0.44$ A for the bulb; its much smaller resistance lets a much larger current flow. (b) The bulb: it draws $0.44$ A, which is within the $1.0$ A rating, while the heater’s $8.8$ A would exceed it. (c) The heater: its $8.8$ A is more than $5$ A, so a $5$ A board would be overloaded, but it is within $15$ A.

    Step-by-step solution →
18. Q198 CBSE 2024
    1. (a)State Ohm’s law. Write formula for the ​equivalent resistance $R_p$ ​of ​the parallel combination of ​three resistors of ​values $R_1$, $R_2$ ​and $R_3$.
    2. (b)Find the resistance ​of the following network ​of resistors :

    Show answer

    Ohm's law is $V=IR$ at constant temperature; the reciprocals of parallel resistances add; the network resistance is $R_{XY}=\frac{3R}{2}$.

    (a) Ohm's law: the potential difference $V$ across the ends of a given metallic wire in an electric circuit is directly proportional to the current $I$ flowing through it, provided its temperature remains the same, so $V\propto I$, that is $V=IR$. For three resistors in parallel, $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$. (b) The two resistors of the block are in parallel: $\frac{1}{R^{\prime }}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}$, so $R^{\prime }=\frac{R}{2}$. This is in series with the first $R$, so the resistance of the network is $R_{XY}=R+\frac{R}{2}=\frac{3R}{2}$ (that is, $1.5R$).

    Step-by-step solution →
19. Q199 CBSE 2024
    (i) State whether the currents and ​potential difference in all the bulbs will be ​same or different when in a circuit three bulbs of :

    1. (a)​same wattage are connected in series.
    2. (b)same wattage are connected in parallel.
    3. (c)different wattage ​are connected in ​series.
    4. (d)different wattage are ​connected in parallel. (ii) Two identical resistors of $24\Omega$ each are connected to a battery of $6$ V. Calculate the ratio of the power consumed by the resulting combinations with (a) minimum resistance and (b) maximum resistance.
    Show answer

    Series bulbs share the current and parallel bulbs share the potential difference; the power ratio (minimum to maximum resistance) is $4:1$.

    (i) Taking the bulbs as rated for the same voltage, so that equal wattage means equal resistance: (a) same wattage in series: current same, potential difference same; (b) same wattage in parallel: potential difference same, current same; (c) different wattage in series: current same, potential difference different; (d) different wattage in parallel: potential difference same, current different. (ii) Minimum resistance (parallel) $=12\Omega$, so $P=V^2/R=36/12=3$ W; maximum resistance (series) $=48\Omega$, so $P=36/48=0.75$ W. Ratio of the powers (a) : (b) $=3:0.75=4:1$.

    Step-by-step solution →
20. Q200 CBSE 2024
    1. (i)Write the relation between ​resistivity ($\rho$) and ​resistance ($R$) for a ​uniform metallic conductor of length ($l$) and area of cross-section ($A$). Use this relation to obtain the SI unit of resistivity.
    2. (ii)The resistance ​of a metallic ​wire ​of $100$ m is $20\Omega$. If the radius of the wire is $5$ ​mm, find the resistivity of the metal of the wire.
    Show answer

    $R=\rho \frac{l}{A}$ gives the SI unit of resistivity as $\Omega \text{m}$; the wire's resistivity is about $1.57\times 10^{-5}\Omega \text{m}$.

    (i) $R=\rho \frac{l}{A}$, so $\rho =\frac{RA}{l}$; the SI unit of resistivity is $\frac{\Omega \times {\text{m}}^2}{\text{m}}=\Omega \text{m}$ (ohm metre). (ii) $A=\pi r^2=3.14\times (5\times 10^{-3}\text{m}{)}^2=7.85\times 10^{-5}{\text{m}}^2$, so $\rho =\frac{20\times 7.85\times 10^{-5}}{100}\approx 1.57\times 10^{-5}\Omega \text{m}$.

    Step-by-step solution →