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 11 of 19

1. Q201 CBSE 2024
   Draw a schematic diagram of ​a circuit consisting of a battery of six $2$ V cells, a $6\Omega$ resistor, a $12\Omega$ resistor and a $18\Omega$ resistor and a plug ​key all ​connected ​in series. ​Calculate the ​following (when key is closed) :

   1. (i)Electric current flowing ​in the circuit.
   2. (ii)​Potential difference across $18\Omega$ resistor.
   3. (iii)Electric power consumed in $18\Omega$ resistor.
   Show answer

   Six 2 V cells give $12$ V across $36\Omega$, so $I=\frac{1}{3}$ A; the $18\Omega$ resistor has $6$ V across it and uses $2$ W.

   Circuit: six $2$ V cells in series (a $12$ V battery), a plug key and the $6\Omega$, $12\Omega$ and $18\Omega$ resistors, all in one loop, as in the figure. Total resistance $R=6+12+18=36\Omega$. (i) $I=V/R=12/36=\frac{1}{3}$ A $\approx 0.33$ A. (ii) $V_{18}=IR=\frac{1}{3}\times 18=6$ V. (iii) $P_{18}=VI=6\times \frac{1}{3}=2$ W.

   Step-by-step solution →
2. Q202 CBSE 2024
   1. (i)Obtain an ​expression for the amount of heat produced in a conductor of resistance $R$ when a steady current $I$ flows through it for a time $t$.
   2. (ii)An electrical heater of ​resistance $10\Omega$ draws a steady current of $5$ A from ​the service mains ​for $2$ hours. Find the power of this heater and the amount of ​energy consumed by ​the heater in kilowatt ​hour.
   Show answer

   Heat produced is $H=I^{2}Rt$; the heater has power $250$ W and consumes $0.5$ kWh in 2 hours.

   (i) In time $t$ a charge $Q=It$ moves through the potential difference $V$, so the energy supplied, which appears as heat, is $H=VQ=VIt$; with $V=IR$ this gives $H=I^{2}Rt$ (Joule's law of heating). (ii) Power $P=I^{2}R=5^2\times 10=250$ W; energy consumed $=0.25\text{kW}\times 2\text{h}=0.5$ kWh.

   Step-by-step solution →
3. Q203 CBSE 2024
   1. (a)List two factors on which the resistance of a ​conductor in the shape ​of a ​straight ​cylinder depends. Write the relation between resistance ($R$) and ​the resistivity ($\rho$) of ​the material of the conductor. Use this relation to obtain the SI unit of resistivity.
   2. (b)Resistance of a metal wire of length ​1 m is $40\Omega$. If the area of cross-section of the wire is $6.5\times 10^{-8}{\text{m}}^2$, find the resistivity of the material of the wire.
   Show answer

   Resistance depends on length and area of cross-section; $R=\rho \frac{l}{A}$ gives the unit $\Omega \text{m}$; $\rho =2.6\times 10^{-6}\Omega \text{m}$.

   (a) The resistance of a cylindrical conductor depends on (1) its length $l$, $R\propto l$, and (2) its area of cross-section $A$, $R\propto \frac{1}{A}$ (for a given material at a given temperature). Relation: $R=\rho \frac{l}{A}$, so $\rho =\frac{RA}{l}$, and the SI unit of resistivity is $\frac{\Omega \times {\text{m}}^2}{\text{m}}=\Omega \text{m}$ (ohm metre). (b) $\rho =\frac{RA}{l}=\frac{40\times 6.5\times 10^{-8}}{1}=2.6\times 10^{-6}\Omega \text{m}$.

   Step-by-step solution →
4. Q204 CBSE 2024
   (i) Define electric power. Express it ​in terms of potential difference ($V$) and resistance ($R$). (ii) An ​electric ​oven is designed to work on the mains voltage of $220$ V. This oven consumes $11$ units ​of electrical energy in $5$ hours. ​Calculate :

   1. (a)​power rating of the oven
   2. (b)current drawn ​by the oven
   3. (c)resistance of the oven when it ​is red hot
   Show answer

   Power is the rate of consuming electric energy, $P=\frac{V^2}{R}$; the oven has power 2200 W, draws 10 A and has resistance $22\Omega$.

   (i) Electric power is the rate at which electric energy is consumed (dissipated) in an electric circuit: $P=VI$. Since $I=V/R$ by Ohm's law, $P=\frac{V^2}{R}$ (also $P=I^{2}R$); its SI unit is the watt. (ii) (a) Power rating: $P=\frac{11\text{kW h}}{5\text{h}}=2.2$ kW $=2200$ W. (b) Current: $I=\frac{P}{V}=\frac{2200}{220}=10$ A. (c) Resistance when red hot: $R=\frac{V}{I}=\frac{220}{10}=22\Omega$.

   Step-by-step solution →
5. Q205 CBSE 2024
   1. (i)The potential ​difference across the ​two ends ​of a circuit component is decreased to one-third of its initial value, while its resistance remains constant. What change will be observed in the current flowing through it ? Name and state the law which helps us to answer this question.
   2. (ii)Draw a schematic ​diagram of ​a circuit consisting of a battery of four $1.5$ V cells, a $5\Omega$ resistor, a $10\Omega$ resistor and a $15\Omega$ resistor and a plug key, all connected in series. Now find (I) the electric current passing through the circuit, and (II) ​potential difference across the $10\Omega$ resistor when the ​plug key is closed.
   Show answer

   The current falls to one-third (Ohm's law); the circuit current is $0.2$ A and the $10\Omega$ resistor has $2$ V across it.

   (i) The current also decreases to one-third of its initial value, $I^{\prime }=I/3$. The law is Ohm's law: the potential difference across the ends of a conductor is directly proportional to the current through it, provided its temperature remains the same; $V=IR$. (ii) Circuit: the four $1.5$ V cells in series (battery of $6$ V), the $5\Omega$, $10\Omega$ and $15\Omega$ resistors and the plug key, all in one loop. (I) $I=\frac{6\text{V}}{30\Omega }=0.2$ A. (II) $V_{10}=0.2\text{A}\times 10\Omega =2$ V.

   Step-by-step solution →
6. Q206 CBSE 2024
   1. (i)Which type ​of circuits — series or parallel, should be used when you have to ​operate different electrical ​gadgets in ​your house ? List two ​reasons for your answer.
   2. (ii)Find the ​current flowing through ​the following electric circuit ​:

   Show answer

   Use parallel circuits at home, as each gadget gets full voltage and works independently; the current in the given circuit is $0.45$ A.

   (i) Parallel circuits. Reasons: each gadget gets the full supply voltage of $220$ V; and each gadget can be switched on or off on its own, so if one fails or is switched off the others still work. (Also, each gadget draws the current it needs, because a parallel circuit divides the current.) (ii) Lower branch $6+6=12\Omega$; in parallel with $6\Omega$ it gives $\frac{6\times 12}{6+12}=4\Omega$; total $R=3+4+3=10\Omega$; so the current is $I=\frac{4.5\text{V}}{10\Omega }=0.45$ A.

   Step-by-step solution →
7. Q207 CBSE 2024
   1. (i)Write the ​relation ​between the electrical resistivity ​‘$\rho$’ and the resistance ‘$R$’ of a cylindrical conductor of length ‘$l$’ ​and area of cross-section ‘$A$’. Use ​this relationship to determine the SI unit of ‘$\rho$’.
   2. (ii)Find the electrical resistivity of the material ​of a wire ​of length $120$ ​cm and diameter $0.4$ mm. The resistance of this wire is $3\Omega$.
   Show answer

   $R=\rho \frac{l}{A}$ gives the unit of resistivity as the ohm metre; the wire's resistivity is $3.14\times 10^{-7}\Omega \text{m}$.

   (i) $R=\rho \frac{l}{A}$, so $\rho =\frac{RA}{l}$. Substituting SI units, the unit of $\rho$ is $\frac{\Omega \times {\text{m}}^2}{\text{m}}=\Omega \text{m}$ (ohm metre). (ii) With $l=1.2$ m, $r=2\times 10^{-4}$ m and $A=\pi r^2=1.256\times 10^{-7}{\text{m}}^2$: $\rho =\frac{3\times 1.256\times 10^{-7}}{1.2}=3.14\times 10^{-7}\Omega \text{m}$.

   Step-by-step solution →
8. Q208 CBSE 2024
   1. (i)State Ohm’s law. ​Draw ​V – I graph for an ohmic conductor.
   2. (ii)Electrical resistance ​of a metal wire of length $1$ km is $200\Omega$. If the area of cross-section of ​the wire is $40\times 10^{-8}{\text{m}}^2$, determine ​the electrical ​resistivity of the material of the wire.
   Show answer

   Ohm's law: $V\propto I$ at constant temperature, so the $V$-$I$ graph is a straight line through the origin; $\rho =8\times 10^{-8}\Omega \text{m}$.

   (i) Ohm’s law: the potential difference $V$ across the ends of a metallic wire is directly proportional to the current $I$ flowing through it, provided its temperature remains the same, so $V=IR$, where $R$ is the resistance of the wire. The V–I graph of an ohmic conductor is a straight line passing through the origin, and its slope $\frac{V}{I}$ equals $R$. (ii) $\rho =\frac{RA}{l}=\frac{200\times 40\times 10^{-8}}{1000}=8\times 10^{-8}\Omega \text{m}$.

   Step-by-step solution →
9. Q209 CBSE 2024
   1. (i)​Write the relation between resistance $R$ and electrical resistivity $\rho$ of the material of a conductor in the shape of cylinder of length $l$ and ​area of cross-section $A$. Hence derive the SI unit of electrical ​resistivity.
   2. (ii)​The resistance of a metal wire of length $3$ m is $60\Omega$. If the area of cross-section of the wire is $4\times 10^{-7}{\text{m}}^2$, calculate ​the electrical resistivity of the wire.
   3. (iii)State ​how would electrical resistivity be affected if ​the wire ​(of part ‘ii’) is stretched so that its length is doubled. Justify your answer.
   Show answer

   $R=\rho \frac{l}{A}$ gives the unit $\Omega \text{m}$; $\rho =8\times 10^{-6}\Omega \text{m}$, and stretching the wire does not change it.

   (i) $R=\rho \frac{l}{A}$, so $\rho =\frac{RA}{l}$. The SI unit of $\rho$ is $\frac{\Omega \times {\text{m}}^2}{\text{m}}=\Omega \text{m}$ (ohm metre). (ii) $\rho =\frac{RA}{l}=\frac{60\times 4\times 10^{-7}}{3}=8\times 10^{-6}\Omega \text{m}$. (iii) The resistivity is not affected; it stays $8\times 10^{-6}\Omega \text{m}$. Resistivity depends only on the nature of the material (and its temperature), not on the length or area of the wire. Doubling the length halves the area of cross-section (the volume is unchanged), so the resistance becomes four times, $240\Omega$, but $\rho =\frac{240\times 2\times 10^{-7}}{6}=8\times 10^{-6}\Omega \text{m}$ is the same.

   Step-by-step solution →
10. Q210 CBSE 2024
    1. (i)When ​is the potential difference between two ​points said to be 1 volt ?
    2. (ii)​A copper wire has a diameter of $0.2$ ​mm and resistivity of $1.6\times 10^{-8}\Omega \text{m}$. ​What will be the length of this wire to ​make its resistance $14\Omega$ ? How much does ​the resistance change, if the diameter of the wire is doubled ?
    Show answer

    One volt is 1 J of work per coulomb; the wire is about 27.5 m long; doubling its diameter gives $3.5\Omega$.

    (i) The potential difference between two points is $1$ volt when $1$ joule of work is done to move a charge of $1$ coulomb from one point to the other ($1\text{V}=1{\text{J C}}^{-1}$). (ii) $A=\pi r^2=\frac{22}{7}\times (1\times 10^{-4}{)}^2\approx 3.14\times 10^{-8}{\text{m}}^2$, so $l=\frac{RA}{\rho }=\frac{14\times 3.14\times 10^{-8}}{1.6\times 10^{-8}}\approx 27.5$ m. If the diameter is doubled, the area becomes $4$ times as large, so the resistance becomes one-fourth: $14/4=3.5\Omega$, a decrease of $10.5\Omega$.

    Step-by-step solution →
11. Q211 CBSE 2024
    The ​values of current $I$ flowing ​in ​a given resistor for the corresponding values of potential ​difference $V$ across the resistor ​are given in the following table :

    1. (i)Plot a ​graph between $V$ and $I$.
    2. (ii)Calculate the resistance of the ​resistor with the help of the graph.
    3. (iii)What does the graph represent ​?
    4. (iv)Why should this graph pass through the origin ?

    Quantity	Reading 1	Reading 2	Reading 3	Reading 4	Reading 5
    I (Amperes)	0.5	1.0	2.0	3.0	3.5
    V (Volts)	1.5	3.0	6.2	9.3	10.8

    Show answer

    The graph is a straight line through the origin, $V=3.1I$, so $R=3.1\Omega$; it represents Ohm's law.

    (i) With $I$ on the horizontal axis and $V$ on the vertical axis, the five points lie very nearly on a straight line through the origin, $V=3.1I$. (ii) Using the points $I=1.0$ A, $V=3.1$ V and $I=3.0$ A, $V=9.3$ V on the line: $R=\frac{\Delta V}{\Delta I}=\frac{9.3-3.1}{3.0-1.0}=\frac{6.2\text{V}}{2.0\text{A}}=3.1\Omega$. (iii) The graph is a straight line through the origin, so $V\propto I$: it represents Ohm’s law, with the resistance of the resistor constant. (iv) With no potential difference across the resistor no current flows ($V=0$ gives $I=0$), so the graph must pass through the origin.

    Step-by-step solution →
12. Q212 CBSE 2024
    1. (i)What is meant by heating ​effect of electric ​current ? Derive an expression for the amount of heat produced ​in a conductor of resistance $R$ through which a current $I$ flows ​for a time $t$.
    2. (ii)Calculate the ​amount of heat produced in a conductor of resistance $10\Omega$ through ​which a ​current of $2$ A flows for $100$ seconds.
    Show answer

    Heat produced is $H=I^{2}Rt$; for 2 A through $10\Omega$ for 100 s, $H=4000$ J.

    (i) When a current flows through a purely resistive conductor, the energy supplied by the source is dissipated entirely as heat in the conductor; this is the heating effect of electric current. If a current $I$ flows through a resistance $R$ for a time $t$ with a potential difference $V$ across it, the charge moved is $Q=It$ and the work done by the source is $W=VQ=VIt$. This work appears as heat, so $H=VIt$, and with Ohm's law $V=IR$ this gives $H=I^{2}Rt$ (Joule's law of heating). (ii) $H=I^{2}Rt=2^2\times 10\times 100=4000$ J, that is $4\times 10^3$ J.

    Step-by-step solution →
13. Q213 CBSE 2024
    1. (i)You have three resistors of resistance $R_1$, $R_2$ and $R_3$. If ​the equivalent resistance of series and parallel combination of the three resistors are $R_s$ and $R_p$ respectively, write the relationship to determine $R_s$ and $R_p$ in terms of $R_1$, $R_2$ and $R_3$.
    2. (ii)On the ​basis of the given circuit ​answer the questions given below : (I) ​Write the ​relationship between the readings of the two ammeters ${\text{A}}_1$ and ${\text{A}}_2$. ​Give reason for your answer. (II) What ​is the equivalent resistance of the parallel combination of three resistors ? (III) What is the value of current flowing in the circuit ​?

    Show answer

    $R_s=R_1+R_2+R_3$; the two ammeters read equally, $R_p=10\Omega$ and the circuit current is $0.1$ A.

    (i) $R_s=R_1+R_2+R_3$ and $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$. (ii) (I) The readings of ${\text{A}}_1$ and ${\text{A}}_2$ are equal. Both ammeters are in the main line, in series with the battery; the current that divides among the three branches joins again before reaching ${\text{A}}_2$, so the same current flows through both. (II) $\frac{1}{R_p}=\frac{1}{30}+\frac{1}{30}+\frac{1}{30}=\frac{1}{10}$, so $R_p=10\Omega$. (III) The total resistance is $10+20=30\Omega$, so $I=\frac{V}{R}=\frac{3}{30}=0.1$ A.

    Step-by-step solution →
14. Q214 CBSE 2024
    When electric current flows in ​a purely resistive circuit ​electrical energy gets fully converted ​into heat energy. The amount of heat produced ($H$) in the circuit ​is ​found to be directly proportional to

    1. (i)the square of current ($I^2$)
    2. (ii)the ​resistance ($R$) of the conductor and ​
    3. (iii)the time ($t$) for which current flows. ​In other words $H=I^{2}Rt$. Electrical devices such an electric fuse, electric heater, electric iron etc. are all based on this effect called heating effect of electric current.

    1. (a)List two properties of heating elements.[1]
    2. (b)List two properties of electric fuse.[1]
    3. (c)(i)Name the principle on which an electric fuse works. Explain how a fuse wire is capable of saving electrical appliances from getting damaged due to accidently produced high currents.[2]
    4. (c)(ii)OR : The power of an electric heater is 1100 W. If the potential difference between the two terminals of the heater is 220 V, find the current flowing in the circuit. What will happen to an electric fuse of rating 5 A connected in this circuit ?[2]
    Show answer

    Heating elements need high resistivity and melting point; a fuse uses the heating effect to break the circuit; or the heater draws $5$ A.

    (a) A heating element should have (1) a high resistivity, so that the heat is produced mainly in the element, and (2) a high melting point, so that it stays red-hot without melting; it should also not oxidise readily at high temperatures. (b) An electric fuse wire has (1) a low melting point and (2) a high resistance per unit length, being much thinner than the connecting wires that carry the same current, so that the heat produced by an excess current is concentrated in the fuse and quickly raises it to its melting point. It is connected in series with the appliance, in the live wire, and has a definite current rating (such as 1 A, 2 A, 3 A, 5 A or 10 A). (c)(i) A fuse works on the heating effect of electric current (Joule's law of heating, $H=I^{2}Rt$). It is connected in series with the appliance, so the whole current flows through it. When an unduly high current flows, the heat produced in the fuse wire rises sharply (as $I^2$), the wire reaches its melting point and melts, and the circuit is broken before the high current can damage the appliance. OR (c)(ii) $I=\frac{P}{V}=\frac{1100}{220}=5$ A. The fuse is rated 5 A and melts only if the current is more than 5 A, so it will not melt and the heater works normally; the fuse is then working at its limit, and any rise in the current above 5 A would melt it.

    Step-by-step solution →
15. Q215 CBSE 2024
    In a domestic circuit five LED bulbs ​are arranged as shown. The source voltage ​is ​220 V ​and the power rating of ​each bulb is marked in the circuit diagram. Based on the following circuit ​diagram, answer the following ​questions :

    1. (a)State what happens when (i) key ${\text{K}}_1$ is closed. (ii) key ${\text{K}}_2$ is closed.[1]
    2. (b)Find the current drawn by the bulb B when it glows.[1]
    3. (c)Calculate (i) the resistance of bulb B, and (ii) total resistance of the combination of four bulbs B, C, D and E.[2]
    4. (c)OR : What would happen to the glow of all the bulbs in the circuit when keys ${\text{K}}_1$ and ${\text{K}}_2$ both are closed and the bulb C suddenly get fused ? Give reason to justify your answer.[2]
    Show answer

    Only A glows with ${\text{K}}_1$, B to E with ${\text{K}}_2$; $I_B=0.2$ A, $R_B=275\Omega$, total $1100\Omega$; if C fuses, A still glows.

    (a) (i) When ${\text{K}}_1$ is closed, only bulb A is connected across the 220 V source; it gets its rated voltage and glows with full brightness (22 W), while B, C, D and E do not glow. (ii) When ${\text{K}}_2$ is closed, bulbs B, C, D and E are connected in series across 220 V; each gets 55 V, its rated voltage, so all four glow normally, while A does not glow. (b) $I_B=\frac{P}{V}=\frac{11}{55}=0.2$ A. (c)(i) $R_B=\frac{V^2}{P}=\frac{55^2}{11}=275\Omega$. (c)(ii) The four bulbs are in series, so $R=4\times 275=1100\Omega$. OR (c) Bulbs B, D and E stop glowing, along with the fused bulb C, because they are in series with C: the break in C opens the only path through them, so no current flows in that branch. Bulb A goes on glowing with the same brightness, because it is on a separate parallel branch that still has the full 220 V across it.

    Step-by-step solution →
16. Q216 CBSE 2024
    Study ​the following ​circuit: On ​the ​basis ​of this circuit, answer the ​following questions:

    1. (a)Find the value of total resistance between the points A and B.[1]
    2. (b)Find the resistance between the points B and C.[1]
    3. (c)(i)Calculate the current drawn from the battery, when the key is closed.[2]
    4. (c)(ii)OR : In the above circuit, the $16\Omega$ resistor or the parallel combination of two resistors of $8\Omega$, which one of the two will have more potential difference across its two ends? Justify your answer.[2]
    Show answer

    $R_{AB}=26\Omega$ and $R_{BC}=4\Omega$; $I=0.2$ A; or the $16\Omega$ resistor has the larger potential difference, 3.2 V against 0.8 V.

    (a) $R_{AB}=4+6+16=26\Omega$. (b) $\frac{1}{R_{BC}}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$, so $R_{BC}=4\Omega$. (c)(i) Total resistance $R=26+4=30\Omega$, so the current drawn from the battery is $I=\frac{V}{R}=\frac{6}{30}=0.2$ A. (c)(ii) The $16\Omega$ resistor. It is in series with the parallel combination, so the same current, 0.2 A, flows through both; by $V=IR$ the potential difference is $0.2\times 16=3.2$ V across the $16\Omega$ resistor but only $0.2\times 4=0.8$ V across the combination, whose resistance is $4\Omega$.

    Step-by-step solution →
17. Q217 CBSE 2023
    If ​four identical resistors, ​of resistance 8 ohm, are first connected in ​series so as to ​give an effective resistance $R_s$, ​and then ​connected ​in parallel so as to give an effective resistance $R_p$, then ​the ratio $\frac{R_s}{R_p}$ is

    1. (a)$32$
    2. (b)$2$
    3. (c)$0.5$
    4. (d)$16$
    Show answer

    (d) $16$

    Step-by-step solution →
18. Q218 CBSE 2023
    Two LED ​bulbs of 12W and 6W are ​connected in series. ​If the ​current through 12W ​bulb is ​0.06A the current through 6W bulb ​will be :

    1. (a)$0.04$ A
    2. (b)$0.06$ A
    3. (c)$0.08$ A
    4. (d)$0.12$ A
    Show answer

    (b) $0.06$ A

    Step-by-step solution →
19. Q219 CBSE 2023
    The expressions that ​relate

    1. (i)$Q$, $I$ and $t$ ​and
    2. (ii)$Q$, $V$ and $W$ ​respectively are (Here ​the symbols have their ​usual ​meanings) ​:

    1. (a)(i) $I=\frac{Q}{t}$ (ii) $W=\frac{V}{Q}$
    2. (b)(i) $Q=I\times t$ (ii) $W=V\times Q$
    3. (c)(i) $Q=\frac{I}{t}$ (ii) $V=\frac{W}{Q}$
    4. (d)(i) $I=\frac{Q}{t}$ (ii) $Q=\frac{V}{W}$
    Show answer

    (b) (i) $Q=I\times t$ (ii) $W=V\times Q$

    Step-by-step solution →
20. Q220 CBSE 2023
    SI ​unit of ​electrical ​resitivity ​is :

    1. (a)ohm metre
    2. (b)ohm per metre
    3. (c)ohm / metre${}^2$
    4. (d)ohm / metre${}^3$
    Show answer

    (a) ohm metre

    Step-by-step solution →