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 13 of 19

1. Q241 CBSE 2023
   There are three ​resistors ​of $10\Omega$, $20\Omega$ and $30\Omega$ joined ​in parallel in a circuit. The ​potential ​difference across the electric circuit is $10$ ​V.

   1. (a)Draw a circuit diagram for ​the above case.
   2. (b)Find the total resistance of the combination of resistors.
   3. (c)Calculate the electric current drawn from the ​same source.
   Show answer

   Three resistors in parallel across 10 V have total resistance $\frac{60}{11}\Omega \approx 5.45\Omega$ and draw about 1.83 A.

   (a) The three resistors are joined between the same two points X and Y, and the $10$ V battery, a plug key and an ammeter are connected in series between X and Y, with a voltmeter across X and Y (see the circuit diagram). (b) $\frac{1}{R_p}=\frac{1}{10}+\frac{1}{20}+\frac{1}{30}=\frac{6+3+2}{60}=\frac{11}{60}$, so $R_p=\frac{60}{11}\Omega \approx 5.45\Omega$. (c) $I=\frac{V}{R_p}=\frac{10}{60/11}=\frac{11}{6}\text{A}\approx 1.83$ A.

   Step-by-step solution →
2. Q242 CBSE 2023
   1. (a)Define electric power and state its SI unit. The commercial unit of electrical energy is known as 'unit'. Write the relation between this 'unit' and joule.
   2. (b)In a house, ​2 bulbs of 50 ​W each are used for 6 hours daily ​and an ​electric geyser of 1 kW is used for 1 hour daily. ​Calculate the total energy consumed in a month of 30 days and its cost at the rate of ₹ 8.00 ​per ​kWh.
   Show answer

   Power is the rate of consuming electrical energy; $1\text{kWh}=3.6\times 10^6$ J; the month's energy is 48 kWh, costing ₹ 384.

   (a) Electric power is the rate at which electrical energy is consumed (or dissipated) in an electric circuit: $P=\frac{W}{t}=VI$. Its SI unit is the watt (W), where $1\text{W}=1{\text{J s}}^{-1}$, the power of a device that carries $1$ A at a potential difference of $1$ V. The commercial 'unit' is the kilowatt hour: $1\text{kWh}=1000\text{W}\times 3600\text{s}=3.6\times 10^6$ J. (b) Bulbs: $2\times 50$ W $=0.1$ kW for $6$ h gives $0.6$ kWh a day. Geyser: $1$ kW for $1$ h gives $1.0$ kWh a day. Total per day $=1.6$ kWh, so the energy consumed in $30$ days is $1.6\times 30=48$ kWh ($48$ units, or $1.728\times 10^8$ J). Cost $=48\times 8.00=384$, that is ₹ $384$.

   Step-by-step solution →
3. Q243 CBSE 2023
   1. (a)An electric iron consumes energy at a rate of 880 W when heating is at the maximum rate and 330 W when the heating is at the minimum. If ​the ​source voltage is 220 ​V, calculate the ​current and resistance in each case.
   2. (b)What is heating effect of electric current ?
   3. (c)​Find an expression for the ​amount of ​heat produced when a current passes through a ​resistor for some time.
   Show answer

   Maximum setting: 4 A and $55\Omega$; minimum: 1.5 A and about $146.67\Omega$; the heat produced is $H=I^{2}Rt$.

   (a) At the maximum setting $I=\frac{P}{V}=\frac{880}{220}=4$ A and $R=\frac{V}{I}=\frac{220}{4}=55\Omega$. At the minimum setting $I=\frac{330}{220}=1.5$ A and $R=\frac{220}{1.5}=146.67\Omega$, that is $\frac{440}{3}\Omega$. (b) When an electric current is passed through a conductor that offers only resistance, the electrical energy supplied by the source is dissipated entirely as heat and the conductor becomes hot. This is the heating effect of electric current, and it is used in the electric iron, heater, toaster, geyser, bulb filament and fuse. (c) A charge $Q=It$ passes through the resistor in time $t$, so the source does work $W=VQ=VIt$ on it. In a purely resistive circuit all of this work appears as heat, and substituting $V=IR$ gives $H=I^{2}Rt$, which is Joule's law of heating.

   Step-by-step solution →
4. Q244 CBSE 2023
   1. (i)How is electric current ​related to the potential difference across the terminals of a conductor ​? Draw a labelled circuit diagram to verify this relationship.
   2. (ii)Why should an ammeter have low resistance ​?
   3. (iii)Two ​V - I graphs A and B for ​series and ​parallel combinations of two resistors are as shown. Giving reason state ​which graph ​shows (a) series, (b) parallel combination of the resistors.

   Show answer

   $I\propto V$; an ammeter needs low resistance so it does not change the current; graph A is the series combination, graph B the parallel.

   (i) By Ohm's law the current is directly proportional to the potential difference across the conductor at constant temperature: $I\propto V$, that is $V=IR$. It is verified with the nichrome wire in series with a battery, plug key, rheostat and ammeter and a voltmeter connected in parallel across the wire; the rheostat is used to take several pairs of readings and $V/I$ comes out constant. (ii) An ammeter is connected in series, so its resistance adds to the circuit and the current becomes $I=V/(R+r)$. Only if $r$ is very small does the ammeter leave the current -- and hence its own reading -- practically unchanged, and only then is the potential difference used up across it negligible. (iii) On the printed axes $I$ is vertical and $V$ horizontal, so the slope is $I/V=1/R$: the steeper line has the smaller resistance. Since $R_s=R_1+R_2$ exceeds either resistor and $R_p$ is less than either, (a) the series combination is graph A (the less steep line) and (b) the parallel combination is graph B (the steeper line).

   Step-by-step solution →
5. Q245 CBSE 2023
   1. (a)​State Ohm's Law.
   2. (b)Name and define the physical quantity determined by the slope of $V-I$ curve given in the diagram. Use this graph to ​find ​the value ​of this physical quantity ​in ​SI units.
   3. (c)Establish the relationship between $1$ kWh and $1$ ​joule.

   Show answer

   Ohm's law: $V\propto I$ at constant temperature; the slope of the $V$-$I$ curve is resistance, $7.5\Omega$; $1\text{kWh}=3.6\times 10^6$ J.

   (a) Ohm's law: the potential difference across the ends of a metallic conductor is directly proportional to the current flowing through it, provided its temperature and other physical conditions remain constant. Hence $V\propto I$, that is $V=IR$. (b) The slope of the $V-I$ curve gives the **resistance** $R$ of the conductor. Resistance is defined as the ratio of the potential difference across a conductor to the current flowing through it, and it measures the opposition offered to the flow of current; its SI unit is the ohm ($\Omega$). From the graph, taking the points $(0.2$ A, $1.5$ V$)$ and $(0.8$ A, $6.0$ V$)$, $R=\frac{\Delta V}{\Delta I}=\frac{6.0-1.5}{0.8-0.2}=\frac{4.5}{0.6}=7.5\Omega$. (c) $1$ kWh $=1$ kW $\times 1$ h $=1000$ W $\times 3600$ s $=3.6\times 10^6$ W s $=3.6\times 10^6$ J. So $1$ kWh $=3.6\times 10^6$ joule.

   Step-by-step solution →
6. Q246 CBSE 2023
   1. (a)​Derive an expression for the amount of ​heat produced when a current flows through a resistor for a given time. Use this expression to state ​Joule's law ​of heating.
   2. (b)$200$ joules of heat energy is produced each ​second in a resistor of $8\Omega$. Find : (i) ​The current flowing through the resistor, and (ii) The potential difference ​across the ends ​of the resistor.
   Show answer

   Heat produced is $H=I^{2}Rt$ (Joule's law); in the $8\Omega$ resistor the current is $5$ A and the potential difference is $40$ V.

   (a) For a current $I$ flowing for a time $t$ through a resistor $R$ across which the potential difference is $V$: the charge carried is $Q=It$, and the work done in taking that charge through the potential difference is $W=VQ=VIt$. With no other energy conversion this whole amount appears as heat, so $H=VIt$; and using Ohm's law $V=IR$, $H=I^{2}Rt$. Joule's law of heating: the heat produced in a resistor is directly proportional to the square of the current for a given resistance, directly proportional to the resistance for a given current, and directly proportional to the time for which the current flows. (b)(i) $H=I^{2}Rt$ gives $200=I^2\times 8\times 1$, so $I^2=25$ and the current is $I=5$ A. (b)(ii) $V=IR=5\times 8=40$ V.

   Step-by-step solution →
7. Q247 CBSE 2023
   1. (a)The ​commercial unit of electric energy is commonly ​known as ‘unit’. ​Write its actual name and establish the relationship between this unit and the ​SI unit of energy.
   2. (b)The current rating of a circuit is $1.0$ A. ​How many LED lamps of power rating $8$ W; $220$ V ​can safely be ​used simultaneously in this circuit ?
   Show answer

   The commercial unit is the kilowatt hour, $1\text{kW h}=3.6\times 10^6$ J; 27 LED lamps can be used safely.

   (a) The commercial unit of electric energy is the **kilowatt hour** (kW h) - the energy used by a $1$ kW appliance in $1$ hour. Since $1\text{kW}=1000$ W and $1\text{h}=3600$ s, $1\text{kW h}=1000\times 3600=3.6\times 10^6$ J. The SI unit of energy is the joule, so one unit of electric energy equals $3.6\times 10^6$ J. (b) Each lamp draws $I=P/V=8/220=0.0364$ A. The lamps are in parallel, so $n\times \frac{8}{220}\le 1.0$, giving $n\le 220/8=27.5$. Since $n$ must be a whole number, $n=27$: 27 LED lamps can safely be used simultaneously ($27$ lamps draw $0.982$ A, while $28$ would draw $1.018$ A and exceed the rating).

   Step-by-step solution →
8. Q248 CBSE 2023
   1. (a)​Calculate the resistance of an ​aluminium wire of length $1$ m ​and area ​of cross-section $2{\text{mm}}^2$. Resistivity of aluminium is $2.63\times 10^{-8}\Omega \text{m}$.
   2. (b)From the values ​given below, ​plot a graph of $I$ versus $V$. Show ​that the data is in ​conformity of the Ohm’s law.

   Readings of current and potential difference

   Current – I (ampere)	0.1	0.2	0.3	0.4
   Potential Difference – V (volt)	1.2	2.4	3.6	4.8

   Show answer

   The wire has resistance $0.013\Omega$; $\frac{V}{I}=12\Omega$ is constant and the graph is a straight line through the origin, so Ohm's law holds.

   (a) $A=2{\text{mm}}^2=2\times 10^{-6}{\text{m}}^2$, so $R=\rho \frac{l}{A}=\frac{2.63\times 10^{-8}\times 1}{2\times 10^{-6}}=1.315\times 10^{-2}\Omega \approx 0.013\Omega$. (b) For every reading $\frac{V}{I}=\frac{1.2}{0.1}=\frac{2.4}{0.2}=\frac{3.6}{0.3}=\frac{4.8}{0.4}=12\Omega$, a constant, and the graph of $I$ against $V$ is a straight line passing through the origin (slope $\frac{1}{12}$ ampere per volt). So $I\propto V$, that is $V=IR$ with $R=12\Omega$: the data are in conformity with Ohm’s law.

   Step-by-step solution →
9. Q249 CBSE 2023
   (a) You have three resistors $R_1$, $R_2$ ​and $R_3$. Write the formula for the equivalent resistance of the combination, when these resistors are joined ​

   1. (i)in series and
   2. (ii)in parallel. (b) Three resistors of $2\Omega$, $3\Omega$ and $4\Omega$ are ​connected in ​series and the combination is connected ​to a battery of $9$ V. An ammeter and a voltmeter are also connected ​in ​the circuit. Draw a circuit diagram to show the connection of the above said devices in a correct order. Determine (i) current shown by the ammeter and (ii) potential difference across the $3\Omega$ resistor.
   Show answer

   Series: $R_s=R_1+R_2+R_3$, parallel: reciprocals add; the ammeter reads $1$ A and the $3\Omega$ resistor has $3$ V across it.

   (a) (i) In series: $R_s=R_1+R_2+R_3$. (ii) In parallel: $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$. (b) The battery, a plug key, the ammeter and the three resistors are joined in series in one loop, and the voltmeter is connected in parallel across the $3\Omega$ resistor (see the circuit diagram). $R_s=2+3+4=9\Omega$. (i) The ammeter shows $I=\frac{9}{9}=1$ A. (ii) The p.d. across the $3\Omega$ resistor is $V=1\times 3=3$ V.

   Step-by-step solution →
10. Q250 CBSE 2023
    1. (i)What is meant by ​resistance of ​a conductor ? Define its SI unit.
    2. (ii)List ​two factors on which the resistance of a rectangular conductor depends.
    3. (iii)How will the resistance ​of a wire be ​affected if its (1) length is doubled, and (2) radius is also ​doubled ? Give ​justification for your answer.
    Show answer

    Resistance opposes current (unit ohm) and depends on length and area; doubling the length and the radius together halves the resistance.

    (i) Resistance is the property of a conductor by which it opposes the flow of electric current through it; numerically $R=V/I$. Its SI unit is the ohm ($\Omega$): one ohm is the resistance of a conductor through which a current of $1$ A flows when a potential difference of $1$ V is applied across its ends, so $1\Omega =1\text{V}/1\text{A}$. (ii) Its length $l$ (with $R\propto l$) and its area of cross-section $A$ (with $R\propto 1/A$), through $R=\rho l/A$. (iii) Since $R=\rho l/\pi r^2$: (1) doubling the length alone gives $R_1=2R$; (2) doubling the radius as well makes the area $\pi (2r{)}^2=4\pi r^2$, four times as large, so $R^{\prime }=\rho (2l)/(4\pi r^2)=R/2$. The resistance of the wire is **halved**, because length multiplies it by $2$ while four times the area divides it by $4$. (If (2) is read as the radius alone being doubled, the area becomes $4A$ and the resistance falls to $R/4$.)

    Step-by-step solution →
11. Q251 CBSE 2023
    In an electric circuit three bulbs ​of 100 W each are connected in series to a ​source. In ​another circuit set of three bulbs of the same wattage are connected in ​parallel to the same source.

    1. (i)Will ​the bulb ​in the two circuits glow with the same brightness ? Justify your answer.
    2. (ii)Now, let one bulb in both the circuits get fused. Will the rest of the bulbs ​continue to glow in each circuit ? Give reason for your answer.
    Show answer

    No: parallel bulbs are about 9 times brighter; if one fuses, the series bulbs go out but the parallel ones keep glowing.

    (i) No. A bulb rated $100$ W has resistance $R=V^2/100$. In the **parallel** circuit the full supply voltage $V$ is across each bulb, so each dissipates $P=V^2/R=100$ W and glows at full brightness. In the **series** circuit the resistance is $3R$, the current is $I=V/3R$ and each bulb dissipates only $P=I^{2}R=V^2/9R=100/9\approx 11.1$ W. The parallel bulbs are therefore about $9$ times brighter than the series ones. (ii) **Series**: no. There is only one path, and a fused bulb breaks it, so no current flows and the other two bulbs also go out. **Parallel**: yes. Each bulb has its own separate path across the source, so the remaining two still have the full $V$ across them and keep glowing with the same brightness as before.

    Step-by-step solution →
12. Q252 CBSE 2023
    Consider ​the ​following electrical circuit ​diagram in which ​nine identical resistors ​of $3\Omega$ each are ​connected as shown. ​If the reading of the ammeter $A_1$ is $1$ ampere, answer the following questions :

    1. (a)What is the relationship between the readings of $A_1$ and $A_3$ ? Give reasons for your answer.[1]
    2. (b)What is the relationship between the readings of $A_2$ and $A_3$ ?[1]
    3. (c)(i)Determine the reading of the voltmeter $V_1$.[2]
    4. (c)(ii)OR : Find the total resistance of the circuit.[2]
    Show answer

    $A_1=A_3=1$ A and $A_2=0.25$ A; $V_1=1.5$ V, or the total resistance is $3.25\Omega$.

    (a) $A_1=A_3=1$ A. Both ammeters lie in the single series line of the circuit, and a parallel cluster only splits the current and then recombines it, so the current entering the network equals the current leaving it. (b) $A_2=\frac{A_3}{4}=0.25$ A, that is $A_3=4A_2$, because the third cluster's four identical branches share the current equally and $A_2$ measures only one of them. (c)(i) The first cluster has resistance $R_1=\frac{3}{2}=1.5\Omega$, so $V_1=I{R}_1=1\times 1.5=1.5$ V. (c)(ii) OR — The three clusters come to $R_1=1.5\Omega$, $R_2=\frac{3}{3}=1\Omega$ and $R_3=\frac{3}{4}=0.75\Omega$, and being in series they add: $R_{\text{total}}=1.5+1+0.75=3.25\Omega$.

    Step-by-step solution →
13. Q253 CBSE 2022
    An electric ​heater rated 1100 W operates ​at 220 V. ​Calculate

    1. (i)​its resistance, and
    2. (ii)the current ​drawn by it.
    Show answer

    The resistance of the heater is $44\Omega$ and it draws a current of 5 A.

    (i) Resistance $R=\frac{V^2}{P}=\frac{220\times 220}{1100}=44\Omega$. (ii) Current $I=\frac{P}{V}=\frac{1100}{220}=5$ A.

    Step-by-step solution →
14. Q254 CBSE 2022
    Calculate ​the ​equivalent ​resistance of ​the following ​electric ​circuit ​:

    Show answer

    The equivalent resistance of the circuit is $40\Omega$.

    The two $10\Omega$ resistors in series make $20\Omega$, which is in parallel with the upper $20\Omega$: $\frac{1}{R_p}=\frac{1}{20}+\frac{1}{20}$, so $R_p=10\Omega$. This is in series with the $10\Omega$ and $20\Omega$ resistors: $R_{\text{eq}}=10+10+20=40\Omega$.

    Step-by-step solution →
15. Q255 CBSE 2022
    State Ohm's law. A student has drawn V – ​I graph ​on ​the basis of his experimental observations. ​Why should this graph be a straight ​line passing ​through ​the origin ?
    Show answer

    At constant temperature $V\propto I$ (Ohm's law); the graph is a straight line through the origin because $R$ is constant and $V=0$ when $I=0$.

    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; that is, $V\propto I$, or $V=IR$, where the resistance $R$ is a constant. The graph should be a straight line because, with $R$ constant, $V=IR$ is a direct proportion of the form $y=mx$: the ratio $V/I$ has the same value, $R$, for every reading, so all the points lie on one straight line whose slope is $R$. It should pass through the origin because when $I=0$, $V=R\times 0=0$ - with no potential difference across the wire there is no current - so the point $(0,0)$ lies on the line.

    Step-by-step solution →
16. Q256 CBSE 2022
    When a 6.0 V battery is connected across a resistor ​there is ​a ​current of 30 ​mA in the circuit. Find the value ​of the ​resistance of the ​resistor ​in ohms.
    Show answer

    The resistance of the resistor is $200\Omega$.

    $R=\frac{V}{I}=\frac{6.0\text{V}}{0.030\text{A}}=200\Omega$. The resistance of the resistor is $200$ ohms.

    Step-by-step solution →
17. Q257 CBSE 2022
    1. (a)Write the relationship ​between ​electrical resistance and electrical resistivity for a ​metallic conductor ​of cylindrical shape. Hence derive the SI unit of electrical resistivity.
    2. (b)Find the resistivity of the material of a metallic conductor ​of length 2m and area ​of ​cross-section $1.4\times 10^{-6}{\text{m}}^2$. ​The resistance of the conductor is 0.04 ohm.
    Show answer

    Resistivity has SI unit $\Omega \text{m}$; for the given conductor $\rho =2.8\times 10^{-8}\Omega \text{m}$.

    (a) For a metallic conductor of length $l$ and area of cross-section $A$, $R=\rho \frac{l}{A}$, where $\rho$ is the electrical resistivity of its material; for a cylinder of radius $r$, $A=\pi r^2$. Hence $\rho =\frac{RA}{l}$, and its SI unit is $\frac{\Omega \times {\text{m}}^2}{\text{m}}=\Omega \text{m}$ (ohm metre). (b) $\rho =\frac{RA}{l}=\frac{0.04\times 1.4\times 10^{-6}}{2}=2.8\times 10^{-8}\Omega \text{m}$.

    Step-by-step solution →
18. Q258 CBSE 2022
    You are provided with a ​resistor, ​a key, an ammeter, a voltmeter, four cells of $1.5$ V each and few connecting wires. Using these circuit components, draw a labelled ​circuit diagram to show ​the setup to ​study the Ohm's law. State the relationship between potential difference ($V$) across the resistor and the current ($I$) ​flowing through it. Also draw $V$-$I$ graph, taking $V$ ​on the X-axis.
    Show answer

    Ohm's law is $V\propto I$ at constant temperature, shown by a straight-line graph through the origin; the voltmeter is connected across the resistor.

    Circuit: the four $1.5$ V cells joined in series (a $6$ V battery), the plug key K, the ammeter A and the resistor $R$ are connected in series in a single loop, and the voltmeter V is connected in parallel across the resistor; the + terminal of each meter is on the side leading to the battery's + terminal (see the circuit diagram). Relationship (Ohm's law): the potential difference $V$ across the resistor is directly proportional to the current $I$ through it, provided its temperature remains the same: $V\propto I$, that is $V=IR$, with $R$ constant. $V$-$I$ graph: with $V$ on the x-axis and $I$ on the y-axis, the graph is a straight line passing through the origin, with slope $\frac{I}{V}=\frac{1}{R}$ (see the graph).

    Step-by-step solution →
19. Q259 CBSE 2022
    1. (a)Calculate the ​resistance ​of a metal wire of length $2\text{m}$ and ​area of cross-section $1.55\times 10^{-6}{\text{m}}^2$. (Resistivity of the metal ​is $2.8\times 10^{-8}\Omega \text{m}$)
    2. (b)Why are alloys ​preferred ​over pure metals to make ​the heating elements of electrical heating devices ?
    Show answer

    The wire's resistance is $0.036\Omega$; alloys are preferred because they have higher resistivity and do not oxidise readily when hot.

    (a) $R=\rho \frac{l}{A}=\frac{2.8\times 10^{-8}\Omega \text{m}\times 2\text{m}}{1.55\times 10^{-6}{\text{m}}^2}\approx 3.6\times 10^{-2}\Omega =0.036\Omega$. (b) The resistivity of an alloy is generally higher than that of its constituent metals, so a compact coil of alloy has a resistance much larger than that of the connecting wires and, carrying the same current, produces nearly all of the heat; and alloys do not oxidise (burn) readily at high temperatures, so the red-hot element lasts.

    Step-by-step solution →
20. Q260 CBSE 2022
    For a ​heater, rated 4 ​kW and 220 V, calculate ​the following :

    1. (a)The current
    2. (b)​Energy consumed in 2 hours
    3. (c)​If 1 kWh is ​priced at ₹ 4.50, then the cost of energy ​consumed
    Show answer

    The heater draws $18.18$ A, consumes $8$ kWh in 2 hours, and the energy costs ₹36.

    (a) $I=\frac{P}{V}=\frac{4000\text{W}}{220\text{V}}=\frac{200}{11}\text{A}\approx 18.18$ A. (b) $E=Pt=4\text{kW}\times 2\text{h}=8$ kWh $=2.88\times 10^7$ J. (c) Cost $=8\text{kWh}\times 4.50$ rupees per kWh $=36$ rupees, that is ₹36.

    Step-by-step solution →