Chapter 11 · Electricity

NCERT Class 10 Science Chapter 11 Exercises

All 18 questions of NCERT Class 10 Science Chapter 11 Exercises (Electricity), in the book's order, each with its answer and a step-by-step solution.

1. Q1
   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
   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) $I{R}^2$

   Step-by-step solution →
3. Q3
   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 →
4. Q4
   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 →
5. Q5
   How ​is a voltmeter connected ​in ​the circuit ​to measure ​the ​potential difference ​between two ​points?
   Show answer

   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 →
6. Q6
   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 →
7. Q7
   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 →
8. Q8
   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.
   Show answer

   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 →
9. Q9
   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 →
10. Q10
    How ​many $176\Omega$ ​resistors (in parallel) ​are required to ​carry $5\text{A}$ on ​a $220\text{V}$ line?
    Show answer

    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 →
11. Q11
    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 →
12. Q12
    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 $5\text{A}$?
    Show answer

    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 →
13. Q13
    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 →
14. Q14
    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 →
15. Q15
    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 →
16. Q16
    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 →
17. Q17
    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 →
18. Q18
    Explain the following.

    1. (a)Why is ​the tungsten used almost exclusively for filament of ​electric lamps?
    2. (b)​Why are the ​conductors of ​electric ​heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
    3. (c)Why is the series ​arrangement not used for domestic circuits?
    4. (d)How ​does the resistance of a wire vary with its area of cross-section?
    5. (e)Why are copper and aluminium wires usually employed for electricity transmission?
    Show answer

    Tungsten has a high melting point; alloys have high resistivity; series wiring shares voltage; $R\propto 1/A$; copper and aluminium lose little heat.

    (a) Tungsten has a very high melting point (about $3380^{\circ }\text{C}$), well above the roughly $2500^{\circ }\text{C}$ at which a filament glows, together with a high resistivity; it retains the heat produced and does not oxidise or vaporise readily, so it can be kept white-hot without melting away. (b) An alloy such as nichrome has a much higher resistivity than the pure metals in it, so it develops far more heat for the same current ($H=I^{2}Rt$), and it does not oxidise or burn readily at high temperature, so the element lasts. (c) In series the same current would pass through every appliance, the supply voltage would be divided so none would get the full $220\text{V}$, the total resistance would be large so the current would be small, and one appliance failing or being switched off would break the circuit and stop all the others; appliances are therefore connected in parallel. (d) Inversely — $R=\rho l/A$, so for a given material and length the resistance is inversely proportional to the area of cross-section: doubling the area halves the resistance. (e) Copper and aluminium have very low resistivity, so the heat lost in the line, $I^{2}Rt$, is small; they are also ductile, strong enough, and far cheaper and more plentiful than silver, with aluminium's lightness suiting long overhead lines.

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