NCERT Solutions Class 10 Science Chapter 11 — Electricity

NCERT Class 10 Science — Electricity, Chapter 11 Exercises. All 18 questions solved.

Also here. Solutions to the in-text questions for this chapter and the activities for this chapter are on their own pages.

The exercises at the end of Chapter 11 fall into four groups. Questions 5 to 8 are about Ohm’s law and what the resistance of a wire depends on. Questions 1, 9 to 11 and 13 join resistors in series and in parallel. Questions 2 to 4, 12 and 14 to 17 are about electrical power, energy and the heating effect of a current. Question 18 asks for explanations from the chapter’s discussion of materials and household wiring. Almost every calculation uses one of three sets of results:

• Ohm’s law and resistivity — $V = IR$, and for a wire of length $l$ and area of cross-section $A$, $R = \rho\dfrac{l}{A}$.
• Resistors in combination — in series $R_s = R_1 + R_2 + R_3$; in parallel $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$.
• Heat, power and energy — $H = I^2Rt$; power $P = VI = I^2R = \dfrac{V^2}{R}$; energy used $=$ power $\times$ time.

Every current in milliamperes, length in millimetres and time in minutes or hours is converted to amperes, metres and seconds before it goes into a formula.

Key insight. Before calculating, ask which quantity is the same for every part of the circuit. In series it is the current, because there is only one path for the charge; in parallel it is the potential difference, because every branch is joined across the same two points. Then use the form of the power formula built on that shared quantity — $P = I^2R$ when the current is common, $P = V^2/R$ when the voltage is common. That single choice turns Questions 4, 14 and 15 into one-line answers, and it is also the reason homes are wired in parallel (Question 18(c)).

Question 1

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 $R’$, then the ratio $R/R’$ is —

Solution. First find the resistance of one piece. Resistance is proportional to length, $R = \rho l/A$, and cutting the wire changes neither the material nor its thickness. Each piece has one-fifth of the length, so each has resistance $R/5$.

Now join the five pieces in parallel. Each contributes $\dfrac{1}{R/5} = \dfrac{5}{R}$ to the sum:

$$\frac{1}{R’} = \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} = \frac{25}{R} \qquad\Rightarrow\qquad R’ = \frac{R}{25}$$

So $\dfrac{R}{R’} = 25$.

Question 2

Which of the following terms does not represent electrical power in a circuit?

Solution. Start from the basic result $P = VI$, which is option (c). Ohm’s law gives $V = IR$; putting this in place of $V$ gives $P = (IR)\times I = I^2R$, which is option (a). Putting $I = V/R$ in place of $I$ instead gives $P = V\times\dfrac{V}{R} = \dfrac{V^2}{R}$, which is option (d). So (a), (c) and (d) are the same power written three ways.

$IR^2$ cannot be reached this way. It equals $(IR)\times R = V\times R$, a volt multiplied by an ohm, which is not a watt.

Question 3

An electric bulb is rated $220\ \text{V}$ and $100\ \text{W}$. When it is operated on $110\ \text{V}$, the power consumed will be —

Solution. The $100\ \text{W}$ is the power only at $220\ \text{V}$. What stays the same when the bulb is put on a different supply is its filament, that is, its resistance (the question treats the resistance as unchanged). So first use the rating to find the resistance, from $P = V^2/R$:

$$R = \frac{V^2}{P} = \frac{(220)^2}{100} = \frac{48400}{100} = 484\ \Omega$$

On $110\ \text{V}$ the same $484\ \Omega$ filament consumes

$$P’ = \frac{V’^2}{R} = \frac{(110)^2}{484} = \frac{12100}{484} = 25\ \text{W}$$

The same result comes quickly from $P = V^2/R$ with $R$ fixed: the power is proportional to $V^2$, so halving the voltage makes the power one-quarter, $100 \div 4 = 25\ \text{W}$.

Question 4

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 —

Solution. The two wires have the same material, length and diameter, so by $R = \rho l/A$ they have the same resistance; call it $R$. In series the combination is $R_s = R + R = 2R$. In parallel, $\dfrac{1}{R_p} = \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{2}{R}$, so $R_p = R/2$.

The quantity that is the same in both cases is the potential difference $V$, not the current, so write the heat in terms of $V$. Putting $I = V/R$ into $H = I^2Rt$ gives $H = \dfrac{V^2}{R}\,t$. For the same $V$ and the same time $t$, the heat is inversely proportional to the resistance:

$$\frac{H_s}{H_p} = \frac{V^2t/R_s}{V^2t/R_p} = \frac{R_p}{R_s} = \frac{R/2}{2R} = \frac{1}{4}$$

Question 5

How is a voltmeter connected in the circuit to measure the potential difference between two points?

Solution. A potential difference is always between two points, so the voltmeter’s two terminals must be joined to those two points, one to each. That places the voltmeter across the component, side by side with it — in parallel. In the figure it measures the potential difference across the resistor $R$ by being connected between $P$ and $Q$. Its positive terminal goes to the point nearer the positive terminal of the cell, here $Q$.

A voltmeter has a very high resistance, so it draws only a tiny current and hardly disturbs the circuit it is measuring. An ammeter is the opposite: it must be in series, because the whole current has to pass through it to be measured.

Question 6

A copper wire has diameter $0.5\ \text{mm}$ and resistivity of $1.6\times10^{-8}\ \Omega\,\text{m}$. 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?

Solution. Length. Rearranging $R = \rho l/A$ gives $l = \dfrac{RA}{\rho}$, so we need the area of cross-section. The area uses the radius, which is half the diameter: $r = 0.25\ \text{mm} = 2.5\times10^{-4}\ \text{m}$.

$$A = \pi r^2 = 3.1416\times(2.5\times10^{-4})^2 = 1.963\times10^{-7}\ \text{m}^2$$

$$l = \frac{RA}{\rho} = \frac{10\times1.963\times10^{-7}}{1.6\times10^{-8}} = 122.7\ \text{m}$$

Doubling the diameter. The area of a circle is $A = \pi r^2 = \pi d^2/4$, so it depends on the square of the diameter. Doubling $d$ multiplies the area by $2^2 = 4$. The length and material are unchanged, and $R$ is inversely proportional to $A$, so the resistance becomes one-quarter:

$$R’ = \frac{10}{4} = 2.5\ \Omega$$

Question 7

The values of current $I$ flowing in a given resistor for the corresponding values of potential difference $V$ across the resistor are given below —

$$\begin{array}{l|ccccc} I\ (\text{amperes}) & 0.5 & 1.0 & 2.0 & 3.0 & 4.0 \\ \hline V\ (\text{volts}) & 1.6 & 3.4 & 6.7 & 10.2 & 13.2 \end{array}$$

Plot a graph between $V$ and $I$ and calculate the resistance of that resistor.

Solution. Plot $V$ on the vertical axis against $I$ on the horizontal axis, as in the chapter’s own $V$–$I$ graph. The five points lie very nearly on a straight line through the origin — which is Ohm’s law, $V \propto I$ — and the small scatter about the line is experimental error. So draw the single straight line through the origin that passes as close as possible to all the points, rather than joining the points one to the next.

The slope of this line is $V/I$, which is the resistance. The best line passes through $(3.0\ \text{A},\ 10.0\ \text{V})$. Reading two points off the line, at $I = 1.0\ \text{A}$ and $I = 4.0\ \text{A}$, and taking the slope triangle between them:

$$R = \frac{\Delta V}{\Delta I} = \frac{13.33 – 3.33}{4.0 – 1.0} = \frac{10.0}{3.0} = 3.33\ \Omega$$

As a check, $V/I$ for the five readings is $3.2$, $3.4$, $3.35$, $3.4$ and $3.3\ \Omega$, whose average is $16.65/5 = 3.33\ \Omega$.

Question 8

When a $12\ \text{V}$ battery is connected across an unknown resistor, there is a current of $2.5\ \text{mA}$ in the circuit. Find the value of the resistance of the resistor.

Solution. Ohm’s law needs the current in amperes. Since $1\ \text{mA} = 10^{-3}\ \text{A}$, the current is $I = 2.5\times10^{-3}\ \text{A}$. Then

$$R = \frac{V}{I} = \frac{12}{2.5\times10^{-3}} = 4800\ \Omega = 4.8\ \text{k}\Omega$$

Question 9

A battery of $9\ \text{V}$ 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?

Solution. In a series circuit there is only one path for the charge, so the same current flows through every resistor. The current in the $12\ \Omega$ resistor is therefore just the current in the circuit. The total resistance in series is the sum:

$$R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4\ \Omega$$

$$I = \frac{V}{R_s} = \frac{9}{13.4} = 0.67\ \text{A}$$

Question 10

How many $176\ \Omega$ resistors (in parallel) are required to carry $5\ \text{A}$ on a $220\ \text{V}$ line?

Solution. First find what resistance the combination must have. To draw $5\ \text{A}$ from a $220\ \text{V}$ line it needs

$$R = \frac{V}{I} = \frac{220}{5} = 44\ \Omega$$

When $n$ equal resistors of $176\ \Omega$ are joined in parallel, each adds $\dfrac{1}{176}$ to the sum, so $\dfrac{1}{R_p} = \dfrac{n}{176}$ and $R_p = \dfrac{176}{n}$. Setting this equal to $44\ \Omega$:

$$\frac{176}{n} = 44 \qquad\Rightarrow\qquad n = \frac{176}{44} = 4$$

Check: each resistor has the full $220\ \text{V}$ across it and carries $220/176 = 1.25\ \text{A}$; four of them carry $4\times1.25 = 5\ \text{A}$.

Question 11

Show how you would connect three resistors, each of resistance $6\ \Omega$, so that the combination has a resistance of (i) $9\ \Omega$, (ii) $4\ \Omega$.

Solution. All three in series give $6 + 6 + 6 = 18\ \Omega$, and all three in parallel give $2\ \Omega$. Both $9\ \Omega$ and $4\ \Omega$ lie between these, so each needs a mixed arrangement. Two useful building blocks: two $6\ \Omega$ resistors in parallel give $\dfrac{1}{R} = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6}$, that is, $3\ \Omega$; and two in series give $12\ \Omega$.

Question 11 (i)

Solution. $9\ \Omega = 6\ \Omega + 3\ \Omega$. Join two of the resistors in parallel, which makes $3\ \Omega$, and connect the third resistor in series with that pair. Series resistances add:

$$R = 6 + 3 = 9\ \Omega$$

Question 11 (ii)

Solution. Join two of the resistors in series, making $12\ \Omega$, and connect the third resistor in parallel with that pair:

$$\frac{1}{R} = \frac{1}{12} + \frac{1}{6} = \frac{1 + 2}{12} = \frac{3}{12} \qquad\Rightarrow\qquad R = 4\ \Omega$$

Question 12

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 $5\ \text{A}$?

Solution. Bulbs connected in parallel across the line each get the full $220\ \text{V}$, so each works at its rated $10\ \text{W}$ and draws

$$I_1 = \frac{P}{V} = \frac{10}{220} = \frac{1}{22}\ \text{A}$$

In parallel the currents of the branches add up to the current in the line. With $n$ bulbs the line current is $n/22$ amperes, and this may be at most $5\ \text{A}$:

$$\frac{n}{22} = 5 \qquad\Rightarrow\qquad n = 110$$

The same answer comes from power. The line can supply at most $P = VI = 220\times5 = 1100\ \text{W}$, and each bulb takes $10\ \text{W}$, so $1100 \div 10 = 110$ bulbs.

Question 13

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?

Solution. In each case, find the resistance connected across the line, then use $I = V/R$ with $V = 220\ \text{V}$.

One coil used separately. $R = 24\ \Omega$:

$$I = \frac{220}{24} = 9.17\ \text{A} \approx 9.2\ \text{A}$$

The two coils in series. $R_s = 24 + 24 = 48\ \Omega$:

$$I = \frac{220}{48} = 4.58\ \text{A} \approx 4.6\ \text{A}$$

The two coils in parallel. $\dfrac{1}{R_p} = \dfrac{1}{24} + \dfrac{1}{24} = \dfrac{2}{24}$, so $R_p = 12\ \Omega$:

$$I = \frac{220}{12} = 18.33\ \text{A} \approx 18.3\ \text{A}$$

The parallel current is exactly twice the single-coil current, as it must be: in parallel each coil has the full $220\ \text{V}$ across it and carries its own $9.17\ \text{A}$, and the line supplies both.

Question 14

Compare the power used in the $2\ \Omega$ resistor in each of the following circuits: (i) a $6\ \text{V}$ battery in series with $1\ \Omega$ and $2\ \Omega$ resistors, and (ii) a $4\ \text{V}$ battery in parallel with $12\ \Omega$ and $2\ \Omega$ resistors.

Question 14 (i)

Solution. In series the resistances add, $R_s = 1 + 2 = 3\ \Omega$, so the current is

$$I = \frac{V}{R_s} = \frac{6}{3} = 2\ \text{A}$$

The same $2\ \text{A}$ flows through the $2\ \Omega$ resistor, since the current is common in series. So use the form of the power formula that contains the current:

$$P = I^2R = (2)^2\times2 = 8\ \text{W}$$

Question 14 (ii)

Solution. “In parallel with $12\ \Omega$ and $2\ \Omega$ resistors” means the two resistors are joined in parallel with each other directly across the battery. Each of them is then connected between the same two terminals, so each has the full $4\ \text{V}$ across it; the $12\ \Omega$ resistor makes no difference to what happens in the $2\ \Omega$ one. With the voltage common, use the form of the power formula that contains $V$:

$$P = \frac{V^2}{R} = \frac{(4)^2}{2} = \frac{16}{2} = 8\ \text{W}$$

(The current through it is $4/2 = 2\ \text{A}$, and $I^2R = 4\times2 = 8\ \text{W}$ agrees.)

Question 15

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 $220\ \text{V}$?

Solution. In parallel across a $220\ \text{V}$ supply, each lamp has its rated voltage across it, so each works at its rated power. The current in each follows from $P = VI$:

$$I_1 = \frac{100}{220} = 0.455\ \text{A}, \qquad I_2 = \frac{60}{220} = 0.273\ \text{A}$$

The line supplies both branches, so the currents add:

$$I = I_1 + I_2 = \frac{100 + 60}{220} = \frac{160}{220} = 0.73\ \text{A}$$

Question 16

Which uses more energy, a $250\ \text{W}$ TV set in $1\ \text{hr}$, or a $1200\ \text{W}$ toaster in $10\ \text{minutes}$?

Solution. Power is the rate of using energy, so the energy used is power multiplied by time. A larger power does not settle the question on its own; the time matters just as much. Put both times in seconds.

TV set: $1\ \text{h} = 3600\ \text{s}$, so

$$E_{\text{TV}} = 250\times3600 = 900\,000\ \text{J} = 9.0\times10^5\ \text{J}$$

Toaster: $10\ \text{min} = 600\ \text{s}$, so

$$E_{\text{toaster}} = 1200\times600 = 720\,000\ \text{J} = 7.2\times10^5\ \text{J}$$

In commercial units these are $0.25\ \text{kW h}$ and $1.2\times\tfrac16 = 0.2\ \text{kW h}$. The toaster has almost five times the power, but it runs for only one-sixth of the time.

Question 17

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.

Solution. The rate at which heat is developed is heat per second, which is the power. With the resistance and the current given, use $P = I^2R$:

$$P = I^2R = (5)^2\times44 = 25\times44 = 1100\ \text{W}$$

So $1100\ \text{J}$ of heat is produced every second. The $2$ hours do not affect the rate; they would be needed only for the total heat, $H = Pt = 1100\times7200 = 7.92\times10^6\ \text{J}$. As a check, the voltage across the heater is $V = IR = 5\times44 = 220\ \text{V}$, the mains voltage, and $VI = 220\times5 = 1100\ \text{W}$ agrees.

Question 18

Explain the following. (a) Why is the tungsten used almost exclusively for filament of electric lamps? (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? (c) Why is the series arrangement not used for domestic circuits? (d) How does the resistance of a wire vary with its area of cross-section? (e) Why are copper and aluminium wires usually employed for electricity transmission?

Question 18 (a)

Why is the tungsten used almost exclusively for filament of electric lamps?

Solution. A filament lamp gives light only because its filament is heated by the current until it glows white-hot. The filament must therefore reach a very high temperature without melting. Tungsten is a strong metal with a very high melting point, $3380\ ^\circ\text{C}$, so it can be kept hot enough to glow brightly and still not melt. Its resistivity is also a few times that of copper, which helps it get hot. The bulb is filled with chemically inactive nitrogen and argon so that the hot filament does not burn away.

Question 18 (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?

Solution. A heating element is meant to turn electrical energy into heat, and the heat produced is $H = I^2Rt$, so for a given current it needs a high resistance. The resistivity of an alloy is generally much higher than that of its constituent metals — nichrome’s is $100\times10^{-6}\ \Omega\,\text{m}$, against $1.62\times10^{-8}\ \Omega\,\text{m}$ for copper — so a short, reasonably thick element made of an alloy has the resistance it needs. The element also has to stay red-hot for long periods, and alloys do not oxidise (burn) readily at high temperatures, whereas a pure metal would.

Question 18 (c)

Why is the series arrangement not used for domestic circuits?

Solution. In a series circuit there is only one path, so the same current passes through every appliance and the supply voltage is shared among them. This causes three problems in a house. If one appliance is switched off or fails, the circuit is broken and every other appliance stops too, so no appliance can be switched on or off by itself. The $220\ \text{V}$ is divided among the appliances, so none gets the voltage it was designed for. And the resistances add, so the current is the same small value for all of them, although a heater needs a far larger current than a bulb. In a parallel arrangement each appliance has the full $220\ \text{V}$ across it, draws the current it needs, and has its own switch.

Question 18 (d)

How does the resistance of a wire vary with its area of cross-section?

Solution. From $R = \rho\dfrac{l}{A}$, for a wire of a given material and length the resistance is inversely proportional to the area of cross-section, $R \propto \dfrac{1}{A}$. A thicker wire offers the charge a wider path, so it has less resistance: doubling the area halves the resistance. Since $A = \pi d^2/4$, doubling the diameter makes the resistance one-quarter, as in Question 6.

Question 18 (e)

Why are copper and aluminium wires usually employed for electricity transmission?

Solution. Transmission lines are very long, and any resistance in them wastes energy as heat, $H = I^2Rt$, before the electricity reaches the consumer. Copper ($1.62\times10^{-8}\ \Omega\,\text{m}$) and aluminium ($2.63\times10^{-8}\ \Omega\,\text{m}$) have very low resistivity, so wires made of them have low resistance and waste little energy. Silver has a slightly lower resistivity still ($1.60\times10^{-8}\ \Omega\,\text{m}$) but is far too costly; aluminium is also light, which suits long overhead lines.

Common mistakes

• Question 1: forgetting that cutting the wire changes each piece’s resistance. Students join five resistances of $R$ in parallel, get $R/5$ and choose option (c). Each piece is one-fifth as long, so it has resistance $R/5$, and five of those in parallel give $R/25$.
• Question 3: assuming power is proportional to voltage. Halving the voltage feels as though it should halve the power, giving $50\ \text{W}$. But halving the voltage also halves the current, so $P = V^2/R$ falls to one-quarter, $25\ \text{W}$. The fixed quantity for a given bulb is its resistance, not its power.
• Question 4: choosing 4 : 1 because “more resistance means more heat”. That is true only when the current is the same, as in $H = I^2Rt$. Here the potential difference is the same, so $H = V^2t/R$, and the combination with less resistance (parallel) produces more heat.
• Question 6: using the diameter as the radius. Taking $r = 0.5\ \text{mm}$ makes the area four times too large and gives $l = 490.9\ \text{m}$. The radius is half the diameter, $0.25\ \text{mm}$, and it must be in metres before squaring. For the second part, doubling the diameter quarters the resistance; it does not halve it, because area depends on the square of the diameter.
• Question 8: leaving the current in milliamperes. $12/2.5$ gives $4.8\ \Omega$, a thousand times too small. The current must first be written as $2.5\times10^{-3}\ \text{A}$.
• Question 14(ii): using the total current for the $2\ \Omega$ resistor. Students find the equivalent resistance ($12\times2/14 = 1.71\ \Omega$) and the total current ($2.33\ \text{A}$), then compute $I^2R = 10.9\ \text{W}$. The total current is shared between the branches; the $2\ \Omega$ resistor carries only its own $2\ \text{A}$. The quick route is that it has the full $4\ \text{V}$ across it, so $P = V^2/R = 8\ \text{W}$.
• Question 16: comparing powers instead of energies, or mixing units. The toaster has the larger power, but energy is power multiplied by time. Writing $1200\times10$ against $250\times1$ also goes wrong, because minutes and hours have been mixed; convert both times to seconds (or both to hours) first.
• Question 17: giving the total heat when the rate is asked. Multiplying by the $2$ hours gives $7.92\times10^6\ \text{J}$, which answers a different question. “Rate at which heat is developed” means heat per second, the power, $1100\ \text{W}$.

Practise next

• The in-text questions of this chapter, twenty-three across seven boxes — the most of any chapter in the book, and between them a complete rehearsal for the questions above, from the definitions through Ohm’s law and combinations of resistors to heating and power.
• Chapter 12, Magnetic Effects of Electric Current — the same currents, now producing magnetic fields and forces, and how domestic circuits are protected with fuses and earthing.
• Chapter 10, The Human Eye and the Colourful World — the chapter before this one, for revising lenses, their power in dioptres and the defects of vision.