NCERT Exemplar Solutions Class 10 Science Electricity — Multiple Choice Questions Q1–18

NCERT Exemplar Problems, Class 10 Science — Electricity, Multiple Choice Questions. All 18 questions solved.

Electricity is the most calculation-heavy unit in the Class 10 exemplar, and nearly every question on this page reduces to one of five relations:

$$V = IR, \qquad R = \frac{\rho l}{A}, \qquad P = VI = I^2R = \frac{V^2}{R}, \qquad Q = It, \qquad H = I^2Rt$$

plus the two combination rules — series adds resistances, $R_s = R_1 + R_2 + \cdots$, and parallel adds their reciprocals, $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots$.

Two consequences of those rules are worth having ready, because several questions turn on them directly. A series combination is always bigger than the biggest resistor in it; a parallel combination is always smaller than the smallest. That single sentence answers Questions 6 and 7 outright.

And the choice of power formula matters. Use $P = V^2/R$ when the voltage is fixed — as it is for anything connected across the mains or across a given battery — and $P = I^2R$ when the current is fixed, as it is for components in series.

Key insight. When several circuits share the same battery, the one with the least resistance draws the most current and dissipates the most heat. Comparing circuits is therefore never about counting resistors; it is about working out the single equivalent resistance of each. That is exactly what Question 2 asks — and it is the question the printed key gets wrong.

Choose the correct answer from the given four options in the following questions:

Question 1

A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure 12.1. The current recorded in the ammeter will be

The three panels show the same four components — one cell, one resistor R, one key and one ammeter — connected in a single series loop. They differ only in the order in which the components are placed round the loop and in which way round the ammeter’s terminals are drawn.

Solution. In a series circuit there is only one path for the charge, so the same current flows through every component, and it is the same wherever in the loop you choose to measure it.

The current is fixed by the whole loop:

$$I = \frac{V}{R}$$

with $V$ the cell’s voltage and $R$ the total resistance. Neither of those changes when the components are rearranged — the same cell and the same resistor are in the loop in all three cases. So the ammeter reads the same value in all three.

The position of the ammeter is irrelevant for the same reason: an ammeter is always placed in series and measures the current through the branch it sits in, and here there is only one branch.

The order of components in a series loop never matters. What would matter is a change of connection type — putting two resistors in parallel instead of in series, for instance — and that is what Question 2 goes on to test.

Question 2

In the following circuits (Figure 12.2), heat produced in the resistor or combination of resistors connected to a 12 V battery will be

Solution. All three circuits are across the same 12 V battery, so the voltage is fixed and the right formula is

$$P = \frac{V^2}{R}$$

Everything therefore depends on the equivalent resistance of each combination, and the smaller that is, the more heat is produced.

(i) A single 2 Ω resistor. $R = 2\ \Omega$, so

$$P = \frac{12^2}{2} = \frac{144}{2} = 72\ \mathrm{W}$$

(ii) Two 2 Ω resistors in series: $R = 2 + 2 = 4\ \Omega$, so

$$P = \frac{144}{4} = 36\ \mathrm{W}$$

(iii) Two 2 Ω resistors in parallel: $\dfrac{1}{R} = \dfrac{1}{2} + \dfrac{1}{2} = 1$, so $R = 1\ \Omega$ and

$$P = \frac{144}{1} = 144\ \mathrm{W}$$

The three powers are $72$ W, $36$ W and $144$ W — a factor of four between the extremes. The maximum is case (iii), which is option (d).

A note on the printed answer. The key gives (a), “same in all the cases”, and that cannot be right. Three different resistances across one and the same voltage must draw three different currents and dissipate three different powers; option (a) would require $2\ \Omega$, $4\ \Omega$ and $1\ \Omega$ to behave identically. The error was checked carefully before being stated — the key page was read as an image as well as from its text layer, and the third circuit was magnified to confirm it really is two resistors in parallel and not a larger network. It is recorded in our issues log as the first outright wrong option we have found in this book’s physics units.

The reasoning to carry away is the general one: parallel combination lowers the resistance, so it raises the power when the voltage is fixed. That is why adding appliances to a household circuit — all wired in parallel — increases the current drawn and can overload the fuse.

Question 3

Electrical resistivity of a given metallic wire depends upon

Solution. The distinction being tested is between resistance and resistivity, and it is the most important one in the chapter.

Resistance $R$ does depend on the dimensions:

$$R = \frac{\rho l}{A}$$

so a longer wire has more resistance and a thicker one has less.

Resistivity $\rho$ is the constant of proportionality in that equation — it is a property of the material itself, not of the particular piece of it. Rearranging, $\rho = RA/l$, and the dimensions have been divided out. Copper has the same resistivity whether it is a thin short wire or a thick long one.

So resistivity depends on the nature of the material, option (d) — and, as Question 13 adds, on the temperature, since heating a metal increases the scattering of its electrons.

The physical meaning is worth stating: resistivity is numerically the resistance of a wire of the material one metre long and one square metre in cross-section. Its unit is the ohm metre ($\Omega\ \mathrm{m}$), which follows from $\rho = RA/l$.

Question 4

A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly

Solution. Two steps: find the charge, then count the electrons in it.

The charge that flows follows from the definition of current, $I = Q/t$:

$$Q = It = 1 \times 16 = 16\ \mathrm{C}$$

The number of electrons is that charge divided by the charge on one electron, $e = 1.6 \times 10^{-19}\ \mathrm{C}$:

$$n = \frac{Q}{e} = \frac{16}{1.6 \times 10^{-19}} = \frac{16}{1.6} \times 10^{19} = 10 \times 10^{19} = 10^{20}$$

which is option (a).

The arithmetic is designed to come out exactly, and that is the clue that the question wants this route: $16 / 1.6 = 10$, and $10 \times 10^{19} = 10^{20}$.

It is worth pausing on the size of the answer. A hundred billion billion electrons pass through the filament in sixteen seconds, for a current most people would call small. That is why charge appears continuous in everyday circuits even though it is carried in discrete units.

Question 5

Identify the circuit (Figure 12.3) in which the electrical components have been properly connected.

Solution. Three rules have to hold at once, and each wrong panel breaks exactly one of them.

1. An ammeter goes in series with the component whose current is to be measured. It has a very low resistance, so it must never be placed across anything.
2. A voltmeter goes in parallel, across the component whose potential difference is to be measured. It has a very high resistance, so it must never be placed in the main loop.
3. Both meters must be connected with the correct polarity — current entering at the $+$ terminal — or the needle deflects backwards.

(i) The voltmeter is in the main loop, in series. Because a voltmeter has an enormous resistance, that would nearly stop the current altogether and the ammeter would read almost zero. Rule 2 broken.

(iii) The ammeter is connected across R, in parallel. An ammeter has almost no resistance, so it would short out the resistor, and a very large current would flow through the meter and probably damage it. Rule 1 broken.

(iv) The meters are in the right places — voltmeter across R, ammeter in the main loop — but the ammeter’s terminals are reversed, with $+$ where $-$ should be. Rule 3 broken.

(ii) Voltmeter across R, ammeter in series in the main loop, and both with the correct polarity. All three rules hold, so (ii) is correct.

The way to remember which meter goes where is to think about the resistance each one has: the ammeter must not disturb the current, so it is made low-resistance and put in the path; the voltmeter must not divert the current, so it is made high-resistance and put across. This is exactly the reasoning Question 21 in the short-answer band asks for.

Question 6

What is the maximum resistance which can be made using five resistors each of 1/5 Ω?

Solution. The maximum resistance from a set of resistors is obtained by connecting them all in series, because series resistances simply add and the total is larger than any individual one.

$$R_s = \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5} = 5 \times \frac{1}{5} = 1\ \Omega$$

which is option (d).

Option (c), 5 Ω, is the answer you get by multiplying by five twice — or by forgetting that each resistor is a fifth of an ohm rather than five ohms. Option (b), 10 Ω, has no route at all.

The general result is worth stating once, because it also settles Question 7: for $n$ identical resistors of value $r$,

$$R_{\max} = nr \quad (\text{all in series}), \qquad R_{\min} = \frac{r}{n} \quad (\text{all in parallel})$$

and every other arrangement lies between them. With $n = 5$ and $r = 1/5$, those limits are $1\ \Omega$ and $1/25\ \Omega$.

Question 7

What is the minimum resistance which can be made using five resistors each of 1/5 Ω?

Solution. The minimum resistance is obtained by connecting them all in parallel, because a parallel combination is always smaller than the smallest resistor in it — each extra branch gives the current another route.

$$\frac{1}{R_p} = 5 \times \frac{1}{1/5} = 5 \times 5 = 25 \qquad\Rightarrow\qquad R_p = \frac{1}{25}\ \Omega$$

which is option (b).

Or use the shortcut for $n$ identical resistors in parallel, $R_p = r/n$:

$$R_p = \frac{1/5}{5} = \frac{1}{25}\ \Omega$$

Option (d), 25 Ω, is the reciprocal of the answer and is what you get by forgetting the final inversion — the commonest slip in every parallel calculation. A quick sanity check catches it: the answer must be smaller than $1/5\ \Omega$, and 25 Ω plainly is not.

Together with Question 6, the five resistors can be arranged to give anything from $1/25\ \Omega$ to $1\ \Omega$ — a range of 25 to 1.

Question 8

The proper representation of series combination of cells (Figure 12.4) obtaining maximum potential is

Solution. Cells in series add their voltages only if they are all connected the same way round — the positive terminal of each joined to the negative terminal of the next, so that all of them drive the current in the same direction.

In the cell symbol, the long thin line is the positive terminal and the short thick line the negative one. So all the cells face the same way when every long line points the same way along the row.

(i) shows all five cells facing the same way, so their voltages add: total $= 5 \times$ the voltage of one cell. That is the maximum obtainable, and it is option (a).

The other three rows have one or more cells reversed. A reversed cell opposes the others, and its voltage is subtracted rather than added. With five identical cells of voltage $E$, reversing one gives $4E – E = 3E$; reversing two gives $E$; and so on. Any arrangement that is not all-the-same-way gives less than $5E$.

There is a practical consequence worth knowing: a reversed cell in a torch or a remote control is not merely wasteful but harmful, because the other cells drive current backwards through it, which heats it and can make it leak.

Question 9

Which of the following represents voltage?

Solution. Start from the definition. Potential difference is the work done per unit charge in moving that charge between two points:

$$V = \frac{W}{Q}$$

Now express the charge in terms of quantities the options use. From $I = Q/t$,

$$Q = I \times t$$

Substituting,

$$V = \frac{W}{I \times t} = \frac{\text{Work done}}{\text{Current} \times \text{Time}}$$

which is option (a).

The other three can be eliminated on units alone, without any physics. Voltage is joules per coulomb, so work must be divided by charge, never multiplied — that rules out (b) and (d) immediately. Option (c) multiplies by time instead of dividing, giving units of $\mathrm{J\,s/A} = \mathrm{V\,s^2}$, which is not a voltage.

Checking the units of the answer confirms it: $\mathrm{J}/(\mathrm{A} \cdot \mathrm{s}) = \mathrm{J}/\mathrm{C} = \mathrm{V}$. ✓

Question 10

A cylindrical conductor of length $l$ and uniform area of cross-section $A$ has resistance $R$. Another conductor of length $2l$ and resistance $R$ of the same material has area of cross section

Solution. Both conductors are of the same material, so both have the same resistivity $\rho$, and both are stated to have the same resistance $R$.

For the first,

$$R = \frac{\rho l}{A}$$

For the second, of length $2l$ and unknown area $A’$,

$$R = \frac{\rho (2l)}{A’}$$

Setting the two equal — which is allowed because the resistances are equal — and cancelling $\rho$:

$$\frac{l}{A} = \frac{2l}{A’} \qquad\Rightarrow\qquad A’ = 2A$$

which is option (c).

The reasoning behind the algebra is worth holding: doubling the length doubles the resistance, so to bring it back down to $R$ the area must also be doubled, since resistance is inversely proportional to area. The two changes cancel.

Option (a), $A/2$, is the answer you get by treating area the same way as length — halving instead of doubling — and it would in fact give a resistance of $4R$.

Question 11

A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances $R_1$, $R_2$ and $R_3$ respectively (Figure 12.5). Which of the following is true?

Solution. The first thing to check is which quantity is on which axis, because it decides whether the slope is $R$ or $1/R$.

Here $I$ is on the vertical axis and $V$ on the horizontal, so

$$\text{slope} = \frac{I}{V} = \frac{1}{R}$$

A steeper line therefore means a larger $1/R$ and so a smaller resistance.

Reading the graph, $R_1$ is the steepest line, $R_2$ the next and $R_3$ the shallowest. So

$$\frac{1}{R_1} > \frac{1}{R_2} > \frac{1}{R_3} \qquad\Rightarrow\qquad R_1 < R_2 < R_3$$

that is, $R_3 > R_2 > R_1$ — option (c).

Option (b) is the exact reverse, and it is what you get by assuming the slope is the resistance — which it would be if $V$ were on the vertical axis. This is the whole point of the question, and the reason to read the axis labels before reading the lines.

That all three graphs are straight lines through the origin is itself worth noting: it is the experimental statement of Ohm’s law, $V \propto I$, which Question 30 in the long-answer band asks you to verify.

Question 12

If the current $I$ through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

Solution. An increase of 100% means the current is doubled: $I’ = 2I$.

The resistor is fixed and the temperature unchanged, so $R$ is constant, and the right power formula is the one in terms of current:

$$P = I^2 R$$

With the new current,

$$P’ = (2I)^2 R = 4I^2R = 4P$$

The power becomes four times its original value. But the question asks for the increase, not the new value:

$$\text{increase} = P’ – P = 4P – P = 3P$$

$$\text{percentage increase} = \frac{3P}{P} \times 100 = 300\%$$

which is option (c).

Option (d), 400%, is the trap, and it is chosen by nearly everyone who works out $P’ = 4P$ correctly and then stops. Four times the original is a 300% increase, because the original 100% was already there. The words increase in are doing the work.

The same distinction applies whenever a quantity is scaled: doubling something is a 100% increase, tripling it a 200% increase, quadrupling a 300% increase.

Question 13

The resistivity does not change if

Solution. Resistivity depends on exactly two things: the material and its temperature. It does not depend on the size or shape of the specimen.

Testing the options against that:

(a) Changing the material changes $\rho$ — copper, nichrome and rubber differ by many orders of magnitude. So it does change.

(b) Changing the temperature changes $\rho$. In a metal, heating makes the atoms vibrate more, the electrons are scattered more often, and the resistivity rises. So it does change.

(d) Changing both certainly changes it.

(c) Changing the shape — drawing the same piece of metal into a longer, thinner wire, say — changes its resistance but not its resistivity. The same atoms are present in the same arrangement, so the material’s intrinsic property is untouched. This is the answer.

The pairing with Question 3 makes the distinction complete: resistance depends on material, temperature and dimensions; resistivity depends on material and temperature only. Everything about size and shape is carried by the $l/A$ factor in $R = \rho l / A$.

Question 14

In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?

Solution. The bulbs are in parallel, so each has the same potential difference across it — the full supply voltage.

With $V$ the same for all three, the power actually dissipated is

$$P = \frac{V^2}{R}$$

and a bulb’s wattage rating tells you what it consumes at that voltage. So at the same voltage, the higher-rated bulb consumes more power and glows more brightly. Brightness follows the rating directly:

$$C\ (100\ \mathrm{W}) > B\ (60\ \mathrm{W}) > A\ (40\ \mathrm{W})$$

Now the options. (a) is false — the ratings differ. (b) is false — A is the dimmest, not the brightest. (d) is false — C is brighter than B, not less. (c) says B is brighter than A, which is true, so it is the answer.

Only one option had to be true, and (c) is a partial statement of the full ordering. It is worth noticing that the question could have offered “brightness of C will be maximum” and it would have been equally correct; the exemplar has picked a less obvious true statement.

The contrast with a series connection is the thing to carry. In series the current is the same, so $P = I^2R$ applies and the bulb with the highest resistance — the lowest wattage rating — glows brightest. Connecting a 40 W and a 100 W bulb in series makes the 40 W one brighter, which is the opposite of the parallel result.

Question 15

In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be

Solution. The resistors are in series, so first find the current, which is the same through both.

$$R_s = 2 + 4 = 6\ \Omega, \qquad I = \frac{V}{R_s} = \frac{6}{6} = 1\ \mathrm{A}$$

Now the heat produced in the 4 Ω resistor alone, by Joule’s law:

$$H = I^2Rt = (1)^2 \times 4 \times 5 = 20\ \mathrm{J}$$

which is option (c).

The formula to use here is $I^2Rt$ and not $V^2t/R$, because the quantity common to both resistors is the current, not the voltage. Using the 6 V of the battery with the 4 Ω resistor would give $6^2 \times 5 / 4 = 45$ J, which is wrong — the 4 Ω resistor does not have 6 V across it.

If you prefer to work with voltage, find the voltage across that resistor: $V_4 = IR = 1 \times 4 = 4\ \mathrm{V}$, and then $H = V_4^2 t / R = 16 \times 5/4 = 20\ \mathrm{J}$. ✓ Same answer.

A useful check: the total heat in 5 s is $I^2 R_s t = 1 \times 6 \times 5 = 30$ J, and it should divide between the resistors in proportion to their resistances — $10$ J in the 2 Ω and $20$ J in the 4 Ω. That is option (d) for the total and option (c) for the part asked about.

Question 16

An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

Solution. Find the current the kettle actually draws, from $P = VI$:

$$I = \frac{P}{V} = \frac{1000}{220} = 4.55\ \mathrm{A}$$

A fuse must carry the working current without melting, so its rating has to be greater than 4.55 A. Of the values offered, 1 A, 2 A and 4 A would all blow as soon as the kettle was switched on. The smallest that works is 5 A, option (d).

The reasoning has two halves and both matter. The rating must be above the working current, or the fuse is useless — it would interrupt normal operation. But it should not be far above, or the fuse will fail to blow during a genuine fault and the wiring will overheat instead. Choosing the next standard rating above the working current is the rule, and 5 A is the standard value just above 4.55 A.

Option (c), 4 A, is the trap for anyone who rounds $4.55$ down. Fuse ratings are never rounded down.

The fuse itself is an application of Joule heating: a thin wire of low melting point in series with the appliance, which melts and breaks the circuit when the current exceeds its rating — which is what Question 23 asks.

Question 17

Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have

Solution. Two rules settle all four options:

In series, there is one path, so the current is the same through every component — but the voltages differ, dividing in proportion to the resistances.

In parallel, every branch is joined to the same two points, so the voltage is the same across each — but the currents differ, dividing in inverse proportion to the resistances.

Now:

• (a) Same current in parallel — false. The branches carry $V/2$ and $V/4$, so the 2 Ω resistor carries twice as much.
• (b) Same current in series — true. This is the answer.
• (c) Same potential difference in series — false. With 2 Ω and 4 Ω the voltages divide as $1:2$.
• (d) Different potential difference in parallel — false. Parallel branches share the same two nodes, so the voltage is necessarily identical.

Notice that the four options pair up: (a) and (b) are the same claim about current applied to the two arrangements, and (c) and (d) the same about voltage. Exactly one of each pair can be true, and knowing which of series and parallel preserves which quantity answers both.

Question 18

Unit of electric power may also be expressed as

Solution. Power is energy per unit time, and its SI unit is the watt. The question asks which of these combinations is also a unit of power rather than of energy.

(a) Volt ampere. From $P = VI$, multiplying volts by amperes gives watts directly:

$$1\ \mathrm{V} \times 1\ \mathrm{A} = 1\ \frac{\mathrm{J}}{\mathrm{C}} \times \frac{\mathrm{C}}{\mathrm{s}} = 1\ \frac{\mathrm{J}}{\mathrm{s}} = 1\ \mathrm{W}$$

So volt ampere is a unit of power. ✓

(b) Kilowatt hour is power multiplied by time, so it is a unit of energy — the commercial unit, one unit on the electricity bill, equal to $3.6 \times 10^6$ J. That is Question 25.

(c) Watt second is also power multiplied by time, so again energy — and it equals exactly one joule.

(d) Joule second is energy multiplied by time, which is neither energy nor power.

The general test is worth keeping: anything of the form power × time is an energy, and anything of the form energy ÷ time is a power. Only (a) is the latter.

Common mistakes

• Trusting the printed key on Question 2. It prints (a), “same in all the cases”, but the three circuits present 2 Ω, 4 Ω and 1 Ω to the same 12 V and dissipate 72 W, 36 W and 144 W. The correct answer is (d).
• Choosing the wrong power formula. Use $P = V^2/R$ when the voltage is shared (parallel branches, mains appliances) and $P = I^2R$ when the current is shared (series). Questions 2, 14 and 15 each depend on picking the right one.
• Reading a V–I graph without checking the axes, in Question 11. With $I$ vertical the slope is $1/R$, so the steepest line is the smallest resistance — the opposite of the usual assumption.
• Answering 400% in Question 12. Doubling the current makes the power four times as large, which is an increase of 300%. The word “increase” is the whole question.
• Forgetting the final inversion in a parallel calculation, in Question 7. $1/R_p = 25$ means $R_p = 1/25\ \Omega$, not 25 Ω. The answer must always be smaller than the smallest branch.
• Rounding a fuse rating down, in Question 16. The kettle draws 4.55 A, so a 4 A fuse blows immediately. Always take the next standard rating above the working current.
• Confusing resistance with resistivity, in Questions 3, 10 and 13. Resistance depends on material, temperature and dimensions; resistivity only on material and temperature.
• Putting the voltmeter in series or the ammeter in parallel, in Question 5. A voltmeter has a very high resistance and goes across; an ammeter has a very low one and goes in the path. Swapping them either stops the current or shorts out the resistor.

Practise next

• Exemplar Short and Long Answer Questions — where the same circuit rules have to be applied to multi-branch networks and the ammeter readings worked out.
• NCERT textbook Chapter 11 — Electricity for the end-of-chapter numericals on Ohm’s law, combinations of resistors and electric power.
• Textbook activities for the V–I experiment of Question 11 and the series and parallel comparisons of Questions 14 and 17.
• Exemplar Magnetic Effects of Electric Current — the next unit, which builds directly on the current and circuit ideas used here.