NCERT Exemplar Solutions Class 10 Science Electricity — Short and Long Answer Questions Q19–35

NCERT Exemplar Problems, Class 10 Science — Electricity, Short Answer Questions 19 to 28 and Long Answer Questions 29 to 35. All 17 questions solved.

This band is where the circuit rules have to be used rather than recited, and nearly every numerical question follows the same three steps: reduce the network to a single equivalent resistance, find the main current, then work back outwards to the individual branches.

The two rules that do the reducing:

$$\text{series: } R_s = R_1 + R_2 + \cdots \qquad\qquad \text{parallel: } \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$$

and the two facts that let you work back outwards:

• In series the current is common; the voltages divide in proportion to the resistances.
• In parallel the voltage is common; the currents divide in inverse proportion.

Key insight. When a question asks whether something changes, the quickest route is usually to ask what has stayed fixed. In Question 26, adding a resistor in parallel across a branch does not alter the voltage across that branch — so nothing inside it changes at all. Spotting the fixed quantity saves recalculating the whole circuit.

Short Answer Questions

Question 19

A child has drawn the electric circuit to study Ohm’s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

Solution. The child’s diagram has two faults, and both concern where the meters are placed.

The voltmeter is in the main loop, in series with the resistor. A voltmeter has a very high resistance, so in series it would almost stop the current. It must be connected in parallel, across the resistor whose potential difference is being measured.

The ammeter is connected across the resistor, in parallel. An ammeter has almost no resistance, so across the resistor it would short it out and carry a dangerously large current. It must be in series in the main loop.

A third point the marking scheme looks for is polarity: current must enter each meter at its positive terminal, so the $+$ of each meter faces the $+$ of the battery.

The corrected circuit:

Now the ammeter reads the current through the resistor and the voltmeter reads the potential difference across it, which is exactly the pair of readings Ohm’s law relates. Varying the number of cells and plotting $V$ against $I$ then gives the straight line of Question 30.

Question 20

Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors?

Solution. Resistor A carries the whole current, while B and C share it between them — and because B and C are equal, each carries exactly half.

The limit is set by whichever resistor reaches 18 W first, so test both.

Resistor A, carrying the full current $I$:

$$P = I^2 R \quad\Rightarrow\quad 18 = I^2 \times 2 \quad\Rightarrow\quad I^2 = 9 \quad\Rightarrow\quad I = 3\ \mathrm{A}$$

Resistor B, carrying $I/2$:

$$18 = \left(\frac{I}{2}\right)^2 \times 2 \quad\Rightarrow\quad \frac{I^2}{4} = 9 \quad\Rightarrow\quad I = 6\ \mathrm{A}$$

So B and C would tolerate a main current of 6 A, but A would have melted at 3 A. The binding constraint is the smaller of the two, and the maximum safe current is

$$I_{\max} = 3\ \mathrm{A} \ \text{through A}, \qquad \frac{3}{2} = 1.5\ \mathrm{A} \ \text{through each of B and C}$$

Checking the powers at that current: A dissipates $3^2 \times 2 = 18$ W, exactly at its limit; B and C dissipate $1.5^2 \times 2 = 4.5$ W each, comfortably inside theirs.

The general principle is worth stating: in a network with a rating limit, you must test every component and take the most restrictive answer. Testing only the parallel pair here would give 6 A and destroy resistor A.

Question 21

Should the resistance of an ammeter be low or high? Give reason.

Solution. Low — ideally zero.

An ammeter is connected in series in the branch whose current it measures, so whatever resistance it has is added to the circuit. If that resistance were substantial, the total resistance would rise and the current would fall — so the meter would change the very quantity it was inserted to measure, and its reading would not be the true current.

To keep that disturbance negligible, the ammeter’s resistance must be as close to zero as possible, and certainly very small compared with the rest of the circuit.

The voltmeter is the exact mirror image, and the two are worth holding together:

Meter	Connected	Resistance must be	So that it does not
Ammeter	in series	as low as possible	reduce the current it measures
Voltmeter	in parallel	as high as possible	divert current from the component

This is also the reason the two must never be interchanged, which is the content of Question 5 and Question 19: an ammeter placed across a resistor short-circuits it, and a voltmeter placed in series nearly stops the current.

Question 22

Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4 Ω resistors? Give reason.

Solution. The circuit is a cell and key in series with an ammeter, then the 2 Ω resistor, then the two 4 Ω resistors in parallel, with the voltmeter connected across the parallel pair.

Will the two potential differences be equal? Yes — and the reason is a coincidence of values rather than a rule.

The parallel combination has resistance

$$\frac{1}{R_p} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \qquad\Rightarrow\qquad R_p = 2\ \Omega$$

which happens to be exactly equal to the 2 Ω resistor in series with it.

The two are in series, so the same current $I$ flows through both, and

$$V_{2\Omega} = I \times 2, \qquad V_{\text{parallel}} = I \times 2$$

Equal resistances carrying equal currents drop equal voltages, so the two potential differences are the same.

The point worth drawing out is that this is not a general fact about series components — in series, voltages divide in proportion to the resistances, so they are equal only when the resistances are equal. Had the parallel pair been two 8 Ω resistors, its resistance would have been 4 Ω and it would have taken twice the voltage of the 2 Ω resistor.

Question 23

How does use of a fuse wire protect electrical appliances?

Solution. A fuse is a short piece of wire of high resistivity and low melting point, connected in series with the appliance it protects.

Because it is in series, the whole of the appliance’s current passes through it. By Joule’s law the heat produced in it each second is

$$P = I^2 R$$

so the heat rises steeply with current — doubling the current produces four times the heating.

Under normal conditions the fuse carries its rated current and warms only slightly. If a fault — a short circuit, or too many appliances on one line — makes the current exceed the rating, the fuse wire’s temperature climbs to its melting point. It melts, the circuit is broken, and the current stops.

That protects the appliance in two ways: the excessive current is cut off before it can burn out the appliance’s own wiring, and the risk of the house wiring overheating and starting a fire is removed.

Two practical points follow. The fuse must be rated just above the appliance’s working current — high enough not to blow in normal use, low enough to act in a fault, which is the calculation of Question 16. And the fuse wire is deliberately the weakest link: it is designed to fail so that nothing more expensive does.

Question 24

What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

Solution.

Resistivity is the property of a material that measures how strongly it opposes the flow of current. From $R = \rho l / A$,

$$\rho = \frac{RA}{l}$$

so resistivity is numerically the resistance of a specimen of that material one metre long with a cross-sectional area of one square metre. Its SI unit is the ohm metre ($\Omega\ \mathrm{m}$), and it depends only on the material and its temperature — not on the size or shape of the piece, as Questions 3 and 13 establish.

Why the current halves. Resistance is directly proportional to length:

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

with $\rho$ and $A$ unchanged. Doubling the length doubles the resistance, so $R’ = 2R$.

The potential difference supplied by the cell is unchanged, so by Ohm’s law

$$I’ = \frac{V}{R’} = \frac{V}{2R} = \frac{I}{2}$$

The current therefore falls to half its value — from 5 A to 2.5 A, which is exactly what the ammeter shows.

The observation is itself evidence for the $R \propto l$ relation, and doubling the length is the cleanest way to demonstrate it experimentally. Question 31 asks for that experiment in full.

Question 25

What is the commercial unit of electrical energy? Represent it in terms of joules.

Solution. The commercial unit is the kilowatt hour (kW h), commonly called one unit on an electricity bill.

The joule is far too small for domestic accounting — a single 100 W bulb burning for an hour uses 360 000 J — so supply companies bill in kilowatt hours instead.

In joules. One kilowatt hour is the energy used by a device of power 1 kilowatt running for 1 hour:

$$1\ \mathrm{kW\,h} = 1000\ \mathrm{W} \times 3600\ \mathrm{s} = 1000 \times 3600\ \mathrm{J}$$

$$\boxed{1\ \mathrm{kW\,h} = 3.6 \times 10^6\ \mathrm{J}}$$

The conversion turns on remembering that an hour is $60 \times 60 = 3600$ seconds, and that a watt is a joule per second — so watt × second = joule.

It is worth being clear that the kilowatt hour is a unit of energy, not of power, even though “kilowatt” appears in its name. It is power multiplied by time. That is exactly the distinction Question 18 tests: volt ampere is power, kilowatt hour and watt second are energy.

Question 26

A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.

Solution.

The resistance of the lamp. The lamp and the 5 Ω conductor are in series across 10 V, carrying 1 A. The total resistance is

$$R_{\text{total}} = \frac{V}{I} = \frac{10}{1} = 10\ \Omega$$

and since the two are in series,

$$R_{\text{lamp}} = R_{\text{total}} – 5 = 10 – 5 = \mathbf{5\ \Omega}$$

Now add 10 Ω in parallel with that series combination. The question is what happens inside the original branch — and the answer is nothing at all.

The new resistor is connected in parallel across the whole series combination, which means it is joined to the same two points as the battery terminals. The potential difference across the original branch is therefore still the battery’s 10 V, exactly as before.

With the same 10 V across the same 10 Ω branch, the current in it is unchanged:

$$I = \frac{10}{10} = 1\ \mathrm{A} \ \text{ still flows through the 5 }\Omega\text{ conductor}$$

and so the potential difference across the lamp is unchanged too:

$$V_{\text{lamp}} = 1 \times 5 = 5\ \mathrm{V}$$

What does change is the current drawn from the battery: the new resistor carries an extra $10/10 = 1$ A of its own, so the battery now supplies 2 A instead of 1 A. But that extra current goes through the new resistor, not through the old branch.

The general rule worth extracting: adding a parallel branch changes the total current but leaves the existing branches untouched, because it does not change the voltage across them. This is precisely why household appliances are wired in parallel — switching one on does not dim the others.

Question 27

Why is parallel arrangement used in domestic wiring?

Solution. Three reasons, and each matters.

Every appliance gets the full supply voltage. In parallel all branches are connected across the same two points, so each receives the whole 220 V it is designed for. In series the supply would divide between the appliances, and each would get only a fraction of its rated voltage and work poorly.

Each appliance can be switched independently. Each branch has its own switch, and opening one breaks only that branch. In series a single switch — or a single failed appliance — would break the whole circuit, and everything in the house would go off together. This is the point Question 29(b) makes about fused bulbs.

Each appliance draws the current it needs. In parallel the branches share the total current in inverse proportion to their resistances, so a heater can draw a large current and a lamp a small one from the same supply. In series they would be forced to carry the same current, which is designed for neither.

There is a fourth, more technical advantage: connecting appliances in parallel lowers the total resistance of the circuit, so the supply can deliver the large total current that a house needs. The cost is that adding appliances increases that total current, which is why a fuse is needed — Question 23.

Question 28

$B_1$, $B_2$ and $B_3$ are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3 A is recorded by the ammeter A.

(i) What happens to the glow of the other two bulbs when the bulb $B_1$ gets fused? (ii) What happens to the reading of $A_1$, $A_2$, $A_3$ and A when the bulb $B_2$ gets fused? (iii) How much power is dissipated in the circuit when all the three bulbs glow together?

Solution. The three bulbs are identical and in parallel, so they share the total current equally. With 3 A in the main ammeter,

$$\text{each bulb carries } \frac{3}{3} = 1\ \mathrm{A}$$

and each of $A_1$, $A_2$, $A_3$ reads 1 A while A reads 3 A.

(i) When $B_1$ fuses. The other two bulbs glow exactly as brightly as before. Each parallel branch is connected directly across the battery, so the voltage across $B_2$ and $B_3$ is unchanged at 4.5 V, and unchanged voltage across unchanged resistance means unchanged current and unchanged power. Only the branch containing the broken bulb stops conducting.

(ii) When $B_2$ fuses. Its own branch carries no current, and the other two are unaffected:

$$A_1 = 1\ \mathrm{A}, \qquad A_2 = 0, \qquad A_3 = 1\ \mathrm{A}, \qquad A = 2\ \mathrm{A}$$

The main ammeter reads the sum of the branch currents, which has fallen from 3 A to 2 A.

(iii) Power with all three glowing.

$$P = VI = 4.5 \times 3 = \mathbf{13.5\ W}$$

Equivalently, each bulb dissipates $4.5 \times 1 = 4.5$ W and there are three of them.

The whole question is an illustration of why domestic wiring is parallel, which is Question 27: one bulb failing leaves the rest exactly as they were.

Long Answer Questions

Question 29

Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.

(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer. (b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.

Solution.

(a) No — the parallel bulbs glow much more brightly.

Let each bulb have resistance $R$ and the supply be $V$.

In series, the total resistance is $3R$, so the current is

$$I_s = \frac{V}{3R}$$

and the power in each bulb is $I_s^2 R = \dfrac{V^2}{9R}$.

In parallel, each bulb has the full $V$ across it, so each carries

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

and dissipates $\dfrac{V^2}{R}$.

Comparing, the current in each series bulb is one third of that in each parallel bulb, and since power goes as the square of the current, each series bulb dissipates only one ninth the power. The parallel bulbs are therefore far brighter.

Notice also that only the parallel bulbs actually run at their rated 100 W — the rating assumes the full supply voltage across the bulb, which is what parallel connection gives.

(b) The series circuit goes completely dark; the parallel bulbs are unaffected.

In series there is only one path. A fused bulb is a break in that path, so the circuit is open, the current falls to zero, and all three bulbs stop glowing — including the two that are perfectly sound. This is the old Christmas-light problem.

In parallel each bulb has its own path to the supply. A break in one branch leaves the other two branches complete and still across the full voltage, so the remaining two continue to glow with exactly the same brightness.

Question 30

State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

Solution.

The law. At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends.

$$V \propto I \qquad\Rightarrow\qquad V = IR$$

where the constant $R$ is the resistance of the conductor.

Verifying it experimentally. Set up the circuit of Question 19 — a battery, key, ammeter and the conductor in series, with a voltmeter across the conductor.

1. Close the key with one cell in the circuit; record the ammeter reading $I$ and the voltmeter reading $V$.
2. Repeat with two cells, then three, then four, recording $V$ and $I$ each time. Changing the number of cells changes the applied voltage.
3. Compute the ratio $V/I$ for each set of readings. It comes out very nearly constant — that constant is $R$.
4. Plot $V$ (horizontal) against $I$ (vertical). The points lie on a straight line through the origin, which is the graphical statement of $V \propto I$. Its slope is $1/R$, as Question 11 uses.

The measurements must be taken quickly, or the current will warm the wire and change its resistance — which is why the law is stated at constant temperature.

Does it hold always? No. Ohm’s law is not a universal law of nature but a description of how some materials behave under some conditions.

• It fails if the temperature changes. A filament lamp’s resistance rises sharply as it heats, so its $V$–$I$ graph curves rather than staying straight.
• It does not apply to non-ohmic components at all — semiconductor diodes, transistors, gas discharge tubes and electrolytes. A diode conducts freely one way and hardly at all the other.
• Even in metals it holds only for moderate currents; very large currents produce heating that breaks the proportionality.

So the honest statement is that Ohm’s law holds for metallic conductors at constant temperature, and those are exactly the conditions of the experiment above.

Question 31

What is electrical resistivity of a material? What is its unit? Describe an experiment to study the factors on which the resistance of conducting wire depends.

Solution.

Resistivity is the property of a material that measures its opposition to current, independent of the shape of the specimen. Rearranging $R = \rho l / A$,

$$\rho = \frac{RA}{l}$$

so resistivity is numerically the resistance of a wire of that material one metre long and one square metre in cross-section. Its unit is the ohm metre ($\Omega\ \mathrm{m}$), which follows directly: $\Omega \times \mathrm{m^2} / \mathrm{m} = \Omega\ \mathrm{m}$.

The experiment. Build the circuit of Question 19 — cell, key, ammeter in series with the test wire, voltmeter across it — and compute $R = V/I$ for each wire tested. Then change one factor at a time.

1. Length. Use wires of the same material and thickness but different lengths $l$, $2l$, $3l$. The resistance is found to double and treble accordingly:

$$R \propto l$$

2. Area of cross-section. Use wires of the same material and length but different thicknesses. A thicker wire gives a smaller resistance, and doubling the area halves it:

$$R \propto \frac{1}{A}$$

3. Nature of the material. Use wires of the same length and thickness but of copper, aluminium and nichrome. The resistances come out markedly different — nichrome’s is many times copper’s — showing the dependence on material, which is carried by $\rho$.

4. Temperature. Heat one wire and repeat the measurement. For a metal the resistance increases with temperature.

Combining the first three gives the relation the experiment establishes:

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

The essential discipline is changing one factor at a time and holding the others fixed; otherwise no single dependence can be isolated.

Question 32

How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?

Solution. The experiment is to measure the current at several different points in the series loop and show that the readings agree.

Method.

1. Connect three resistors $R_1$, $R_2$ and $R_3$ in series with a battery, a key and an ammeter.
2. Place the ammeter first between the battery and $R_1$. Close the key and record the reading, $I_1$.
3. Open the key and move the ammeter to sit between $R_1$ and $R_2$. Close the key again and record $I_2$.
4. Repeat with the ammeter between $R_2$ and $R_3$, and then between $R_3$ and the battery, recording $I_3$ and $I_4$.

Observation. All four readings are the same:

$$I_1 = I_2 = I_3 = I_4$$

Inference. The current has the same value at every point in the loop, so the same current flows through every part of a series circuit.

Two precautions make the result trustworthy. The battery and the resistors must not be changed between readings, or the current would differ for a different reason. And the readings should be taken promptly, so that heating does not alter the resistances during the experiment.

The underlying reason is that a series circuit offers only one path: charge cannot accumulate anywhere or leak away, so whatever flows past one point in a second must flow past every other point in that second.

Question 33

How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?

Solution. The experiment is the mirror image of Question 32: instead of moving an ammeter round a series loop, move a voltmeter across each parallel branch in turn.

Method.

1. Connect three resistors $R_1$, $R_2$ and $R_3$ in parallel across a battery, with a key in the main circuit.
2. Connect the voltmeter across $R_1$. Close the key and record the reading, $V_1$.
3. Move the voltmeter across $R_2$, close the key, record $V_2$.
4. Move it across $R_3$, record $V_3$.
5. Finally connect it across the battery terminals and record $V$.

Observation. All four readings are the same:

$$V_1 = V_2 = V_3 = V$$

Inference. The same potential difference exists across each resistor in a parallel arrangement, and it equals the voltage of the battery.

The reason is geometrical: in a parallel arrangement every branch is joined to the same two points. The potential difference between a given pair of points has one value, so every component connected between them must have that same voltage across it — regardless of its resistance.

Note what does not stay the same: the currents in the branches differ, dividing in inverse proportion to the resistances. That contrast is the whole of Question 17, and it is why domestic appliances are wired in parallel — each gets the full voltage while drawing only the current it needs.

Question 34

What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

Solution.

The effect. When a current flows through a conductor, electrical energy is converted into heat. This is Joule’s heating effect, and the heat produced is

$$H = I^2Rt$$

— proportional to the square of the current, to the resistance, and to the time. The squaring is the important part: doubling the current produces four times the heat.

The heat arises because the moving electrons collide with the atoms of the conductor and give up energy to them, which appears as increased vibration — that is, as a rise in temperature.

Demonstrating it.

1. Connect a coil of nichrome wire in series with a battery, an ammeter and a key, and immerse the coil in a known mass of water in a calorimeter with a thermometer.
2. Close the key, note the current $I$ and let it flow for a measured time $t$, stirring gently.
3. Record the rise in temperature. The water is found to warm, showing that electrical energy has become heat.
4. Repeat with a larger current for the same time: the temperature rise is much greater — roughly four times for twice the current, confirming $H \propto I^2$.
5. Repeat with a higher-resistance coil at the same current and time: the heating is greater, confirming $H \propto R$.

Four applications.

• Electric heater, iron, geyser and toaster — a nichrome element of high resistance and high melting point is deliberately made to glow hot.
• Electric bulb — the tungsten filament is heated to around $2500^\circ\mathrm{C}$ until it emits light.
• Electric fuse — a low-melting-point wire designed to melt and break the circuit when the current is excessive, as in Question 23.
• Electric kettle and cooking appliances — the same principle used to boil water and cook food.

Question 35

Find out the following in the electric circuit given in Figure 12.9

(a) Effective resistance of two 8 Ω resistors in the combination (b) Current flowing through 4 Ω resistor (c) Potential difference across 4 Ω resistance (d) Power dissipated in 4 Ω resistor (e) Difference in ammeter readings, if any.

Solution. Reduce the network first, then work outwards.

(a) The two 8 Ω resistors are in parallel:

$$\frac{1}{R_p} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \qquad\Rightarrow\qquad R_p = \mathbf{4\ \Omega}$$

(For two equal resistors in parallel the result is always half of one of them — a useful check.)

(b) That 4 Ω combination is in series with the 4 Ω resistor, so the total resistance of the circuit is

$$R_{\text{total}} = 4 + 4 = 8\ \Omega$$

and the current drawn from the 8 V cell is

$$I = \frac{V}{R_{\text{total}}} = \frac{8}{8} = \mathbf{1\ A}$$

Since the 4 Ω resistor is in the main loop, the whole of this current flows through it.

(c) The potential difference across the 4 Ω resistor:

$$V_{4\Omega} = IR = 1 \times 4 = \mathbf{4\ V}$$

(The remaining 4 V is across the parallel combination, and $4 + 4 = 8$ V checks against the cell. ✓)

(d) The power dissipated in it:

$$P = I^2R = (1)^2 \times 4 = \mathbf{4\ W}$$

(e) No difference — both ammeters read 1 A.

$A_1$ sits between B and C, and $A_2$ sits between the cell and D. Both are in the main loop, on the same single path, before the current divides between the two 8 Ω branches and after it recombines. In a series path the current is the same everywhere — Question 32 — so the two readings are identical and the difference is zero.

Common mistakes

• Testing only one component against a power rating, in Question 20. Resistor A carries the whole current and melts at 3 A, while B and C would each have allowed 6 A. The limit is always the most restrictive component.
• Recalculating the whole circuit in Question 26. The added resistor is across the battery terminals, so the voltage across the original branch is unchanged and nothing inside it changes. Ask what has stayed fixed before computing anything.
• Swapping the meters, in Questions 19 and 21. The ammeter goes in series and must have a low resistance; the voltmeter goes in parallel and must have a high one. Reversing them either shorts the resistor or stops the current.
• Assuming series and parallel bulbs are equally bright, in Question 29. Three identical bulbs in series each get a third of the current and so a ninth of the power of the same bulbs in parallel.
• Forgetting that $H \propto I^2$, in Questions 20 and 34. Doubling the current quadruples the heat. It is the squaring that makes fuses and heating elements work.
• Quoting the kilowatt hour as a unit of power, in Question 25. It is power multiplied by time, so it measures energy — $3.6 \times 10^6$ J.
• Expecting the two ammeters to differ in Question 35. Both sit in the main loop, so both read the full 1 A. Only a meter placed inside one of the parallel branches would read less.
• Stating Ohm’s law without “at constant temperature”, in Question 30. That condition is what makes the law true, and its failure is exactly why a filament lamp’s $V$–$I$ graph is curved.

Practise next

• Exemplar Multiple Choice Questions — eighteen questions on the same relations, including the one place where this unit’s printed key is wrong.
• NCERT textbook Chapter 11 — Electricity for the end-of-chapter numericals on combinations of resistors, electric power and heating effects.
• Textbook activities for the experiments of Questions 30 to 33 — verifying Ohm’s law and the series and parallel rules is done there rather than described.
• Exemplar Magnetic Effects of Electric Current — the next unit, which takes the same currents and asks what they do magnetically.