Exercise — Electricity

18 questions✓ Free · step-by-step

11 markNCERT Cl-10 Science, Exercise 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′, then the ratio R/R′ is – (a) 1/25 (b) 1/5 (c) 5 (d) 25
Hint. Each piece has a fifth of the original resistance — then combine five of those in parallel.
Step 1 — Find each piece's resistance. Each of the 5 equal parts has resistance R/5.

Step 2 — Combine them in parallel. 1/R′ = 5/(R/5) = 25/R, so R′ = R/25.

Step 3 — Find the ratio. R/R′ = R/(R/25) = 25.

✦ Answer: (d) 25.
Where students slip. Stopping after finding R′ = R/25 and answering '1/25' instead of the requested ratio R/R′ — the question asks for R/R′, which inverts to 25, not 1/25.
21 markNCERT Cl-10 Science, Exercise Q2
Which of the following terms does not represent electrical power in a circuit? (a) I²R (b) IR² (c) VI (d) V²/R
Hint. Check each option against P = VI and Ohm's law substitutions — one of them doesn't come from a valid substitution at all.
Step 1 — Derive the valid power formulas. From P = VI and V = IR: P = I²R and P = V²/R are both valid, along with P = VI itself.

Step 2 — Check the remaining option. IR² doesn't arise from any valid substitution — swapping V for IR in P = VI gives I²R, not IR².

✦ Answer: (b) IR².
Where students slip. Assuming any combination of I, V and R with the right 'feel' must be a power formula — only the three combinations derivable from P = VI and Ohm's law are actually valid.
32 marksNCERT Cl-10 Science, Exercise Q3
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be – (a) 100 W (b) 75 W (c) 50 W (d) 25 W
Hint. First find the bulb's fixed resistance from its rated values, then use that resistance at the new voltage.
Step 1 — Find the bulb's resistance from its rating. R = V²/P = 220²/100 = 48400/100 = 484 Ω.

Step 2 — Find the power at the new voltage. P′ = V′²/R = 110²/484 = 12100/484 = 25 W, since resistance stays fixed while only the voltage changes.

✦ Answer: (d) 25 W.
Where students slip. Assuming power scales directly with voltage (halving V halves P, giving 50 W) — power actually scales with the square of voltage for fixed resistance, so halving V quarters P, not halves it.
42 marksNCERT Cl-10 Science, Exercise 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 – (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
Hint. Since the two wires are identical, call each resistance r, and compare the combined resistance in each arrangement.
Step 1 — Find the combined resistance in each case. Series: R_s = 2r. Parallel: R_p = r/2.

Step 2 — Compare heat at the same voltage and time. Since H = V²t/R for fixed V and t, heat is inversely proportional to resistance: H_series/H_parallel = R_p/R_s = (r/2)/(2r) = 1/4.

✦ Answer: (c) 1:4.
Where students slip. Comparing the resistances directly (2r vs r/2, a ratio of 4:1) and forgetting that heat is inversely proportional to resistance — the heat ratio is the inverse of the resistance ratio, giving 1:4, not 4:1.
51 markNCERT Cl-10 Science, Exercise Q5
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Hint. The voltmeter needs to see the same two points as the component it's measuring, not sit in the current's only path.
Step 1 — Recall how a voltmeter must be connected. A voltmeter is always connected in parallel across the two points, since it must sense the same two points as the component it's measuring rather than sit in the current's path.

✦ Answer: In parallel, across the two points of interest.
Where students slip. Connecting the voltmeter in series (like an ammeter) — that would place it directly in the current path, which is not how a voltmeter measures potential difference.
63 marksNCERT Cl-10 Science, Exercise Q6
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Hint. Get the cross-sectional area from the diameter first, then rearrange R = ρl/A to solve for length.
Step 1 — Find the cross-sectional area. A = πd²/4 = π(5 × 10⁻⁴)²/4 ≈ 1.9635 × 10⁻⁷ m².

Step 2 — Solve for length. l = RA/ρ = (10 × 1.9635 × 10⁻⁷)/(1.6 × 10⁻⁸) ≈ 122.7 m.

Step 3 — Find the effect of doubling the diameter. Since A ∝ d², doubling the diameter quadruples the area, and since R ∝ 1/A, the resistance drops to a quarter: R′ = 10/4 = 2.5 Ω.

✦ Answer: Length ≈ 122.7 m. Doubling the diameter reduces the resistance to 2.5 Ω (a quarter of the original).
Where students slip. Assuming doubling the diameter halves the resistance — resistance depends on area, which scales with diameter squared, so doubling the diameter quarters the resistance, not halves it.
73 marksNCERT Cl-10 Science, Exercise 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 – I (amperes): 0.5, 1.0, 2.0, 3.0, 4.0; V (volts): 1.6, 3.4, 6.7, 10.2, 13.2. Plot a graph between V and I and calculate the resistance of that resistor.
Hint. The resistance is the slope of the straight-line graph — you can estimate it from the ratio V/I at each data point too.
Step 1 — Compute V/I at each data point. 1.6/0.5 = 3.2; 3.4/1.0 = 3.4; 6.7/2.0 = 3.35; 10.2/3.0 = 3.4; 13.2/4.0 = 3.3.

Step 2 — Average these to estimate the slope. The values cluster around 3.3–3.4 Ω, consistent with a straight line through the origin (as Ohm's law predicts).

Step 3 — State the resistance. Since the graph's slope gives resistance directly, R ≈ 3.3 Ω.

✦ Answer: R ≈ 3.3 Ω (a straight line through the origin with this slope).
Where students slip. Picking just one data point's V/I ratio as 'the' answer instead of using the overall slope/average — small experimental scatter is expected, and the resistance is best read from the line's overall slope, not a single pair of readings.
82 marksNCERT Cl-10 Science, Exercise 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.
Hint. Convert the current to amperes before applying Ohm's law.
Step 1 — Convert units. 2.5 mA = 2.5 × 10⁻³ A.

Step 2 — Apply Ohm's law. R = V/I = 12/(2.5 × 10⁻³) = 4800 Ω.

✦ Answer: 4800 Ω (4.8 kΩ).
Where students slip. Using 2.5 directly as amperes without converting from milliamperes — this would give an answer 1000 times too small.
92 marksNCERT Cl-10 Science, Exercise Q9
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Hint. In series, one total-resistance calculation and one Ohm's law application tells you the current everywhere.
Step 1 — Add up the series resistances. R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω.

Step 2 — Apply Ohm's law. I = V/R = 9/13.4 ≈ 0.67 A.

Step 3 — Apply this to the 12 Ω resistor. Since it's a series circuit, the same current flows through every resistor, including the 12 Ω one.

✦ Answer: ≈ 0.67 A flows through the 12 Ω resistor (and through every other resistor in the chain).
Where students slip. Trying to find a separate current for the 12 Ω resistor using only its own resistance — in series, there's only one current for the whole loop, found from the total resistance, not from any single resistor in isolation.
102 marksNCERT Cl-10 Science, Exercise Q10
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Hint. Find the total resistance the parallel combination needs to have, then work out how many equal resistors give that.
Step 1 — Find the required total resistance. R_total = V/I = 220/5 = 44 Ω.

Step 2 — Relate this to n identical resistors in parallel. For n resistors of 176 Ω each, R_total = 176/n.

Step 3 — Solve for n. 44 = 176/n, so n = 176/44 = 4.

✦ Answer: 4 resistors.
Where students slip. Dividing 5 A by 176 Ω directly — the correct route is finding the required total resistance from V and I first, then relating it to how many equal resistors in parallel produce that value.
113 marksNCERT Cl-10 Science, Exercise Q11
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
Hint. For each target, try combining two resistors one way and adding the third the other way.
Step 1 — (i) Target 9 Ω. Two 6 Ω resistors in parallel give 1/R = 1/6 + 1/6 = 1/3, so R = 3 Ω. Adding the third 6 Ω resistor in series gives 3 + 6 = 9 Ω.

Step 2 — (ii) Target 4 Ω. Two 6 Ω resistors in series give 12 Ω. Putting this in parallel with the third 6 Ω resistor: 1/R = 1/12 + 1/6 = 1/4, so R = 4 Ω.

✦ Answer: (i) Two resistors in parallel, in series with the third. (ii) Two resistors in series, in parallel with the third.
Where students slip. Using the same arrangement pattern for both targets — (i) and (ii) need opposite structures (parallel-then-series vs. series-then-parallel), since the two target values sit on either side of 6 Ω itself.
122 marksNCERT Cl-10 Science, Exercise Q12
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Hint. Find how much current one bulb draws, then see how many of those fit into the maximum allowed.
Step 1 — Find one bulb's current. I_bulb = P/V = 10/220 A.

Step 2 — Divide the maximum current by this. n = 5/(10/220) = 5 × 220/10 = 110.

✦ Answer: 110 bulbs.
Where students slip. Dividing the maximum current by the power directly (5/10) instead of by each bulb's actual current draw — the per-bulb current must first be found from P/V, not from the power figure alone.
133 marksNCERT Cl-10 Science, Exercise Q13
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Hint. Work out the combined resistance for each of the three arrangements first, then apply Ohm's law each time.
Step 1 — Separately (either coil alone). I = V/R = 220/24 ≈ 9.17 A.

Step 2 — In series. R = 24 + 24 = 48 Ω, so I = 220/48 ≈ 4.58 A.

Step 3 — In parallel. Since the two resistances are equal, R = 24/2 = 12 Ω, so I = 220/12 ≈ 18.33 A.

✦ Answer: Separately ≈ 9.17 A; in series ≈ 4.58 A; in parallel ≈ 18.33 A.
Where students slip. Using the same combined-resistance formula for all three cases — each arrangement (single coil, series, parallel) has its own distinct total resistance, and mixing them up changes every subsequent current value.
143 marksNCERT Cl-10 Science, Exercise Q14
Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Hint. In (i) find the shared series current first; in (ii) remember the parallel resistor already has the full battery voltage across it.
Step 1 — (i) Series circuit. R = 1 + 2 = 3 Ω, so I = 6/3 = 2 A (same current through both resistors). Power in the 2 Ω resistor: P = I²R = 2² × 2 = 8 W.

Step 2 — (ii) Parallel circuit. Each resistor gets the full 4 V directly. Power in the 2 Ω resistor: P = V²/R = 4²/2 = 8 W.

Step 3 — Compare. Since both give exactly 8 W, the power used in the 2 Ω resistor is the same in both circuits.

✦ Answer: 8 W in both cases — the power in the 2 Ω resistor is identical in the series and parallel circuits described.
Where students slip. Assuming the parallel case must give more power just because parallel circuits generally draw more total current — that's true for the circuit as a whole, but this specific 2 Ω resistor happens to dissipate the same power in both setups here.
152 marksNCERT Cl-10 Science, Exercise Q15
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Hint. Find each lamp's own current, then add them for the total drawn from the line.
Step 1 — Find each lamp's current. I₁ = 100/220 ≈ 0.4545 A; I₂ = 60/220 ≈ 0.2727 A.

Step 2 — Add them. I_total = 0.4545 + 0.2727 ≈ 0.727 A.

✦ Answer: ≈ 0.727 A is drawn from the line.
Where students slip. Adding the wattages first and dividing once (160/220) instead of adding the two currents — both routes actually give the same numeric answer here since the lamps share the same voltage, but treating current addition as the underlying rule (not a coincidence) matters for cases where voltages differ.
162 marksNCERT Cl-10 Science, Exercise Q16
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Hint. Convert both times to the same unit before multiplying by power.
Step 1 — TV energy. Energy = 250 W × 1 h = 250 Wh.

Step 2 — Toaster energy. 10 minutes = 1/6 h, so energy = 1200 × (1/6) = 200 Wh.

Step 3 — Compare. 250 Wh > 200 Wh.

✦ Answer: The TV uses more energy (250 Wh vs. 200 Wh for the toaster).
Where students slip. Assuming the higher-power toaster must use more energy — power alone doesn't determine energy; the much shorter running time (10 minutes vs. 1 hour) means the toaster actually uses less energy overall.
172 marksNCERT Cl-10 Science, Exercise Q17
An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Hint. 'Rate' of heat production is just another name for power — the 2 hours isn't needed to answer this specific question.
Step 1 — Recognise that 'rate of heat production' means power. Power P = I²R.

Step 2 — Substitute the given values. P = 5² × 44 = 25 × 44 = 1100 W, since power depends on the square of the current, a small change in current would raise this heating rate much faster than the same proportional change in resistance would.

✦ Answer: Heat is developed at a rate of 1100 W (1100 J/s).
Where students slip. Multiplying by the 2 hours to get a total energy figure and reporting that as the 'rate' — the question specifically asks for the rate of heat production, which is the power itself, independent of how long the heater runs.
185 marksNCERT Cl-10 Science, Exercise Q18
Explain the following. (a) Why is 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?
Hint. Each part has its own single key fact — melting point, oxidation resistance, shared current, an inverse relationship, and low resistivity respectively.
Step 1 — (a) Tungsten in filaments. Tungsten has a very high melting point (3380°C), so it can be heated enough to glow white-hot without melting.

Step 2 — (b) Alloys in heating devices. Alloys have higher resistivity than their constituent pure metals (producing more heat) and resist oxidising even at high temperatures, making them more durable for repeated heating.

Step 3 — (c) Series not used domestically. In series, every appliance shares the same current, and if one fails or is switched off, the whole circuit breaks — appliances needing different currents also couldn't operate correctly together this way. Parallel wiring avoids both problems.

Step 4 — (d) Resistance vs. area. Resistance is inversely proportional to the cross-sectional area (R ∝ 1/A) — a thicker wire has lower resistance than a thinner one of the same length and material.

Step 5 — (e) Copper and aluminium for transmission. Both have very low resistivity among affordable, abundant metals, minimising energy lost as heat over long transmission distances.

✦ Answer: (a) Very high melting point (3380°C). (b) Higher resistivity and resistance to oxidising at high temperature. (c) Series wiring shares one current and fails entirely if one device fails; parallel avoids this. (d) Resistance is inversely proportional to cross-sectional area. (e) Their low resistivity minimises transmission energy loss, at an affordable cost.
Where students slip. Answering (d) with 'resistance increases with area' — it's the opposite: greater cross-sectional area means lower resistance, since R ∝ 1/A.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Science textbook, Reprint 2026-27 (jesc111.pdf) — seven in-text question sets (23 questions total, not 16 as some older manifests claim) plus one end-of-chapter Exercise (18 questions, correctly counted). Unchanged by rationalisation. Table 11.2's resistivity values (used for several answers) were read directly from the book, including nichrome's 100 × 10⁻⁶ Ω·m.. Questions are referenced from the NCERT textbook for identification.

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