Electricity is worth 7–9 marks in the Class 10 Science paper, and the questions are almost entirely from the list below. Solve each one before reading the answer, with the formula written first — the formula line is usually a mark on its own.
Short answer (1–2 marks)
Q1. Define one ampere. — One coulomb of charge flowing per second through a conductor: 1 A = 1 C/s.
Q2. A charge of 60 C flows through a bulb in 2 minutes. Find the current. — I = Q/t = 60/120 = 0.5 A. (Convert minutes to seconds.)
Q3. What is the resistance of a conductor? Give its SI unit. — The property that opposes the flow of current; R = V/I; unit ohm (Ω).
Q4. How does the resistance of a wire change if its length is doubled and its area halved? — R = ρL/A: doubling L doubles R, halving A doubles it again: 4 times.
Q5. Why is tungsten used for bulb filaments? — High melting point (about 3380 °C) and high resistivity, so it glows without melting.
Q6. Why are copper and aluminium used for wiring? — Low resistivity (good conductors) and cheap enough to use in quantity.
Numericals (3 marks)
Q7. A wire of resistance 20 Ω is connected to a 10 V battery. Find the current, and the heat produced in 5 minutes.
I = V/R = 10/20 = 0.5 A. H = I²Rt = 0.25 × 20 × 300 = 1500 J.
Q8. Three resistors of 4 Ω, 6 Ω and 12 Ω are connected in parallel. Find the equivalent resistance.
1/R = 1/4 + 1/6 + 1/12 = 3/12 + 2/12 + 1/12 = 6/12 ⇒ R = 2 Ω. The parallel result is always smaller than the smallest resistor — a quick check.
Q9. The same three resistors are connected in series across 22 V. Find the current and the potential difference across the 6 Ω resistor.
R = 4 + 6 + 12 = 22 Ω. I = 22/22 = 1 A. V across 6 Ω = 1 × 6 = 6 V.
Q10. An electric bulb is rated 100 W, 220 V. Find its resistance and the current it draws.
R = V²/P = 220²/100 = 48400/100 = 484 Ω. I = P/V = 100/220 = 0.45 A.
Q11. Two bulbs of 40 W and 60 W (both 220 V) are connected in series across 220 V. Which glows brighter?
Resistances: R₄₀ = 220²/40 = 1210 Ω; R₆₀ = 220²/60 ≈ 807 Ω. In series the current is the same, and power = I²R, so the 40 W bulb (higher resistance) glows brighter. In parallel it would be the 60 W bulb — the usual follow-up.
Q12. A heater of 1000 W runs for 3 hours a day for 30 days. Find the energy in kWh and the cost at ₹6 per unit.
Energy = 1 kW × 3 h × 30 = 90 kWh (90 units). Cost = 90 × 6 = ₹540.
Long answer (5 marks)
Q13. State Ohm's law. Describe an experiment to verify it and draw the V–I graph.
At constant temperature, current through a conductor is proportional to the potential difference across it: V = IR. Circuit: a cell (or battery with rheostat), key, ammeter in series with the resistor, voltmeter across the resistor. Vary the current with the rheostat, note V and I for four or five readings, and plot V against I. A straight line through the origin verifies the law; its slope is R.
Q14. Derive the expression for the equivalent resistance of three resistors in series and in parallel.
Series: same current I; V = V₁ + V₂ + V₃ = I(R₁ + R₂ + R₃), so R = R₁ + R₂ + R₃. Parallel: same voltage V; I = I₁ + I₂ + I₃ = V(1/R₁ + 1/R₂ + 1/R₃), so 1/R = 1/R₁ + 1/R₂ + 1/R₃. Draw both circuits and label the currents and voltages — the diagram is a mark.
Q15. Explain the heating effect of current. State two applications and why a fuse wire has low melting point.
Work done to push charge Q through a potential difference V is W = VQ = VIt = I²Rt; in a pure resistor it appears as heat (Joule's law). Applications: electric iron and heater, fuse. A fuse is a short wire of low melting point (tin–lead alloy) in series with the circuit: when the current exceeds the rated value, I²Rt heats it past its melting point, it breaks, and the circuit is cut before the appliance is damaged.
What to revise if you got these wrong
- Q2, Q7, Q12: unit conversions — minutes to seconds, watts to kilowatts, hours as hours in kWh.
- Q8–Q9: series adds, parallel adds reciprocals; check the parallel answer is below the smallest resistor.
- Q10–Q11: the three power formulas P = VI = I²R = V²/R, and which one to use when V is fixed (V²/R) versus when I is fixed (I²R).
- Q13–Q14: draw the circuit; label; state the condition "constant temperature".