9 min readConcept sheet1-mark MCQ, 2-mark questions on bulbs in series and parallel, and 3-mark numericals and derivations on transmission loss

Electrical energy and power

Power used in a resistor is P = VI = I²R = V²/R. Power is sent at high voltage to cut I²R loss in long transmission lines.

PowerP = VI = I²R = V²/R; SI unit watt (W), 1 W = 1 J s⁻¹

EnergyW = Pt = VIt; commercial unit 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J

Transmission lossP_c = P²R_c/V²; P = power sent, R_c = resistance of the cables, V = transmission voltage

Derivation: power dissipated in a conductor

A conductor AB carries a steady current I. Point A is at a higher potential than B, and V = V(A) − V(B).

1
In a time Δt, a charge ΔQ = IΔt moves from A to B.
2
Its potential energy falls by ΔU = ΔQ × V = IVΔt.
3
The moving charges collide with the ions and hand over this energy. The ions vibrate more, so the conductor heats up.
4
Energy given out in time Δt: ΔW = IVΔt.
5
Power = energy per unit time: P = ΔW/Δt = VI.
6★
Using V = IR: P = I²R = V²/R. So, the power dissipated in a resistor is P = VI = I²R = V²/R.

★ key step

Why power is sent at high voltage

A power station sends power P to a town through cables of total resistance R_c.

1
If the transmission voltage is V, the current in the cables is I = P/V.
2
Power lost as heat in the cables: P_c = I²R_c = P²R_c/V².
3
For fixed P and R_c, the loss varies as 1/V². Raising V ten times cuts the loss a hundred times.
4
So power is sent at very high voltages, and step-down transformers lower it to 220 V near homes.

Which formula to use: I²R or V²/R?

Series (same current)Use P = I²R. The larger resistance takes the larger share of power. With a 60 W and a 100 W bulb, both rated 220 V, in series, the 60 W bulb (higher R) glows brighter.

Parallel (same voltage)Use P = V²/R. The smaller resistance takes more power. On 220 V mains, a 100 W bulb glows brighter than a 60 W bulb, as each gets its rated voltage.

For a bulb rated P at V, first find R = V²/P, then pick the formula that fits the circuit.

Worked example: a bulb on half voltage

Problem

A bulb is rated 220 V, 100 W. Find (a) its resistance, (b) the current it draws at 220 V, and (c) its power on a 110 V supply, assuming its resistance stays the same.

Working

(a) R = V²/P = (220 V)² ÷ 100 W = 48400 ÷ 100 = 484 Ω.
(b) I = P/V = 100 W ÷ 220 V ≈ 0.45 A.
(c) P' = V'²/R = (110 V)² ÷ 484 Ω = 12100 ÷ 484 = 25 W.
Check: halving V gives one quarter of the power, and 100 W ÷ 4 = 25 W.

Answer

(a) 484 Ω; (b) about 0.45 A; (c) 25 W

Worked example: transmission loss

Problem

A station sends 2.0 MW of power through cables of total resistance 5.0 Ω. Find the power lost in the cables if the power is sent at (a) 20 kV and (b) 200 kV.

Working

(a) I = P/V = (2.0 × 10⁶ W) ÷ (2.0 × 10⁴ V) = 100 A.
Loss P_c = I²R_c = (100 A)² × 5.0 Ω = 5.0 × 10⁴ W = 50 kW, which is 2.5% of the power sent.
(b) I = (2.0 × 10⁶ W) ÷ (2.0 × 10⁵ V) = 10 A.
Loss P_c = (10 A)² × 5.0 Ω = 500 W, only 0.025% of the power sent.
Ten times the voltage gives one hundredth of the loss, as P_c ∝ 1/V².

Answer

(a) 50 kW; (b) 500 W

In real life

Your electricity bill counts units, and 1 unit = 1 kWh. A 1500 W geyser running for 2 hours uses 3 units. High-voltage towers across India carry power at hundreds of kilovolts, so that less is lost as heat on the way.

Exam angle

1 marks: “Express 1 kWh in joules.” 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

2 marks: “Two bulbs of 60 W and 100 W, both rated 220 V, are connected across 220 V (a) in parallel and (b) in series. Which glows brighter in each case?” R = V²/P: about 807 Ω for 60 W and 484 Ω for 100 W. (a) Parallel, same V, P = V²/R: the 100 W bulb. (b) Series, same I, P = I²R: the 60 W bulb.

3 marks: “Show that the power lost in transmission cables is P²R_c/V². Why is electric power sent over long distances at high voltage?” I = P/V, so the loss is I²R_c = P²R_c/V². For fixed P and R_c the loss falls as 1/V², so a high voltage means a small current and a small heating loss.

Remember it

Same I (series): use I²R, and the bigger R gets hotter. Same V (parallel): use V²/R, and the smaller R gets hotter. Line loss ∝ 1/V².