The power output of an electrical supply is 2.4 kW at a potential difference (p.d.) of 240 V.

Question 15

The power output of an electrical supply is 2.4 kW at a potential difference (p.d.) of 240 V. The two wires between the supply and a kettle each have a resistance of 0.50 Ω, as shown.

What is the power supplied to the kettle and what is the p.d. across the kettle?

power / kW     p.d. / V

A         2.3                   230

B         2.3                   235

C         2.4                   230

D         2.4                   235

Reference: Past Exam Paper – June 2018 Paper 12 Q32

Solution:

Answer: A.

Total power supplied = Power supplied to kettle + Power dissipated in wires

The 2 wires and the kettle are connected in series. So, the same current I flows through them.

Power P = VI

At the supply, P = 2.4 kW and V = 240 V

Current I = P / V = 2400 / 240 = 10 A

Power dissipated in each wire = I²R = 10² × 0.5 = 50 W

Total power dissipated in both wires = 2 × 50 = 100 W = 0.1 kW

Total power supplied = Power supplied to kettle + Power dissipated in wires

Power supplied to kettle = Total power supplied – Power dissipated in wires

Power supplied to kettle = 2.4 – 0.1 = 2.3 kW

e.m.f. from supply = p.d. across wires + p.d. across kettle

V = IR

p.d. across 1 wire = IR = 10 × 0.5 = 5 V

p.d. across both wires = 2 × 5 = 10 V

e.m.f. from supply = p.d. across wires + p.d. across kettle

p.d. across kettle = 240 – 10 = 230 V