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Wednesday, October 31, 2018

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

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