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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