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JAMB Physics 2007 Question 26

Question 26

Electrical Energy and Power

A 40 kW electric cable is used to transmit electricity through a resistor of resistance 2.0 Ω at 800 V. The power loss as internal energy is

  1. A.5.0 × 102 W.
  2. B.4.0 × 103 W.
  3. C.5.0 × 103 W.
  4. D.4.0 × 102 W.
Show the answer and explanation

Answer: C (5.0 × 103 W.)

Explanation

Current in the cable I = P/V = 40 000 ÷ 800 = 50 A. Power lost as heat = I2R = 2500 × 2.0 = 5000 W = 5.0 × 103 W.