A 40KW electric cable was uses to transmit electricity through a resistor of resistance 2.00Ω at 800V. The power loss as internal energy is

Assessment: JAMB UTME - Physics - 2007 Subject: Physics

Question 1 Report

4.0 x 102W

5.0 x 102W

4.0 x 103W

5.0 x 103W

Answer Details

In general, Power = IV;

⟹ 40 K W = I V Therefore 40000 = 1 \times 800 ⟹ I = \frac{40000}{800} = 50 A, tune the current through

Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W

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