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JAMB UTME - Physics - 2007

Question 1 Report

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
Answer Details
In general, Power = IV; 40 K W = I V Therefore  40000 = 1 × 800 I = 40000 800 = 50 A , tune the current through

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