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

A. 4.0 x 10²W
B. 5.0 x 10²W
C. 4.0 x 10³W
D. 5.0 x 10³W

Correct Answer:

Option D – 5.0 x 10³W

Explanation

In general, Power = IV; ⟹ 40KW = IV; Therefore 40000 = 1×800⟹I=40000/800=50A, tune the current through

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