A 40KW electric cable was used to transmit electricity through a resistor of resistance 2.00\Omega 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
Verified Correct Answer: D.
Physical Principles & Numerical Solution
In general, Power = IV; \( \implies 40KW = IV \\
\text{Therefore } 40000 = 1 \times 800\\
\implies I = \frac{40000}{800} = 50A,\text{tune the current through} \)
Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W
Source: Myschool Past Questions Archive (Record #41388)