JAMB - Physics (2007 - No. 44)
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
4.0 x 102W
5.0 x 102W
4.0 x 103W
5.0 x 103W
Explanation
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
\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
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