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;⟹40KW=IVTherefore 40000=1×800⟹I=40000800=50A,tune the current through$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}40KW=IV\phantom{\rule{0ex}{0ex}}\text{Therefore}40000=1\times 800\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}I=\frac{40000}{800}=50A,\text{tune the current through}$

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