A \(40 \mathrm{~kW}\) electricity cable is used to transmit electricity through a resistor of resistance \(2.0 \Omega\) at \(800 \mathrm{~V}\). The power loss as internal energy is ____________
A. \(5.0 \times 10^{3} \mathrm{~W}\) B. \(4.0 \times 10^{3} \mathrm{~W}\) C. \(5.0 \times 10^{2} \mathrm{~W}\) D. \(4.0 \times 10^{2} \mathrm{~W}\)
Correct Answer: A
Explanation
Given, power \(=\) IV \(\mathrm{V}=\mathrm{IR}\) \(\mathrm{P}=\mathrm{I}^{2} \mathrm{R}\) Since \(\mathrm{P}=\) IV \(\Rightarrow \mathrm{P}=\mathrm{IV}^{\mathrm{P}} / \mathrm{V}\) \(I=\frac{40000}{500}=50 \mathrm{AMPS}\) Power loss to internal energy \(=\mathrm{I}^{2} \mathrm{R}\) \(=50.^{2}, 2=2500 \times 2=5000 \mathrm{w}\) \(\Rightarrow 5 .-\times 10^{3} \mathrm{w}\)
