62.5A, 4.0\(\Omega\)
16.0A, 62.5\(\Omega\)
62.5A, 16.0\(\Omega\)
4.0A, 62.5\(\Omega\)
4.0A, 250\(\Omega\)
Correct answer is D
R = \(\frac{V^2}{R}\) = IV ; R = \(\frac{V^2}{P}\) and I = \(\frac{P}{V}\)
R = \(\frac{(250)^2}{100}\) and I = R = \(\frac{1000}{250}\)
R = 62.5\(\Omega\) and I = 4A