For DC voltage:
\[ R = \frac{V}{I} = \frac{100}{5} = 20 \, \Omega \]
For AC voltage:
\[ X_L = 20\sqrt{3} \, \Omega \]
\[ Z = \sqrt{X_L^2 + R^2} = \sqrt{(20\sqrt{3})^2 + 20^2} = \sqrt{1200 + 400} = 40 \, \Omega \]
Power dissipated in the circuit:
\[ P = I_\text{rms}^2 R = \left( \frac{V_\text{rms}}{Z} \right)^2 \times R \]
\[ P = \left( \frac{200}{\sqrt{2} \times 40} \right)^2 \times 20 \]
\[ P = \left( \frac{200}{40\sqrt{2}} \right)^2 \times 20 = 250 \, \text{W} \]
AB is a part of an electrical circuit (see figure). The potential difference \(V_A - V_B\), at the instant when current \(i = 2\) A and is increasing at a rate of 1 amp/second is:
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).