When a $d c$ voltage of $100 \mathrm{~V}$ is applied to an inductor, a $d c$ current of $5 \mathrm{~A}$ flows through it. When an ac voltage of $200 \mathrm{~V}$ peak value is connected to inductor, its inductive reactance is found to be $20 \sqrt{3} \Omega$. The power dissipated in the circuit is _________ W.

<p>To determine the power dissipated in the circuit, follow these steps:</p> <ol> <li>Calculate the resistance (R) using the DC current:</li> </ol> <p>$ R = \frac{100 \, \text{V}}{5 \, \text{A}} = 20 \, \Omega $</p> <ol> <li>Determine the impedance (Z) by considering both resistance (R) and inductive reactance ($X_L$):</li> </ol> <p>$ Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + (20\sqrt{3})^2} = 40 \, \Omega $</p> <ol> <li>Compute the peak current ($I_0$) using the AC peak voltage ($V_0$):</li> </ol> <p>$ I_0 = \frac{V_0}{Z} = \frac{200 \, \text{V}}{40 \, \Omega} = 5 \, \text{A} $</p> <ol> <li>Calculate the power (P) using the RMS values of voltage and current, and consider the phase angle ($\cos \phi$):</li> </ol> <p>$ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi = \frac{V_0 \cdot I_0}{2} \times \frac{R}{Z} = \frac{200 \cdot 5}{2} \times \frac{20}{40} = 250 \, \text{W} $</p>