Question

A capacitor of reactance \( 4 \sqrt{3} \, \Omega \) and a resistor of resistance \( 4 \, \Omega \) are connected in series with an AC source of peak value \( 8 \sqrt{2} \, \text{V} \). The power dissipation in the circuit is ___________ W.

Answer 1

Correct Answer: 4

To calculate power dissipation in the circuit, the impedance \( Z \) of the series resistor and capacitor combination is determined first. The impedance \( Z \) is defined as: \[ Z = \sqrt{R^2 + X_C^2} \] Given \( R = 4 \, \Omega \) and \( X_C = 4 \sqrt{3} \, \Omega \), the impedance is computed as: \[ Z = \sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \, \Omega \] The peak voltage is \( V_0 = 8 \sqrt{2} \, \text{V} \). The RMS voltage \( V_{\text{RMS}} \) is calculated using: \[ V_{\text{RMS}} = \frac{V_0}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \, \text{V} \] The RMS current \( I_{\text{RMS}} \) is: \[ I_{\text{RMS}} = \frac{V_{\text{RMS}}}{Z} = \frac{8}{8} = 1 \, \text{A} \] Power dissipation in the resistor, \( P \), is calculated as: \[ P = I_{\text{RMS}}^2 \cdot R = 1^2 \cdot 4 = 4 \, \text{W} \] The calculated power dissipation is \( 4 \, \text{W} \), which is within the specified range of 4 to 4. The final answer is \( \textbf{4 W} \).