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 determine the power dissipation in the circuit, we start by calculating the impedance \( Z \) of the series combination of the resistor and the capacitor. The impedance \( Z \) is given by: 

\[ Z = \sqrt{R^2 + X_C^2} \] 
where \( R = 4 \, \Omega \) and \( X_C = 4 \sqrt{3} \, \Omega \). Compute \( Z \): 
\[ Z = \sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \, \Omega \] 
The peak voltage \( V_0 = 8 \sqrt{2} \, \text{V} \). To find the RMS voltage \( V_{\text{RMS}} \), use: 
\[ V_{\text{RMS}} = \frac{V_0}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \, \text{V} \] 
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 given by: 
\[ P = I_{\text{RMS}}^2 \cdot R = 1^2 \cdot 4 = 4 \, \text{W} \] 
The computed power dissipation is \( 4 \, \text{W} \), which falls within the given range of 4 to 4. Thus, the final answer is \( \textbf{4 W} \).

Answer 2

The impedance $Z$ of the circuit is: 
\(Z = \sqrt{R^2 + X_C^2} = \sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \, \Omega.\)

The RMS voltage is: 
\(V_\text{rms} = \frac{V_\text{peak}}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \, \text{V}.\) 
The RMS current is: \(I_\text{rms} = \frac{V_\text{rms}}{Z} = \frac{8}{8} = 1 \, \text{A}.\)
Power dissipation in the resistor is: \(P = I_\text{rms}^2 R = 1^2 \times 4 = 4 \, \text{W}.\)