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

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: $P = I_\text{rms}^2 R = 1^2 \times 4 = 4 \, \text{W}.$

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: $P = I_\text{rms}^2 R = 1^2 \times 4 = 4 \, \text{W}.$