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 $8 \sqrt{2} \mathrm{~V}$. The power dissipation in the circuit is __________ W.

<p>To calculate the power dissipation in the circuit, we follow a systematic approach. We're provided with the reactance of the capacitor ($X_C = 4 \sqrt{3} \Omega$), the resistance ($R = 4 \Omega$), and the peak value of the AC voltage source ($V_{peak} = 8 \sqrt{2} V$). The power dissipated in an AC circuit is primarily through the resistive component, as inductors and capacitors store and release energy but do not dissipate it as heat.</p> <p>First, we need to determine the effective impedance of the series circuit, which combines the resistance (R) and the capacitive reactance (X_C) in a series configuration. We calculate the impedance (Z) using the formula:</p> <p>$Z = \sqrt{R^2 + X_C^2}$</p> <p>Plugging in the given values:</p> <p>$Z = \sqrt{(4)^2 + (4 \sqrt{3})^2}$</p> <p>$Z = \sqrt{16 + 48} = \sqrt{64} = 8 \Omega$</p> <p>Next, we convert the peak voltage to RMS (root mean square) voltage because power calculations in AC circuits are performed using RMS values. The formula to convert peak voltage ($V_{peak}$) to RMS voltage ($V_{RMS}$) is:</p> <p>$V_{RMS} = \frac{V_{peak}}{\sqrt{2}}$</p> <p>Plugging in the given peak voltage value:</p> <p>$V_{RMS} = \frac{8 \sqrt{2}}{\sqrt{2}} = 8 \, \mathrm{V}$</p> <p>Now, to find the RMS current ($I_{RMS}$) in the circuit, we use Ohm's law as applied to AC circuits, which is $I_{RMS} = \frac{V_{RMS}}{Z}$:</p> <p>$I_{RMS} = \frac{8}{8} = 1 \, \mathrm{A}$</p> <p>Finally, the power dissipated in the circuit is calculated using the formula for power in resistive components of an AC circuit, which is $P = I_{RMS}^2 \times R$:</p> <p>$P = (1)^2 \times 4 = 4 \, \mathrm{W}$</p> <p>Therefore, the power dissipation in the circuit is <strong>4 W</strong>.</p>