QJEEM24Phd7c3: JEE Main 2024 Physics - Class 12 JEEM24Phd7c3 Solution: A capacitor of reactance and a resistor of resistance are…

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.

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:

$Z = \sqrt{R^2 + X_C^2}$

Plugging in the given values:

$Z = \sqrt{(4)^2 + (4 \sqrt{3})^2}$

$Z = \sqrt{16 + 48} = \sqrt{64} = 8 \Omega$

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:

$V_{RMS} = \frac{V_{peak}}{\sqrt{2}}$

Plugging in the given peak voltage value:

$V_{RMS} = \frac{8 \sqrt{2}}{\sqrt{2}} = 8 \, \mathrm{V}$

Now, to find the RMS current ( $I_{RMS}$ ) in the circuit, we use Ohm's law as applied to AC circuits, which is $I_{RMS} = \frac{V_{RMS}}{Z}$ :

$I_{RMS} = \frac{8}{8} = 1 \, \mathrm{A}$

Finally, the power dissipated in the circuit is calculated using the formula for power in resistive components of an AC circuit, which is $P = I_{RMS}^2 \times R$ :

$P = (1)^2 \times 4 = 4 \, \mathrm{W}$

Therefore, the power dissipation in the circuit is 4 W.