To find the peak current in the circuit, we need to calculate the capacitive reactance and then use it to find the current.$\\$Given:$\\$• Capacitance $C = 10 \mu F = 10 \times 10^{-6} F \\$• Voltage $V = 210 V \\$• Frequency $f = 50 Hz \\$• $\pi = 3.14 \\$The capacitive reactance $X_c$ is given by:$\\ X_c = \frac{1}{2 \pi f C} \\$Substitute the known values:$\\ X_c = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} \\ X_c = \frac{1}{3.14 \times 10^{-3}} \\ X_c \approx 318.47 \, \Omega \\$The peak current $I_0$ is given by:$\\ I_0 = \frac{V_0}{X_c} \\$The peak voltage $V_0$ is related to the RMS voltage $V$ by:$\\ V_0 = V \sqrt{2} \\ V_0 = 210 \sqrt{2} \\ V_0 \approx 210 \times 1.414 \\ V_0 \approx 297 \, V \\$Now, calculate the peak current:$\\ I_0 = \frac{297}{318.47} \\ I_0 \approx 0.93 \, A \\$Therefore, the peak current in the circuit is approximately $0.93 \, A$. This corresponds to Option 4.