$A$ resistor of $200 \; \Omega$ and a capacitor of $15.0 \; \mu F$ are connected in series to a $220 \; V, 50 \; Hz$ $ac$ source.
$(a)$ Calculate the current in the circuit.
$(b)$ Calculate the voltage $(rms)$ across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes,resolve the paradox.

(A) $R = 200 \; \Omega, C = 15.0 \; \mu F = 15.0 \times 10^{-6} \; F$
$V = 220 \; V, \nu = 50 \; Hz$
$(a)$ To calculate the current,we need the impedance $Z$ of the circuit:
$X_C = \frac{1}{2 \pi \nu C} = \frac{1}{2 \times 3.14 \times 50 \times 15.0 \times 10^{-6}} \approx 212.3 \; \Omega$
$Z = \sqrt{R^2 + X_C^2} = \sqrt{200^2 + 212.3^2} \approx 291.67 \; \Omega$
$I = \frac{V}{Z} = \frac{220}{291.67} \approx 0.754 \; A$
$(b)$ The voltage across the resistor is $V_R = I R = 0.754 \times 200 = 150.8 \; V$.
The voltage across the capacitor is $V_C = I X_C = 0.754 \times 212.3 = 160.1 \; V$.
The algebraic sum is $150.8 + 160.1 = 310.9 \; V$,which is greater than the source voltage of $220 \; V$. This is not a paradox because $V_R$ and $V_C$ are not in phase. $V_R$ is in phase with the current,while $V_C$ lags the current by $90^{\circ}$. The total voltage is the phasor sum: $V = \sqrt{V_R^2 + V_C^2} = \sqrt{150.8^2 + 160.1^2} \approx 220 \; V$.