Q.A 15.0 $\mu F$ capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current?

[!TLDR] $X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi(50)(15\times10^{-6})}\approx212.1\,\Omega$; $i_{rms}=V/X_C=1.037$ A, $i_0=\sqrt2\,i_{rms}=1.465$ A. Doubling f halves $X_C$ and doubles the current. [!ANSWER] $X_C=212.1\,\Omega$, $i_{rms}=1.037$ A, $i_0=1.465$ A; doubling f halves $X_C$ and doubles the current.

Given C = 15.0 $\mu F$, $V_{rms}=220$ V, f = 50 Hz. $X_C=\dfrac{1}{2\pi fC}=\dfrac{1}{2\times3.1416\times50\times15\times10^{-6}}=\dfrac{1}{4.712\times10^{-3}}\approx212.1\,\Omega$.\n\n$i_{rms}=\dfrac{V_{rms}}{X_C}=\dfrac{220}{212.1}\approx1.037$ A, and $i_0=\sqrt2\,i_{rms}=1.414\times1.037\approx1.466$ A ($\approx1.465$ A).\n\nIf the frequency is doubled ($f\to2f$), since $X_C=\dfrac{1}{2\pi fC}\propto\dfrac{1}{f}$, the reactance is HALVED. With the same rms voltage, the rms current $i_{rms}=V_{rms}/X_C$ is then DOUBLED. [!ANSWER] $X_C\approx212.1\,\Omega$, $i_{rms}\approx1.037$ A, $i_0\approx1.465$ A; doubling the frequency halves $X_C$ and doubles the current.

Compute $X_C=1/(2\pi fC)$, then $i_{rms}=V_{rms}/X_C$ and $i_0=\sqrt2\,i_{rms}$; use the inverse proportionality $X_C\propto1/f$ to reason about the effect of doubling f.

Forgetting that $i_0$ and $i_{rms}$ must both be reported (the question asks for both), or reasoning incorrectly that doubling frequency doubles $X_C$ (it is inversely, not directly, proportional for a capacitor).