Step 1: Recall the formula for capacitive reactance.
In an AC circuit with only a capacitor, the opposition offered to current flow is called capacitive reactance, given by $X_C = \frac{1}{\omega C}$, where $\omega$ is the angular frequency and $C$ is the capacitance. This is the AC equivalent of resistance for a capacitor.
Step 2: Find the angular frequency.
The angular frequency is related to the supply frequency $f$ by $\omega = 2\pi f$. Here $f = 50$ Hz, so \[ \omega = 2\pi \times 50 = 100\pi \text{ rad s}^{-1} \]
Step 3: Convert capacitance to SI units.
The given capacitance is $\frac{50}{\pi}$ microfarads. Since $1\ \mu\text{F} = 10^{-6}$ F, \[ C = \frac{50}{\pi} \times 10^{-6} \text{ F} = \frac{50 \times 10^{-6}}{\pi} \text{ F} \]
Step 4: Calculate the capacitive reactance.
Substituting into the formula, \[ X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times \dfrac{50 \times 10^{-6}}{\pi}} \] The $\pi$ in the numerator and denominator cancel, leaving \[ X_C = \frac{1}{100 \times 50 \times 10^{-6}} = \frac{1}{5 \times 10^{-3}} = 200\ \Omega \]
Step 5: Apply Ohm's law for AC circuits.
In an AC circuit, the rms current is related to the rms voltage and reactance by $I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}$. This is analogous to Ohm's law ($I = V/R$) but applies to the capacitive reactance.
Step 6: Calculate the rms current.
Given $V_{\text{rms}} = 250$ V and $X_C = 200\ \Omega$, \[ I_{\text{rms}} = \frac{250}{200} = 1.25\ \text{A} \] \[ \boxed{I_{\text{rms}} = 1.25\ \text{A}} \]