An alternating emf $\mathrm{E}=110 \sqrt{2} \sin 100 \mathrm{t}$ volt is applied to a capacitor of $2 \mu \mathrm{F}$, the rms value of current in the circuit is ________ $\mathrm{mA}$.

<p>To determine the RMS (Root Mean Square) value of the current in the circuit, we start by analyzing the given emf and the capacitive reactance.</p> <p>The given alternating emf is:</p> <p> $\mathrm{E} = 110 \sqrt{2} \sin 100 \mathrm{t} \, \text{volts}$ </p> <p>Here, the peak voltage (or maximum voltage) $\mathrm{E_{max}}$ is:</p> <p> $\mathrm{E_{max}} = 110 \sqrt{2} \, \text{volts}$ </p> <p>Next, the RMS value of the voltage, $\mathrm{E_{rms}}$, is obtained by dividing the peak voltage by $\sqrt{2}$:</p> <p> $$\mathrm{E_{rms}} = \frac{\mathrm{E_{max}}}{\sqrt{2}} = \frac{110 \sqrt{2}}{\sqrt{2}} = 110 \, \text{volts}$$ </p> <p>We are given a capacitor with a capacitance $C = 2 \mu \mathrm{F} = 2 \times 10^{-6} \, \text{F}$ and we need to determine the RMS current. The capacitive reactance $\mathrm{X_C}$ is given by:</p> <p> $\mathrm{X_C} = \frac{1}{\omega C}$ </p> <p>where $\omega$ is the angular frequency. From the given formula for emf, we see that:</p> <p> $\omega = 100 \, \text{rad/s}$ </p> <p>Therefore, the capacitive reactance is:</p> <p> $$\mathrm{X_C} = \frac{1}{100 \times (2 \times 10^{-6})} = \frac{1}{200 \times 10^{-6}} = 5000 \, \Omega$$ </p> <p>Now, we can calculate the RMS value of the current $\mathrm{I_{rms}}$ using Ohm's law for AC circuits, which states:</p> <p> $\mathrm{I_{rms}} = \frac{\mathrm{E_{rms}}}{\mathrm{X_C}}$ </p> <p>Substituting the known values:</p> <p> $\mathrm{I_{rms}} = \frac{110}{5000} = 0.022 \, \text{A} = 22 \, \text{mA}$ </p> <p>Hence, the RMS value of the current in the circuit is $22 \, \text{mA}$.</p>