QJEEM24Phdc8e: JEE Main 2024 Physics - Class 12 JEEM24Phdc8e Solution: An alternating emf volt is applied to a capacitor of , the rms…

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.

The given alternating emf is:

$\mathrm{E} = 110 \sqrt{2} \sin 100 \mathrm{t} \, \text{volts}$

Here, the peak voltage (or maximum voltage) $\mathrm{E_{max}}$ is:

$\mathrm{E_{max}} = 110 \sqrt{2} \, \text{volts}$

Next, the RMS value of the voltage, $\mathrm{E_{rms}}$ , is obtained by dividing the peak voltage by $\sqrt{2}$ :

$\mathrm{E_{rms}} = \frac{\mathrm{E_{max}}}{\sqrt{2}} = \frac{110 \sqrt{2}}{\sqrt{2}} = 110 \, \text{volts}$

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:

$\mathrm{X_C} = \frac{1}{\omega C}$

where $\omega$ is the angular frequency. From the given formula for emf, we see that:

$\omega = 100 \, \text{rad/s}$

Therefore, the capacitive reactance is:

$\mathrm{X_C} = \frac{1}{100 \times (2 \times 10^{-6})} = \frac{1}{200 \times 10^{-6}} = 5000 \, \Omega$

Now, we can calculate the RMS value of the current $\mathrm{I_{rms}}$ using Ohm's law for AC circuits, which states:

$\mathrm{I_{rms}} = \frac{\mathrm{E_{rms}}}{\mathrm{X_C}}$

Substituting the known values:

$\mathrm{I_{rms}} = \frac{110}{5000} = 0.022 \, \text{A} = 22 \, \text{mA}$

Hence, the RMS value of the current in the circuit is $22 \, \text{mA}$ .