Question 79: AC fundamentals

Question

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 ________ $$\mathrm{mA}$$.

Accepted Answer

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:
$$\mathrm{E} = 110 \sqrt{2} \sin 100 \mathrm{t} \, 	ext{volts}$$
Here, the peak voltage (or maximum voltage) $$\mathrm{E_{max}}$$ is:
$$\mathrm{E_{max}} = 110 \sqrt{2} \, 	ext{volts}$$
Next, the RMS value of the voltage, $$\mathrm{E_{rms}}$$, is obtained by dividing the peak voltage by $$\sqrt{2}$$:
$$\mathrm{E_{rms}} = \frac{\mathrm{E_{max}}}{\sqrt{2}} = \frac{110 \sqrt{2}}{\sqrt{2}} = 110 \, 	ext{volts}$$
We are given a capacitor with a capacitance $$C = 2 \mu \mathrm{F} = 2 	imes 10^{-6} \, 	ext{F}$$ 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:
$$\omega = 100 \, 	ext{rad/s}$$
Therefore, the capacitive reactance is:
$$\mathrm{X_C} = \frac{1}{100 	imes (2 	imes 10^{-6})} = \frac{1}{200 	imes 10^{-6}} = 5000 \, \Omega$$
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:
$$\mathrm{I_{rms}} = \frac{110}{5000} = 0.022 \, 	ext{A} = 22 \, 	ext{mA}$$
Hence, the RMS value of the current in the circuit is $$22 \, 	ext{mA}$$.

AC fundamentals

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

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