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Step-by-Step Solution
Step 1: Identify the given parameters
The alternating voltage is given as
$e = 200\sqrt{2}\,\sin(100t)\,\text{volts}$.
From this expression:
The maximum (peak) voltage is $e_{0} = 200\sqrt{2}$ volts.
The angular frequency is $\omega = 100 \,\text{rad s}^{-1}$.
The capacitance of the capacitor is $C = 1\,\mu\text{F} = 1 \times 10^{-6}\,\text{F}$.
Step 2: Write down the formula for capacitive reactance
The capacitive reactance $X_{\mathrm{C}}$ is given by:
$X_{\mathrm{C}} = \frac{1}{\omega C}.$
Step 3: Calculate the capacitive reactance
Substitute $\omega = 100\,\text{rad s}^{-1}$ and $C = 1 \times 10^{-6}\,\text{F}$ into the reactance formula:
$X_{\mathrm{C}}
= \frac{1}{(100)\,(1 \times 10^{-6})}
= \frac{1}{1 \times 10^{-4}}
= 10^{4}\,\Omega
= 10000\,\Omega.$
Step 4: Relate RMS voltage and peak voltage
The RMS value of the sinusoidal voltage is:
$V_{\mathrm{rms}} = \frac{e_{0}}{\sqrt{2}}.$
Here,
$e_{0} = 200\sqrt{2}\,\text{V}$,
so
$V_{\mathrm{rms}}
= \frac{200\sqrt{2}}{\sqrt{2}}
= 200\,\text{V}.$
Step 5: Find the RMS current in the circuit
For a purely capacitive AC circuit, the RMS current is given by:
$i_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{X_{\mathrm{C}}}.$
Substituting the values:
$i_{\mathrm{rms}}
= \frac{200}{10^{4}}
= 0.02\,\text{A}
= 20\,\text{mA}.$
Step 6: State the final answer
The RMS value of the current in the circuit is 20 mA.
