A 10 μF capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (π = 3.14):

The correct answer is 0.93 A.

Step-by-step Explanation:

From the circuit diagram and the problem description, we are given the following values:
Capacitance, $C=10\mathrm{\mu F}=10\times {10}^{-6}F$
RMS Voltage of the AC source, $V_{\mathrm{rms}}=210V$
Frequency of the AC source, $f=50\mathrm{Hz}$
Value of $\pi =3.14$

1. Find the Capacitive Reactance ($X_C$):
The capacitive reactance of the circuit opposes the flow of alternating current and is given by the formula:
$X_C=\frac{1}{2\pi fC}$
Substituting the given values into the formula:
$X_C=\frac{1}{2\times 3.14\times 50\times 10\times {10}^{-6}}$
$X_C=\frac{1}{314\times {10}^{-5}}=\frac{{10}^5}{314}$
$X_C\approx 318.47\Omega$

2. Find the RMS Current ($I_{\mathrm{rms}}$):
Using Ohm's law for an AC capacitive circuit, the RMS current is:
$I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}$
Substitute the values of $V_{\mathrm{rms}}$ and $X_C$:
$I_{\mathrm{rms}}=\frac{210}{318.47}\approx 0.659A$

3. Calculate the Peak Current ($I_0$):
The relation between peak current ($I_0$) and RMS current ($I_{\mathrm{rms}}$) is given by:
$I_0=\sqrt{2}\times I_{\mathrm{rms}}$
Substitute the value of $I_{\mathrm{rms}}$ and $\sqrt{2}\approx 1.414$:
$I_0=1.414\times 0.659\approx 0.932A$
Thus, the peak current in the circuit is approximately 0.93 A.