A `6 mu F ` capacitor is connected to an AC source of frequency 100 Hz . Calculate the rms value of voltage if current in the circuit is 1.2 A. Also calculate the average energy stored in the capacitor.

Answer

To solve the problem step by step, we will first calculate the RMS voltage across the capacitor and then find the average energy stored in the capacitor.

### Step 1: Calculate the Capacitive Reactance (Xc)

The formula for capacitive reactance (Xc) is given by:

\[ X_c = \frac{1}{2 \pi f C} \]

Where:
- \( f = 100 \, \text{Hz} \) (frequency)
- \( C = 6 \, \mu F = 6 \times 10^{-6} \, F \) (capacitance)

Substituting the values into the formula:

\[ X_c = \frac{1}{2 \pi (100) (6 \times 10^{-6})} \]

Calculating this:

\[ X_c = \frac{1}{2 \times 3.14 \times 100 \times 6 \times 10^{-6}} \]

\[ X_c = \frac{1}{0.0003768} \approx 265.4 \, \Omega \]

### Step 2: Calculate the RMS Voltage (Vrms)

The RMS voltage across the capacitor can be calculated using the formula:

\[ V_{rms} = I_{rms} \times Z \]

Where:
- \( I_{rms} = 1.2 \, A \) (current)
- \( Z = X_c \) (impedance for a purely capacitive circuit)

Substituting the values:

\[ V_{rms} = 1.2 \times 265.4 \]

Calculating this:

\[ V_{rms} = 318.48 \, V \]

### Step 3: Calculate the Average Energy Stored in the Capacitor (E)

The average energy stored in a capacitor is given by the formula:

\[ E = \frac{1}{2} C V_{rms}^2 \]

Substituting the values:

\[ E = \frac{1}{2} (6 \times 10^{-6}) (318.48)^2 \]

Calculating \( (318.48)^2 \):

\[ (318.48)^2 \approx 101,439.47 \]

Now substituting back into the energy formula:

\[ E = \frac{1}{2} \times 6 \times 10^{-6} \times 101,439.47 \]

Calculating this:

\[ E \approx 0.31 \, J \]

### Final Answers

1. RMS Voltage \( V_{rms} \approx 318.48 \, V \)
2. Average Energy Stored \( E \approx 0.31 \, J \)

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