In a series `RC` circuit with an AC source, `R = 300 Omega, C = 25 muF, epsilon_0= 50 V` and `v = 50/(pi) Hz`. Find the peak current and the average power dissipated in the circuit.

Options

A. the `rms` current in the circuit is `0.1 A`
B. the `rms` potential difference across the capacitor is `50 V`
C. the `rms` potential difference across the capacitor is `14.1 V`
D. the `rms` current in the circuit is `0.14 A`

Answer

To solve the problem step by step, we will calculate the peak current and the average power dissipated in the given series RC circuit with an AC source.

### Given Data:
- Resistance, \( R = 300 \, \Omega \)
- Capacitance, \( C = 25 \, \mu F = 25 \times 10^{-6} \, F \)
- Peak voltage, \( \epsilon_0 = 50 \, V \)
- Frequency, \( f = \frac{50}{\pi} \, Hz \)

### Step 1: Calculate the capacitive reactance \( X_C \)

The formula for capacitive reactance is given by:
\[ X_C = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \).

First, we calculate \( \omega \):
\[ \omega = 2\pi f = 2\pi \left(\frac{50}{\pi}\right) = 100 \, rad/s \]

Now substituting \( \omega \) and \( C \) into the formula for \( X_C \):
\[ X_C = \frac{1}{\omega C} = \frac{1}{100 \times 25 \times 10^{-6}} = \frac{1}{0.0025} = 400 \, \Omega \]

### Step 2: Calculate the impedance \( Z \)

The total impedance \( Z \) in a series RC circuit is given by:
\[ Z = \sqrt{R^2 + X_C^2} \]
Substituting the values of \( R \) and \( X_C \):
\[ Z = \sqrt{300^2 + 400^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega \]

### Step 3: Calculate the peak current \( I_0 \)

The peak current can be calculated using Ohm's law:
\[ I_0 = \frac{E_0}{Z} \]
Substituting the values:
\[ I_0 = \frac{50}{500} = 0.1 \, A \]

### Step 4: Calculate the average power \( P_{avg} \)

The average power in an AC circuit is given by:
\[ P_{avg} = E_{rms} \cdot I_{rms} \cdot \cos \phi \]
where \( E_{rms} = \frac{E_0}{\sqrt{2}} \) and \( I_{rms} = \frac{I_0}{\sqrt{2}} \).

First, we need to calculate \( \cos \phi \):
\[ \cos \phi = \frac{R}{Z} = \frac{300}{500} = 0.6 \]

Now substituting \( E_{rms} \), \( I_{rms} \), and \( \cos \phi \) into the power formula:
\[ P_{avg} = \left(\frac{50}{\sqrt{2}}\right) \cdot \left(\frac{0.1}{\sqrt{2}}\right) \cdot 0.6 \]
Calculating this step by step:
\[ P_{avg} = \frac{50 \cdot 0.1 \cdot 0.6}{2} = \frac{3}{2} = 1.5 \, W \]

### Final Answers:
- Peak Current \( I_0 = 0.1 \, A \)
- Average Power \( P_{avg} = 1.5 \, W \)