In a series `RC` circuit with an `AC` source (peak voltage `E_(0)=50 V` and `f=50//pi Hz),R=300 Omega`.`C=25 muF`.Then:

Options

A. the peak current is `0.1 A`
B. the peak current is `0.7 A`
C. the average power dissipated is `1.5 W`
D. the average power dissipated is `3 W`

Answer

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

### Given Data:
- Peak Voltage, \( E_0 = 50 \, V \)
- Frequency, \( f = \frac{50}{\pi} \, Hz \)
- Resistance, \( R = 300 \, \Omega \)
- Capacitance, \( C = 25 \, \mu F = 25 \times 10^{-6} \, F \)

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

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

### Step 2: Calculate the impedance \( Z \)
The 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 \( I_0 \) 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 \( P \)
The average power \( P \) in an AC circuit is given by:
\[ P = V_{rms} \times I_{rms} \times \cos \phi \]
Where \( V_{rms} = \frac{E_0}{\sqrt{2}} \) and \( I_{rms} = \frac{I_0}{\sqrt{2}} \).

First, calculate \( V_{rms} \):
\[ V_{rms} = \frac{50}{\sqrt{2}} \approx 35.36 \, V \]

Next, calculate \( I_{rms} \):
\[ I_{rms} = \frac{0.1}{\sqrt{2}} \approx 0.0707 \, A \]

Now, calculate \( \cos \phi \):
\[ \cos \phi = \frac{R}{Z} = \frac{300}{500} = 0.6 \]

Now substitute these values into the power formula:
\[ P = 35.36 \times 0.0707 \times 0.6 \]
Calculating this gives:
\[ P \approx 1.5 \, W \]

### Final Answers:
- Peak Current \( I_0 = 0.1 \, A \)
- Average Power \( P \approx 1.5 \, W \)

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