`2.5/(pi) muF` capacitor and `3000-`ohm resistance are joined in series to an `AC` source of `200 "volts"` and `50sec^(-1)` frequency. The power factor of the circuit and the power dissipated in it will respectively

Options

A. 0.6, 0.06W
B. 0.06, 0.6W
C. 0.6, 4.8W
D. 4.8, 06W

Answer

To solve the problem, we need to find the power factor and the power dissipated in a circuit consisting of a capacitor and a resistor connected in series to an AC source.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Capacitance, \( C = \frac{2.5}{\pi} \, \mu F = \frac{2.5}{\pi} \times 10^{-6} \, F \)
- Resistance, \( R = 3000 \, \Omega \)
- Voltage, \( V_{rms} = 200 \, V \)
- Frequency, \( f = 50 \, Hz \)

2. **Calculate the Capacitive Reactance (\( X_C \)):**
The formula for capacitive reactance is:
\[ X_C = \frac{1}{C \omega} \]
where \( \omega = 2\pi f \).
First, calculate \( \omega \):
\[ \omega = 2\pi \times 50 = 100\pi \, rad/s \]
Now, substitute \( C \) and \( \omega \) into the formula for \( X_C \):
\[ X_C = \frac{1}{\left(\frac{2.5}{\pi} \times 10^{-6}\right)(100\pi)} = \frac{1}{2.5 \times 10^{-4}} = 4000 \, \Omega \]

3. **Calculate the Power Factor (\( \cos \phi \)):**
The power factor is given by:
\[ \text{Power Factor} = \cos \phi = \cos(\tan^{-1}(\frac{X_C}{R})) \]
Substitute \( X_C \) and \( R \):
\[ \tan^{-1}(\frac{4000}{3000}) = \tan^{-1}(\frac{4}{3}) \]
Now calculate \( \cos \phi \):
\[ \cos \phi = \frac{R}{\sqrt{R^2 + X_C^2}} = \frac{3000}{\sqrt{3000^2 + 4000^2}} = \frac{3000}{\sqrt{9000000 + 16000000}} = \frac{3000}{5000} = 0.6 \]

4. **Calculate the Impedance (\( Z \)):**
The impedance in an RC circuit is given by:
\[ Z = \sqrt{R^2 + X_C^2} \]
Substitute the values:
\[ Z = \sqrt{3000^2 + 4000^2} = \sqrt{9000000 + 16000000} = \sqrt{25000000} = 5000 \, \Omega \]

5. **Calculate the RMS Current (\( I_{rms} \)):**
The RMS current can be calculated using:
\[ I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{5000} = 0.04 \, A \]

6. **Calculate the Power Dissipated (\( P \)):**
The power dissipated in the circuit is given by:
\[ P = V_{rms} \times I_{rms} \times \cos \phi \]
Substitute the values:
\[ P = 200 \times 0.04 \times 0.6 = 4.8 \, W \]

### Final Answers:
- Power Factor: \( 0.6 \)
- Power Dissipated: \( 4.8 \, W \)