A circuit draws a power of 550 watt from a source of 220 volt, `50 Hz`. The power factor of the circuit is 0.8 and the current lags in phase behind the potential difference. To make the power factor of the circuit as 1.0, The capacitance should be connected in series with it is

Options

A. `75 mu F`
B. `60 mu F`
C. `50 mu F`
D. `65 mu F`

Explanation

To solve the problem step by step, we will calculate the necessary capacitance to make the power factor of the circuit equal to 1.0.

### Step 1: Calculate the impedance (Z) of the circuit
The formula for power in an AC circuit is given by:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos \phi \]
We can rearrange this to find the impedance:
\[ Z = \frac{V_{rms}^2}{P \cdot \cos \phi} \]

Given:
- \( P = 550 \, \text{W} \)
- \( V_{rms} = 220 \, \text{V} \)
- \( \cos \phi = 0.8 \)

Substituting the values:
\[ Z = \frac{(220)^2}{550 \cdot 0.8} \]
\[ Z = \frac{48400}{440} \]
\[ Z = 110 \, \Omega \]

### Step 2: Calculate the resistance (R) of the circuit
Using the relationship between power factor, resistance, and impedance:
\[ \cos \phi = \frac{R}{Z} \]
We can rearrange this to find R:
\[ R = Z \cdot \cos \phi \]
Substituting the values:
\[ R = 110 \cdot 0.8 \]
\[ R = 88 \, \Omega \]

### Step 3: Calculate the inductive reactance (X_L)
Using the impedance formula:
\[ Z^2 = R^2 + X_L^2 \]
We can rearrange this to find \( X_L \):
\[ X_L = \sqrt{Z^2 - R^2} \]
Substituting the values:
\[ X_L = \sqrt{(110)^2 - (88)^2} \]
\[ X_L = \sqrt{12100 - 7744} \]
\[ X_L = \sqrt{4356} \]
\[ X_L = 66 \, \Omega \]

### Step 4: Calculate the capacitive reactance (X_C) needed for power factor 1
At power factor 1, the inductive reactance (X_L) must equal the capacitive reactance (X_C):
\[ X_C = X_L = 66 \, \Omega \]

### Step 5: Calculate the capacitance (C)
The capacitive reactance is given by:
\[ X_C = \frac{1}{\omega C} \]
Where \( \omega = 2 \pi f \) and \( f = 50 \, \text{Hz} \):
\[ \omega = 2 \pi \cdot 50 = 100 \pi \, \text{rad/s} \]
Now substituting for \( C \):
\[ C = \frac{1}{\omega X_C} \]
Substituting the values:
\[ C = \frac{1}{100 \pi \cdot 66} \]
\[ C = \frac{1}{6600 \pi} \]
Calculating the value:
\[ C \approx \frac{1}{20735.6} \approx 4.82 \times 10^{-5} \, \text{F} \]
Converting to microfarads:
\[ C \approx 48.2 \, \mu F \]

### Final Answer
To make the power factor of the circuit equal to 1.0, a capacitance of approximately **48.2 microfarads** should be connected in series with it.