An AC circuit drawing power from a source of angular frequency `"50 rad s"^(-1)` has a power factor of 0.6. In this condition, a resistance of `100Omega` is present and the current is lagging behind the voltage. If a capacitor is connected in series, then the required capacitance that will result in a power factor of unity is

Options

A. `30muF`
B. `150muF`
C. `50muF`
D. `200muF`

Answer

To solve the problem, we need to determine the required capacitance that will result in a power factor of unity (1) when a capacitor is connected in series with a circuit that has a power factor of 0.6, a resistance of 100Ω, and an angular frequency of 50 rad/s.

### Step-by-Step Solution:

1. **Understand the Power Factor**:
The power factor (PF) is given by the formula:
\[ \text{PF} = \cos(\phi) = \frac{R}{Z} \]
where \( R \) is the resistance and \( Z \) is the impedance.

2. **Calculate the Impedance**:
Given that the power factor is 0.6 and the resistance \( R = 100 \, \Omega \):
\[ 0.6 = \frac{100}{Z} \implies Z = \frac{100}{0.6} \approx 166.67 \, \Omega \]

3. **Relate Impedance to Reactance**:
The impedance \( Z \) can also be expressed as:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where \( X_L \) is the inductive reactance and \( X_C \) is the capacitive reactance. Since the current lags behind the voltage, we have \( X_L > X_C \).

4. **Set Up the Equation**:
Plugging in the values we have:
\[ 166.67 = \sqrt{100^2 + (X_L - X_C)^2} \]
Squaring both sides:
\[ 166.67^2 = 10000 + (X_L - X_C)^2 \]
\[ (X_L - X_C)^2 = 166.67^2 - 10000 \]

5. **Calculate \( X_L - X_C \)**:
\[ 166.67^2 \approx 27777.78 \]
\[ (X_L - X_C)^2 = 27777.78 - 10000 = 17777.78 \]
\[ X_L - X_C \approx \sqrt{17777.78} \approx 133.33 \, \Omega \]

6. **Condition for Power Factor of Unity**:
For the power factor to be unity, we need:
\[ X_L = X_C \]
Therefore, we can express this as:
\[ X_L - X_C = 0 \implies X_C = X_L \]

7. **Calculate the Required Capacitive Reactance**:
From the previous step, we have:
\[ X_C = X_L - (X_L - X_C) = X_L - 133.33 \]
Since \( X_C = \frac{1}{\omega C} \), we can set:
\[ X_C = \frac{1}{50C} \]

8. **Set Up the Equation for Capacitance**:
Setting the two expressions for \( X_C \) equal gives:
\[ \frac{1}{50C} = X_L - 133.33 \]
Rearranging gives:
\[ C = \frac{1}{50(X_L - 133.33)} \]

9. **Substituting \( X_L \)**:
To find \( X_L \), we can use the relation:
\[ X_L = \sqrt{Z^2 - R^2} = \sqrt{166.67^2 - 100^2} \approx \sqrt{17777.78} \approx 133.33 \, \Omega \]
Thus, substituting \( X_L \) back into the capacitance equation:
\[ C = \frac{1}{50 \times 133.33} \]
\[ C \approx \frac{1}{6666.67} \approx 1.5 \times 10^{-4} \, F \]

10. **Convert to Microfarads**:
\[ C \approx 150 \, \mu F \]

### Final Answer:
The required capacitance that will result in a power factor of unity is approximately **150 microfarads**.