A circuit draws 330 W from a 110 V , 60 Hz AC line. The power factor is 0.6 and the current lags the voltage. The capacitance of a series capacitor that will result in a power factor of unity is equal to

Options

A. 31 `mu`F
B. 54 `mu`F
C. 151 `mu`F
D. 201 `mu`F

Explanation

To solve the problem step by step, we will follow a structured approach:

### Step 1: Calculate the Resistance (R)
Given:
- Power (P) = 330 W
- Voltage (V) = 110 V

Using the formula for power in an AC circuit:
\[ P = V \cdot I \cdot \text{Power Factor} \]
We can express the current (I) as:
\[ I = \frac{P}{V \cdot \text{Power Factor}} \]
The power factor is given as 0.6. Therefore, we can calculate the current:
\[ I = \frac{330}{110 \cdot 0.6} = \frac{330}{66} = 5 \text{ A} \]

Now, we can find the resistance using Ohm's law:
\[ R = \frac{V}{I} = \frac{110}{5} = 22 \, \Omega \]

### Step 2: Relate Power Factor to Resistance and Reactance
The power factor (PF) is given by:
\[ \text{PF} = \cos \phi = \frac{R}{Z} \]
Where \(Z\) is the impedance. Rearranging gives:
\[ Z = \frac{R}{\text{PF}} = \frac{22}{0.6} \approx 36.67 \, \Omega \]

### Step 3: Calculate the Inductive Reactance (X_L)
Using the relationship between resistance, reactance, and impedance:
\[ Z^2 = R^2 + X_L^2 \]
Substituting the known values:
\[ (36.67)^2 = (22)^2 + X_L^2 \]
Calculating:
\[ 1344.89 = 484 + X_L^2 \]
\[ X_L^2 = 1344.89 - 484 = 860.89 \]
\[ X_L = \sqrt{860.89} \approx 29.35 \, \Omega \]

### Step 4: Determine the Required Capacitive Reactance (X_C)
To achieve a power factor of unity, we need:
\[ X_L = X_C \]
Thus:
\[ X_C = 29.35 \, \Omega \]

### Step 5: Calculate the Capacitance (C)
The capacitive reactance is given by:
\[ X_C = \frac{1}{2 \pi f C} \]
Rearranging gives:
\[ C = \frac{1}{2 \pi f X_C} \]
Substituting \(f = 60 \, \text{Hz}\) and \(X_C = 29.35 \, \Omega\):
\[ C = \frac{1}{2 \pi \cdot 60 \cdot 29.35} \]
Calculating:
\[ C = \frac{1}{2 \cdot 3.14 \cdot 60 \cdot 29.35} \approx \frac{1}{2 \cdot 3.14 \cdot 1761} \approx \frac{1}{11079.48} \approx 9.03 \times 10^{-5} \, \text{F} = 90.3 \, \mu\text{F} \]

### Final Answer
The capacitance of the series capacitor that will result in a power factor of unity is approximately:
\[ C \approx 90.3 \, \mu\text{F} \]