A capacitor and resistor of 12`Omega` are connected in series with an AC supply 100 V, 50 Hz . Calculate the capacitance of the capacitor if power factor of the RC circuit is 0.8.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given values
We have:
- Resistance, \( R = 12 \, \Omega \)
- Voltage, \( V = 100 \, V \) (RMS)
- Frequency, \( f = 50 \, Hz \)
- Power factor, \( \text{pf} = 0.8 \)

### Step 2: Relate power factor to impedance
The power factor is defined as:
\[ \text{pf} = \cos(\phi) = \frac{R}{Z} \]
where \( Z \) is the impedance of the circuit. Rearranging gives:
\[ Z = \frac{R}{\text{pf}} = \frac{12}{0.8} = 15 \, \Omega \]

### Step 3: Relate impedance to resistive and reactive components
In an RC circuit, the impedance \( Z \) can be expressed as:
\[ Z = \sqrt{R^2 + X_C^2} \]
where \( X_C \) is the capacitive reactance. We can set up the equation:
\[ 15 = \sqrt{12^2 + X_C^2} \]

### Step 4: Solve for \( X_C \)
Squaring both sides:
\[ 15^2 = 12^2 + X_C^2 \]
\[ 225 = 144 + X_C^2 \]
\[ X_C^2 = 225 - 144 = 81 \]
\[ X_C = \sqrt{81} = 9 \, \Omega \]

### Step 5: Relate capacitive reactance to capacitance
The capacitive reactance is given by:
\[ X_C = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \). Thus:
\[ 9 = \frac{1}{2\pi \cdot 50 \cdot C} \]

### Step 6: Solve for \( C \)
Rearranging gives:
\[ C = \frac{1}{2\pi \cdot 50 \cdot 9} \]
Calculating this:
\[ C = \frac{1}{2 \cdot 3.14 \cdot 50 \cdot 9} \approx \frac{1}{2826} \approx 3.53 \times 10^{-4} \, \text{F} \]

### Step 7: Convert capacitance to microfarads
To convert farads to microfarads:
\[ C \approx 3.53 \times 10^{-4} \, \text{F} = 353 \, \mu F \]

### Final Answer
The capacitance of the capacitor is approximately \( 353 \, \mu F \).

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