What inductance should be connected in series with a capacitor of 20 mF and resistance of 10 `Omega` to make power factor unity ? Frequency of supply is 50 Hz.

Options

A. 2 H
B. 1 H
C. 0.5 H
D. 0.25 H

Explanation

To solve the problem of finding the inductance that should be connected in series with a capacitor of 20 mF and a resistance of 10 Ω to make the power factor unity at a frequency of 50 Hz, we can follow these steps:

### Step 1: Understand the Condition for Unity Power Factor
For the power factor to be unity, the inductive reactance (XL) must equal the capacitive reactance (XC). This means:
\[ XL - XC = 0 \]
or
\[ XL = XC \]

### Step 2: Calculate the Capacitive Reactance (XC)
The capacitive reactance (XC) is given by the formula:
\[ XC = \frac{1}{\omega C} \]
where:
- \( \omega = 2\pi f \) (angular frequency)
- \( C \) is the capacitance in farads.

Given:
- \( f = 50 \, \text{Hz} \)
- \( C = 20 \, \text{mF} = 20 \times 10^{-3} \, \text{F} \)

First, calculate \( \omega \):
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

Now, substitute the values to find \( XC \):
\[ XC = \frac{1}{100\pi \times 20 \times 10^{-3}} \]

### Step 3: Calculate the Value of XC
Calculating \( XC \):
\[ XC = \frac{1}{100\pi \times 0.02} \]
\[ XC = \frac{1}{2\pi} \]
\[ XC \approx 0.159 \, \Omega \]

### Step 4: Calculate the Inductive Reactance (XL)
Since \( XL = XC \), we can use the same value for inductive reactance:
\[ XL = XC \approx 0.159 \, \Omega \]

### Step 5: Calculate the Inductance (L)
Inductive reactance is given by:
\[ XL = \omega L \]
Rearranging gives:
\[ L = \frac{XL}{\omega} \]

Substituting the values:
\[ L = \frac{0.159}{100\pi} \]

### Step 6: Calculate the Value of L
Calculating \( L \):
\[ L = \frac{0.159}{100\pi} \approx \frac{0.159}{314.16} \]
\[ L \approx 0.000506 \, \text{H} \]
\[ L \approx 0.506 \, \text{mH} \]

### Final Answer
The inductance that should be connected in series is approximately:
\[ L \approx 0.506 \, \text{mH} \]

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