An LCR series circuit consists of a resistance of a `10 Omega` a capacitance of reactance `60 Omega` and an inductor coil The circuit is found to resonate when put across a `300 V ,100 ` Hz supply The inductance of the coil is `(taken pi =3)` .

Options

A. 0.1 H
B. 0.01 H
C. 0.2 H
D. 0.02 H

Explanation

To solve the problem, we need to find the inductance \( L \) of the coil in an LCR series circuit that resonates at a given frequency. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand Resonance Condition
In an LCR circuit at resonance, the inductive reactance \( X_L \) is equal to the capacitive reactance \( X_C \). This can be expressed as:
\[ X_L = X_C \]
Given that \( X_C = 60 \, \Omega \), we have:
\[ X_L = 60 \, \Omega \]

### Step 2: Relate Inductive Reactance to Inductance
The inductive reactance \( X_L \) is related to the inductance \( L \) and the angular frequency \( \omega \) by the formula:
\[ X_L = \omega L \]
where \( \omega = 2 \pi f \) and \( f \) is the frequency in hertz.

### Step 3: Calculate Angular Frequency
The frequency \( f \) is given as \( 100 \, \text{Hz} \). Therefore, we can calculate \( \omega \):
\[ \omega = 2 \pi f = 2 \pi \times 100 \]
Using \( \pi \approx 3 \):
\[ \omega \approx 2 \times 3 \times 100 = 600 \, \text{rad/s} \]

### Step 4: Substitute Values to Find Inductance
Now we can substitute \( X_L \) and \( \omega \) into the equation for inductive reactance:
\[ 60 = 600 L \]
To find \( L \), rearrange the equation:
\[ L = \frac{60}{600} = 0.1 \, \text{H} \]

### Conclusion
The inductance of the coil is:
\[ L = 0.1 \, \text{H} \]

### Final Answer
The correct option is \( 0.1 \, \text{H} \).

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