In an LCR circuit, the capacitance is made one-fourth, when in resonance. Then what should be the change in inductance, so that the circuit remains in resonance?

Options

A. 4 times
B. `1/4` times
C. 8 time
D. 2 times

Answer

To solve the problem, we need to understand the resonance condition in an LCR circuit. The resonance occurs when the inductive reactance (X_L) equals the capacitive reactance (X_C).

### Step-by-Step Solution:

1. **Understanding Resonance Condition**:
The resonance condition in an LCR circuit is given by:
\[ X_L = X_C \]
where \( X_L = \omega L \) and \( X_C = \frac{1}{\omega C} \).

2. **Setting the Equation**:
At resonance:
\[ \omega L = \frac{1}{\omega C} \]
Rearranging gives:
\[ L = \frac{1}{\omega^2 C} \]

3. **Change in Capacitance**:
According to the problem, the capacitance \( C \) is made one-fourth:
\[ C' = \frac{C}{4} \]

4. **Finding New Inductance**:
We need to find the new inductance \( L' \) such that the circuit remains in resonance with the new capacitance \( C' \):
\[ L' = \frac{1}{\omega^2 C'} \]
Substituting \( C' \):
\[ L' = \frac{1}{\omega^2 \left(\frac{C}{4}\right)} = \frac{4}{\omega^2 C} \]

5. **Relating New Inductance to Original Inductance**:
Since \( L = \frac{1}{\omega^2 C} \), we can express \( L' \) in terms of \( L \):
\[ L' = 4L \]

6. **Conclusion**:
Therefore, to keep the circuit in resonance after reducing the capacitance to one-fourth, the inductance must be increased to four times its original value:
\[ L' = 4L \]

### Final Answer:
The inductance should be changed to four times its original value.

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