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` times
D. `2` times

Explanation

To solve the problem, we need to analyze the relationship between inductance (L) and capacitance (C) in a resonant LCR circuit. The resonance condition in an LCR circuit is given by the equation:

\[ X_L = X_C \]

Where:
- \( X_L = \omega L \) (inductive reactance)
- \( X_C = \frac{1}{\omega C} \) (capacitive reactance)

At resonance, we have:

\[ \omega L = \frac{1}{\omega C} \]

From this, we can derive the relationship:

\[ \omega^2 = \frac{1}{LC} \]

Now, according to the problem, the capacitance is changed to one-fourth of its original value:

\[ C' = \frac{C}{4} \]

We need to find the new inductance \( L' \) that will keep the circuit in resonance with the new capacitance \( C' \).

1. **Substituting the new capacitance into the resonance condition:**

The new resonance condition can be expressed as:

\[ \omega^2 = \frac{1}{L'C'} \]

Substituting \( C' \):

\[ \omega^2 = \frac{1}{L' \left(\frac{C}{4}\right)} \]

This simplifies to:

\[ \omega^2 = \frac{4}{L'C} \]

2. **Setting the two expressions for \( \omega^2 \) equal to each other:**

From the original resonance condition, we have:

\[ \omega^2 = \frac{1}{LC} \]

Equating the two expressions for \( \omega^2 \):

\[ \frac{1}{LC} = \frac{4}{L'C} \]

3. **Solving for the new inductance \( L' \):**

We can cancel \( C \) from both sides (assuming \( C \neq 0 \)):

\[ \frac{1}{L} = \frac{4}{L'} \]

Rearranging gives:

\[ L' = 4L \]

Thus, the new inductance \( L' \) must be four times the original inductance \( L \) to maintain resonance with the new capacitance.

**Final Answer:**
The change in inductance should be \( L' = 4L \).

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