The time constant of L-R circuit is doubled if

Options

A. Both Land R become two times
B. L becomes four times and R becomes two times
C. L becomes two times and R becomes four times
D. L becomes two times and R becomes eight times

Answer

To solve the problem regarding the time constant of an L-R circuit and how it can be doubled, we need to understand the formula for the time constant in an L-R circuit.

### Step-by-Step Solution:

1. **Understand the Time Constant Formula**:
The time constant (τ) of an L-R circuit is given by the formula:
\[ \tau = \frac{L}{R} \]
where \(L\) is the inductance and \(R\) is the resistance.

2. **Identify the Condition for Doubling**:
We need to find conditions under which the time constant τ is doubled. This means we want:
\[ \tau' = 2\tau \]
Therefore, we need:
\[ \frac{L'}{R'} = 2 \cdot \frac{L}{R} \]

3. **Evaluate Each Option**:
We will evaluate each option provided in the question to see which one satisfies the condition of doubling the time constant.

- **Option 1**: Both \(L\) and \(R\) become two times.
\[ \tau' = \frac{2L}{2R} = \frac{L}{R} = \tau \]
This does not double the time constant. **(Incorrect)**

- **Option 2**: \(L\) becomes 4 times, \(R\) becomes 2 times.
\[ \tau' = \frac{4L}{2R} = 2 \cdot \frac{L}{R} = 2\tau \]
This doubles the time constant. **(Correct)**

- **Option 3**: \(L\) becomes 2 times, \(R\) becomes 4 times.
\[ \tau' = \frac{2L}{4R} = \frac{L}{2R} = \frac{1}{2}\tau \]
This does not double the time constant. **(Incorrect)**

- **Option 4**: \(L\) becomes 2 times, \(R\) becomes 8 times.
\[ \tau' = \frac{2L}{8R} = \frac{L}{4R} = \frac{1}{4}\tau \]
This does not double the time constant. **(Incorrect)**

4. **Conclusion**:
The only option that results in the time constant being doubled is **Option 2**.

### Final Answer:
The time constant of the L-R circuit is doubled if \(L\) becomes 4 times and \(R\) becomes 2 times. Thus, the correct answer is Option 2.

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