What is the value of inductance L for which the current is maximum in a series LCR circuit with `C = 10 muF and omega = 1000 s^(-1)` ?

Options

A. 1 mH
B. Cannot be calculated unless R is know
C. 10 mH
D. 100 mH

Answer

To find the value of inductance \( L \) for which the current is maximum in a series LCR circuit, we can use the resonance condition. At resonance, the inductive reactance equals the capacitive reactance, and the current is at its maximum.

### Step-by-Step Solution:

1. **Understand the Resonance Condition**:
The current in an LCR circuit is maximum at resonance. The condition for resonance is given by:
\[ \omega^2 = \frac{1}{LC} \]
where:
- \( \omega \) is the angular frequency,
- \( L \) is the inductance,
- \( C \) is the capacitance.

2. **Rearranging the Formula**:
We can rearrange the resonance condition to solve for \( L \):
\[ L = \frac{1}{\omega^2 C} \]

3. **Substituting the Given Values**:
We are given:
- \( C = 10 \, \mu F = 10 \times 10^{-6} \, F \)
- \( \omega = 1000 \, s^{-1} \)

Now, substituting these values into the formula:
\[ L = \frac{1}{(1000)^2 \times (10 \times 10^{-6})} \]

4. **Calculating \( L \)**:
First, calculate \( (1000)^2 \):
\[ (1000)^2 = 1000000 \]
Now, substitute this back into the equation:
\[ L = \frac{1}{1000000 \times 10 \times 10^{-6}} = \frac{1}{1000000 \times 0.00001} = \frac{1}{0.01} = 100 \, H \]

5. **Final Result**:
Therefore, the inductance \( L \) is:
\[ L = 0.1 \, H \text{ or } 100 \, mH \]

### Conclusion:
The value of inductance \( L \) for which the current is maximum in the series LCR circuit is \( 0.1 \, H \) or \( 100 \, mH \).