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 follow these steps:

### Step 1: Understand the Condition for Maximum Current
In a series LCR circuit, the current is maximum when the impedance \( Z \) is minimum. The condition for minimum impedance occurs at resonance, where the inductive reactance \( X_L \) is equal to the capacitive reactance \( X_C \).

### Step 2: Write the Expressions for Inductive and Capacitive Reactance
The inductive reactance \( X_L \) is given by:
\[ X_L = \omega L \]
The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{\omega C} \]

### Step 3: Set the Inductive Reactance Equal to the Capacitive Reactance
At resonance, we set \( X_L \) equal to \( X_C \):
\[ \omega L = \frac{1}{\omega C} \]

### Step 4: Solve for Inductance \( L \)
Rearranging the equation gives:
\[ L = \frac{1}{\omega^2 C} \]

### Step 5: Substitute the Given Values
We are given:
- \( C = 10 \, \mu F = 10 \times 10^{-6} \, F \)
- \( \omega = 1000 \, s^{-1} \)

Substituting these values into the equation:
\[ L = \frac{1}{(1000)^2 \times (10 \times 10^{-6})} \]

### Step 6: Calculate the Value of \( L \)
Calculating the denominator:
\[ (1000)^2 = 1000000 \]
\[ 1000000 \times 10 \times 10^{-6} = 10 \]
Thus,
\[ L = \frac{1}{10} = 0.1 \, H \]

### Step 7: Convert to Millihenry
Since \( 1 \, H = 1000 \, mH \):
\[ L = 0.1 \, H = 100 \, mH \]

### Final Answer
The value of inductance \( L \) for which the current is maximum in the series LCR circuit is:
\[ \boxed{100 \, mH} \]

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