In a series LCR circuit with variable inductor 80 nF capacitor and 70 ohm resistance an AC source is connected . Calculate the value of inductor if current drawn into the circuit is maximum. The frequency of AC source is 1 kHz.

Answer

To solve the problem of finding the value of the inductor in a series LCR circuit where the current is maximum, 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 impedance \( Z \) is given by the formula:
\[ Z = \sqrt{(X_L - X_C)^2 + R^2} \]
where \( X_L \) is the inductive reactance, \( X_C \) is the capacitive reactance, and \( R \) is the resistance.

### Step 2: Set the condition for minimum impedance
For the impedance to be minimum, the inductive reactance \( X_L \) must equal the capacitive reactance \( X_C \):
\[ X_L = X_C \]

### Step 3: Write the formulas for reactances
The inductive reactance \( X_L \) is given by:
\[ X_L = \omega L \]
The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{\omega C} \]
where \( \omega = 2\pi f \) and \( f \) is the frequency of the AC source.

### Step 4: Substitute the expressions for reactances
Setting \( X_L \) equal to \( X_C \):
\[ \omega L = \frac{1}{\omega C} \]

### Step 5: Solve for inductance \( L \)
Rearranging the equation gives:
\[ L = \frac{1}{\omega^2 C} \]
Substituting \( \omega = 2\pi f \):
\[ L = \frac{1}{(2\pi f)^2 C} \]

### Step 6: Substitute the known values
Given:
- \( f = 1 \, \text{kHz} = 1000 \, \text{Hz} \)
- \( C = 80 \, \text{nF} = 80 \times 10^{-9} \, \text{F} \)

Now substituting these values into the formula:
\[ L = \frac{1}{(2\pi \times 1000)^2 \times (80 \times 10^{-9})} \]

### Step 7: Calculate the value of \( L \)
First, calculate \( (2\pi \times 1000)^2 \):
\[ (2\pi \times 1000)^2 = (2000\pi)^2 = 4000000\pi^2 \]
Now, substituting this back into the formula for \( L \):
\[ L = \frac{1}{4000000\pi^2 \times 80 \times 10^{-9}} \]

Calculating \( L \):
\[ L = \frac{1}{320000000\pi^2 \times 10^{-9}} = \frac{10^9}{320000000\pi^2} \]

Using \( \pi^2 \approx 9.87 \):
\[ L \approx \frac{10^9}{320000000 \times 9.87} \approx \frac{10^9}{3168000000} \approx 0.317 \, \text{H} \]

### Final Answer
Thus, the value of the inductor \( L \) is approximately:
\[ L \approx 0.317 \, \text{H} \]

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