A series LCR circuit containing a resistance of `120 Omega` has angular resonance frequency `4 xx 10^(5) rad s^(-1)`. At resonance the vlotage across resistance and inductance are 60V and 40 V, repectively,
the value of inductance L is

Options

A. `0.1 mH`
B. `0.2 mH`
C. `0.35 mH`
D. `0.4 mH`

Explanation

To find the value of inductance \( L \) in the given series LCR circuit, we can follow these steps:

### Step 1: Write down the given values
We have the following values from the problem statement:
- Resistance \( R = 120 \, \Omega \)
- Angular resonance frequency \( \omega = 4 \times 10^5 \, \text{rad/s} \)
- Voltage across the resistance \( V_R = 60 \, \text{V} \)
- Voltage across the inductance \( V_L = 40 \, \text{V} \)

### Step 2: Calculate the current \( I \) through the circuit
At resonance, the voltage across the resistance can be expressed using Ohm's law:
\[ V_R = I \cdot R \]
Substituting the known values:
\[ 60 = I \cdot 120 \]
To find \( I \):
\[ I = \frac{60}{120} = \frac{1}{2} \, \text{A} \]

### Step 3: Use the voltage across the inductance to find \( L \)
The voltage across the inductance \( V_L \) can be expressed as:
\[ V_L = I \cdot \omega L \]
Substituting the known values:
\[ 40 = \left(\frac{1}{2}\right) \cdot (4 \times 10^5) \cdot L \]
This simplifies to:
\[ 40 = 2 \times 10^5 \cdot L \]

### Step 4: Solve for \( L \)
Rearranging the equation to isolate \( L \):
\[ L = \frac{40}{2 \times 10^5} = \frac{40}{200000} = 0.0002 \, \text{H} \]
Converting to millihenries:
\[ L = 0.0002 \, \text{H} = 0.2 \, \text{mH} \]

### Final Answer
The value of inductance \( L \) is \( 0.2 \, \text{mH} \).

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