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 capacitance C is

Options

A. `(1)/(32)(mu)F`
B. `(1)/(16)(mu)F`
C. `32(mu)F`
D. `16(mu)F`

Explanation

To find the value of capacitance \( C \) in a series LCR circuit, we can follow these steps:

### Step 1: Given Values
We have the following values from the problem:
- 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 \( I \)
At resonance, the voltage across the resistance can be expressed using Ohm's Law:
\[ V_R = I \cdot R \]
Substituting the known values:
\[ 60 = I \cdot 120 \]
Solving for \( I \):
\[ I = \frac{60}{120} = \frac{1}{2} \, \text{A} \]

### Step 3: Calculate the Inductance \( L \)
The voltage across the inductor can be expressed as:
\[ V_L = I \cdot \omega L \]
Rearranging this gives:
\[ L = \frac{V_L}{I \cdot \omega} \]
Substituting the known values:
\[ L = \frac{40}{\left(\frac{1}{2}\right) \cdot (4 \times 10^5)} \]
Calculating:
\[ L = \frac{40}{2 \times 4 \times 10^5} = \frac{40}{8 \times 10^5} = \frac{5}{10^5} = 5 \times 10^{-6} \, \text{H} = 20 \times 10^{-6} \, \text{H} \]

### Step 4: Use Resonance Condition to Find Capacitance \( C \)
At resonance, the relationship between inductance \( L \) and capacitance \( C \) is given by:
\[ \omega L = \frac{1}{\omega C} \]
Rearranging gives:
\[ C = \frac{1}{\omega^2 L} \]
Substituting the known values:
\[ C = \frac{1}{(4 \times 10^5)^2 \cdot (20 \times 10^{-6})} \]
Calculating \( (4 \times 10^5)^2 \):
\[ (4 \times 10^5)^2 = 16 \times 10^{10} \]
Thus,
\[ C = \frac{1}{16 \times 10^{10} \cdot 20 \times 10^{-6}} = \frac{1}{320 \times 10^4} \]
Calculating:
\[ C = \frac{1}{320} \times 10^{-4} = 3.125 \times 10^{-6} \, \text{F} = 31.25 \, \mu\text{F} \]

### Final Answer
The value of capacitance \( C \) is:
\[ C = \frac{1}{32} \, \mu\text{F} \]