A series LCR circuit containing a resistance of 120 ohm has angular resonance frequency `4 xx 10^(3)rad s^(-1)`. At resonance, the voltage across resistance and inductance are 60V and 40 V respectively. The values of Land C are respectively

Options

A. `20 mH,25//8 muF`
B. `2 mH,1//35 muF`
C. `20 mH,1//40 muF`
D. `2 mH,25//8 muF`

Answer

To solve the problem step by step, we will follow the information provided in the question and the video transcript.

### Step 1: Identify the Given Values
- Resistance \( R = 120 \, \Omega \)
- Angular resonance frequency \( \omega = 4 \times 10^3 \, \text{rad/s} \)
- Voltage across resistance \( V_R = 60 \, V \)
- Voltage across inductor \( V_L = 40 \, V \)

### Step 2: Calculate the Current in the Circuit
Using Ohm's law for the resistor:
\[ V_R = I \cdot R \]
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:
\[ V_L = I \cdot X_L \]
Where \( X_L = \omega L \). Thus:
\[ V_L = I \cdot \omega L \]
Rearranging for \( L \):
\[ L = \frac{V_L}{I \cdot \omega} \]
Substituting the known values:
\[ L = \frac{40}{\left(\frac{1}{2}\right) \cdot (4 \times 10^3)} \]
Calculating:
\[ L = \frac{40}{2000} = 0.02 \, \text{H} = 20 \, \text{mH} \]

### Step 4: Calculate the Capacitance \( C \)
At resonance, the relationship between inductance and capacitance is given by:
\[ \omega L = \frac{1}{\omega C} \]
Rearranging gives:
\[ C = \frac{1}{\omega^2 L} \]
Substituting the values:
\[ C = \frac{1}{(4 \times 10^3)^2 \cdot (20 \times 10^{-3})} \]
Calculating:
\[ C = \frac{1}{16 \times 10^6 \cdot 20 \times 10^{-3}} = \frac{1}{320000} = 3.125 \times 10^{-6} \, \text{F} = 31.25 \, \mu\text{F} \]

### Final Values
- Inductance \( L = 20 \, \text{mH} \)
- Capacitance \( C = 31.25 \, \mu\text{F} \)

### Summary
The values of inductance \( L \) and capacitance \( C \) are:
- \( L = 20 \, \text{mH} \)
- \( C = 31.25 \, \mu\text{F} \)