An LCR circuit with a resistance `50 Omega` has a resonant angular frequency `2xx 10^(3)` rad/s. At resonance, the voltage across the resistance and inductance are 25 V and 20 V respectively. Then
The value of capacitance is:

Options

A. `25 mu F`
B. `1 muF`
C. `2mu F`
D. `12.5 mu F`

Explanation

To find the value of capacitance in the given LCR circuit, we can follow these steps:

### Step 1: Understand the resonance condition
At resonance in an LCR circuit, the inductive reactance (X_L) is equal to the capacitive reactance (X_C). This can be expressed mathematically as:
\[ X_L = X_C \]

### Step 2: Write the expressions for inductive and capacitive reactance
The inductive reactance is given by:
\[ X_L = \omega L \]
And the capacitive reactance is given by:
\[ X_C = \frac{1}{\omega C} \]

### Step 3: Set the inductive and capacitive reactance equal to each other
From the resonance condition:
\[ \omega L = \frac{1}{\omega C} \]

### Step 4: Rearrange the equation to find capacitance
Cross-multiplying gives:
\[ \omega^2 LC = 1 \]
Thus, we can express capacitance as:
\[ C = \frac{1}{\omega^2 L} \]

### Step 5: Calculate the current in the circuit
The current (I) in the circuit can be calculated using the voltage across the resistor (V_R) and the resistance (R):
\[ I = \frac{V_R}{R} \]
Given \( V_R = 25 \, V \) and \( R = 50 \, \Omega \):
\[ I = \frac{25}{50} = 0.5 \, A \]

### Step 6: Calculate the inductive reactance
The inductive reactance can be calculated using the voltage across the inductor (V_L) and the current (I):
\[ X_L = \frac{V_L}{I} \]
Given \( V_L = 20 \, V \):
\[ X_L = \frac{20}{0.5} = 40 \, \Omega \]

### Step 7: Calculate the inductance (L)
Using the formula for inductive reactance:
\[ X_L = \omega L \]
We can rearrange to find L:
\[ L = \frac{X_L}{\omega} \]
Substituting \( X_L = 40 \, \Omega \) and \( \omega = 2 \times 10^3 \, \text{rad/s} \):
\[ L = \frac{40}{2 \times 10^3} = 0.02 \, H \]

### Step 8: Substitute L into the capacitance formula
Now substitute \( L \) back into the capacitance formula:
\[ C = \frac{1}{\omega^2 L} \]
Substituting \( \omega = 2 \times 10^3 \) and \( L = 0.02 \):
\[ C = \frac{1}{(2 \times 10^3)^2 \times 0.02} \]
Calculating \( (2 \times 10^3)^2 = 4 \times 10^6 \):
\[ C = \frac{1}{4 \times 10^6 \times 0.02} = \frac{1}{8 \times 10^4} \]
Thus:
\[ C = 0.125 \times 10^{-4} \, F = 12.5 \, \mu F \]

### Final Answer
The value of capacitance is:
\[ C = 12.5 \, \mu F \]

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