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 impedance at resonance is:

Options

A. `50 Omega`
B. `16 Omega`
C. `64 Omega`
D. `25 Omega`

Explanation

To solve the problem, we need to determine the impedance at resonance in an LCR circuit. Let's break down the steps:

### Step-by-Step Solution:

1. **Understand the Given Values**:
- Resistance \( R = 50 \, \Omega \)
- Resonant angular frequency \( \omega_0 = 2 \times 10^3 \, \text{rad/s} \)
- Voltage across the resistance \( V_R = 25 \, V \)
- Voltage across the inductance \( V_L = 20 \, V \)

2. **Recall the Concept of Impedance at Resonance**:
- At resonance in an LCR circuit, the inductive reactance \( X_L \) is equal to the capacitive reactance \( X_C \). Therefore, the impedance \( Z \) at resonance simplifies to:
\[ Z = R \]
- This means that the impedance at resonance is equal to the resistance of the circuit.

3. **Calculate the Impedance**:
- Since we have established that at resonance:
\[ Z = R = 50 \, \Omega \]

4. **Conclusion**:
- The impedance at resonance is \( 50 \, \Omega \).

### Final Answer:
The impedance at resonance is \( 50 \, \Omega \).