200 V ac source is fed to series LCR circuit having `X_(L)=50 Omega, X_(C )=50 Omega` and `R = 25 Omega`. Potential drop across the inductor is

Options

A. 100 V
B. 200 V
C. 400 V
D. 10 V

Explanation

To find the potential drop across the inductor in a series LCR circuit with given values, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- AC Source Voltage (V) = 200 V
- Inductive Reactance (X_L) = 50 Ω
- Capacitive Reactance (X_C) = 50 Ω
- Resistance (R) = 25 Ω

2. **Calculate the Impedance (Z)**:
The impedance in a series LCR circuit is given by the formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values:
\[ Z = \sqrt{25^2 + (50 - 50)^2} = \sqrt{625 + 0} = \sqrt{625} = 25 \, \Omega \]

3. **Calculate the Current (I)**:
The current in the circuit can be calculated using Ohm's law:
\[ I = \frac{V}{Z} \]
Substituting the values:
\[ I = \frac{200 \, V}{25 \, \Omega} = 8 \, A \]

4. **Calculate the Voltage Drop Across the Inductor (V_L)**:
The voltage drop across the inductor is given by the formula:
\[ V_L = I \times X_L \]
Substituting the values:
\[ V_L = 8 \, A \times 50 \, \Omega = 400 \, V \]

5. **Conclusion**:
The potential drop across the inductor is **400 V**.