In a series resonant LCR circuit the voltage across R is 100 volts and R = `1 k(Omega) with C =2(mu)F`. The resonant frequency `(omega)` is `200 rad//s`. At resonance the voltage across L is

Options

A. `4xx10^(-3)V`
B. `2.5xx10^(-2)V`
C. `40V`
D. `250V`

Answer

To solve the problem, we need to find the voltage across the inductor (L) in a series resonant LCR circuit. We are given the following information:

- Voltage across the resistor (R) = 100 V
- Resistance (R) = 1 kΩ = 1000 Ω
- Capacitance (C) = 2 μF = 2 × 10^(-6) F
- Resonant frequency (ω) = 200 rad/s

### Step-by-Step Solution:

1. **Calculate the Current (I) in the Circuit:**
The current in the circuit can be calculated using Ohm's law:
\[ I = \frac{V_R}{R} \]
Where \( V_R \) is the voltage across the resistor.
\[ I = \frac{100 \, \text{V}}{1000 \, \Omega} = 0.1 \, \text{A} \]

2. **Calculate the Capacitive Reactance (Xc):**
The capacitive reactance \( X_c \) can be calculated using the formula:
\[ X_c = \frac{1}{\omega C} \]
Substituting the given values:
\[ X_c = \frac{1}{200 \, \text{rad/s} \times 2 \times 10^{-6} \, \text{F}} = \frac{1}{4 \times 10^{-4}} = 2500 \, \Omega \]

3. **Calculate the Voltage across the Capacitor (Vc):**
The voltage across the capacitor can be calculated using:
\[ V_c = I \times X_c \]
Substituting the values we have:
\[ V_c = 0.1 \, \text{A} \times 2500 \, \Omega = 250 \, \text{V} \]

4. **Voltage across the Inductor (V_L):**
At resonance, the voltage across the inductor \( V_L \) is equal to the voltage across the capacitor \( V_c \):
\[ V_L = V_c = 250 \, \text{V} \]

### Final Answer:
The voltage across the inductor \( V_L \) is **250 V**.