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

Explanation

To find the voltage across the inductor (V_L) in a series resonant LCR circuit, we can follow these steps:

### Step 1: Calculate the Current in the Circuit
The current (I) flowing through the resistor (R) can be calculated using Ohm's law:
\[ I = \frac{V_R}{R} \]
Where:
- \( V_R = 100 \, \text{volts} \)
- \( R = 1 \, \text{k}\Omega = 1000 \, \Omega \)

Substituting the values:
\[ I = \frac{100}{1000} = 0.1 \, \text{A} \]

### Step 2: Calculate the Reactance of the Inductor (X_L)
At resonance, the reactance of the inductor (X_L) is equal to the reactance of the capacitor (X_C). The formula for the reactance of the inductor is:
\[ X_L = \omega L \]
And for the capacitor:
\[ X_C = \frac{1}{\omega C} \]
At resonance:
\[ X_L = X_C \]
Thus:
\[ \omega L = \frac{1}{\omega C} \]

### Step 3: Calculate the Inductive Reactance (X_L)
Rearranging the equation gives:
\[ X_L = \frac{1}{\omega C} \]
Substituting the values:
- \( \omega = 200 \, \text{rad/s} \)
- \( C = 2 \, \mu\text{F} = 2 \times 10^{-6} \, \text{F} \)

Calculating \( X_L \):
\[ X_L = \frac{1}{200 \times 2 \times 10^{-6}} = \frac{1}{0.0004} = 2500 \, \Omega \]

### Step 4: Calculate the Voltage Across the Inductor (V_L)
The voltage across the inductor can be calculated using Ohm's law:
\[ V_L = I \times X_L \]
Substituting the values:
\[ V_L = 0.1 \, \text{A} \times 2500 \, \Omega = 250 \, \text{volts} \]

### Final Answer
The voltage across the inductor (V_L) is **250 volts**.

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