In a series resonant `R L C` circuit, the voltage across resistance `R` is 100V and the value of `R=1000 ohm`. The capacitance of the capacitor is `2 xx 10^(-6)F` and angular frequency of `A C` is `200 (rad) s^(-1)`. The potential (in volt) across the inductance coil is

Answer

To solve the problem of finding the potential across the inductance coil in a series resonant RLC circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Voltage across resistance, \( V_R = 100 \, V \)
- Resistance, \( R = 1000 \, \Omega \)
- Capacitance, \( C = 2 \times 10^{-6} \, F \)
- Angular frequency, \( \omega = 200 \, \text{rad/s} \)

2. **Calculate the Current in the Circuit:**
The current \( I \) through the resistor can be calculated using Ohm's Law:
\[ I = \frac{V_R}{R} = \frac{100 \, V}{1000 \, \Omega} = 0.1 \, A \]

3. **Calculate the Capacitive Reactance \( X_C \):**
The capacitive reactance \( X_C \) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{200 \times 2 \times 10^{-6}} = \frac{1}{4 \times 10^{-4}} = 2500 \, \Omega \]

4. **Determine the Inductive Reactance \( X_L \):**
In a series resonant circuit, the inductive reactance \( X_L \) is equal to the capacitive reactance \( X_C \):
\[ X_L = X_C = 2500 \, \Omega \]

5. **Calculate the Voltage Across the Inductor \( V_L \):**
The voltage across the inductor can be calculated using the formula:
\[ V_L = I \cdot X_L \]
Substituting the values:
\[ V_L = 0.1 \, A \times 2500 \, \Omega = 250 \, V \]

### Final Answer:
The potential across the inductance coil is \( V_L = 250 \, V \).

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