In a series `L-C-R` circuit, currenrt in the circuit is `11A` when the applied voltage is `220 V`. voltage across the caspcitor is `200 V`. If the value of resistor is `20 Omega`, then the voltage across the unknown inductor is

Options

A. zero
B. `200V`
C. `20V`
D. none of these

Explanation

To solve the problem, we need to find the voltage across the unknown inductor in a series L-C-R circuit. We are given the following information:

- Current in the circuit (I) = 11 A
- Applied voltage (V) = 220 V
- Voltage across the capacitor (V_C) = 200 V
- Resistance (R) = 20 Ω

### Step-by-step Solution:

1. **Calculate the Impedance (Z) of the Circuit:**
The impedance (Z) in a series L-C-R circuit can be calculated using the formula:
\[ Z = \frac{V}{I} \]
Substituting the values:
\[ Z = \frac{220 \, \text{V}}{11 \, \text{A}} = 20 \, \Omega \]

2. **Determine the Condition of Resonance:**
Since the impedance (Z) is equal to the resistance (R), we can conclude that the circuit is at resonance. This implies that the inductive reactance (X_L) is equal to the capacitive reactance (X_C):
\[ X_L = X_C \]

3. **Calculate the Capacitive Reactance (X_C):**
The capacitive reactance can be calculated using the formula:
\[ X_C = \frac{V_C}{I} \]
Substituting the values:
\[ X_C = \frac{200 \, \text{V}}{11 \, \text{A}} \approx 18.18 \, \Omega \]

4. **Find the Inductive Reactance (X_L):**
Since we established that \(X_L = X_C\) at resonance:
\[ X_L = 18.18 \, \Omega \]

5. **Calculate the Voltage across the Inductor (V_L):**
The voltage across the inductor can be calculated using the formula:
\[ V_L = X_L \cdot I \]
Substituting the values:
\[ V_L = 18.18 \, \Omega \cdot 11 \, \text{A} \approx 200 \, \text{V} \]

### Conclusion:
The voltage across the unknown inductor is approximately **200 V**.