In a series LCR circuit, the voltage across the resistance, capacitance and inductance is 10 V each. If the capacitance is short circuited, the voltage across the inductance will be

Options

A. 10V
B. `10sqrt(2)V`
C. `(10)/(sqrt(2))V`
D. 20V

Answer

To solve the problem, we need to analyze the given series LCR circuit and determine the voltage across the inductance when the capacitance is short-circuited.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- In the series LCR circuit, the voltage across the resistance (R), capacitance (C), and inductance (L) is given as 10 V each.

2. **Identify the Impedance**:
- The total voltage across the circuit can be expressed using the impedance of the circuit. In a series LCR circuit, the impedance (Z) is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
- Where \(X_L\) is the inductive reactance and \(X_C\) is the capacitive reactance.

3. **Calculate the Equivalent Resistance**:
- Since the voltage across each component is 10 V, we can express the current (I) in terms of the resistance:
\[ V = I \cdot R \implies I = \frac{10}{R} \]

4. **Short-Circuit the Capacitance**:
- When the capacitance is short-circuited, the capacitive reactance \(X_C\) becomes 0. Thus, the impedance of the circuit simplifies to:
\[ Z = \sqrt{R^2 + X_L^2} \]

5. **Recalculate the Current**:
- The new current in the circuit can be calculated using the voltage across the resistance:
\[ I = \frac{V}{Z} = \frac{10}{\sqrt{R^2 + X_L^2}} \]

6. **Determine the Voltage Across the Inductance**:
- The voltage across the inductance \(V_L\) can be calculated as:
\[ V_L = I \cdot X_L \]
- Substituting the expression for I gives:
\[ V_L = \left(\frac{10}{\sqrt{R^2 + X_L^2}}\right) \cdot X_L \]

7. **Substituting Values**:
- Since initially, the voltage across the inductance was 10 V, we can relate \(X_L\) to the resistance R:
\[ X_L = R \implies V_L = \frac{10 \cdot R}{\sqrt{R^2 + R^2}} = \frac{10R}{\sqrt{2R^2}} = \frac{10R}{R\sqrt{2}} = \frac{10}{\sqrt{2}} \approx 7.07 V \]

8. **Final Answer**:
- The voltage across the inductance when the capacitance is short-circuited is:
\[ V_L = \frac{10}{\sqrt{2}} \text{ V} \]