In a series LCR the voltage across resistance, capacitance and inductance is 10V each. If the capacitance is shor t circulated, the voltage across the inducatance will be

Options

A. `(10)/(sqrt2)V`
B. 10V
C. `10sqrt2V`
D. 20V

Answer

To solve the problem step-by-step, we will analyze the situation in a series LCR circuit where the voltage across resistance (VR), capacitance (VC), and inductance (VL) is initially 10V each. We will then determine the voltage across the inductance when the capacitance is short-circuited.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- In a series LCR circuit, the voltage across each component (resistor, capacitor, and inductor) is given as 10V.
- Therefore, we have:
\[ V_R = V_L = V_C = 10V \]

2. **Applying the Phasor Relationship:**
- In a series LCR circuit, the total voltage applied (V) can be expressed using the phasor relationship:
\[ V = \sqrt{V_R^2 + (V_L - V_C)^2} \]
- Since \( V_L \) and \( V_C \) are equal (both are 10V), we can simplify this to:
\[ V = \sqrt{V_R^2 + 0^2} = V_R = 10V \]

3. **Short-Circuiting the Capacitance:**
- When the capacitance is short-circuited, it effectively removes the capacitor from the circuit.
- The impedance of the circuit now consists only of the resistor (R) and the inductor (L).

4. **Calculating the New Impedance:**
- The new impedance (Z) of the circuit when the capacitor is short-circuited is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
- Since \( V_R = V_L = 10V \), we can also say:
\[ R = X_L = 10 \, \text{(since voltage across R and L are equal)} \]
- Therefore, the new impedance becomes:
\[ Z = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \]

5. **Calculating the New Current:**
- The new current (I') in the circuit can be calculated using Ohm's law:
\[ I' = \frac{V}{Z} = \frac{10V}{10\sqrt{2}} = \frac{1}{\sqrt{2}} \, \text{A} \]

6. **Finding the Voltage Across the Inductor:**
- The voltage across the inductor (V_L) can be calculated using:
\[ V_L = I' \cdot X_L \]
- Substituting the values:
\[ V_L = \left(\frac{1}{\sqrt{2}}\right) \cdot 10 = 10\sqrt{2} \, \text{V} \]

### Final Answer:
The voltage across the inductance when the capacitance is short-circuited is:
\[ \boxed{10\sqrt{2} \, \text{V}} \]