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

Options

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

Explanation

To solve the problem, we will analyze the series L-C-R circuit step by step.

### Step 1: Understand the Given Information
We know that in the series L-C-R circuit:
- Voltage across the resistor (V_R) = 10 V
- Voltage across the capacitor (V_C) = 10 V
- Voltage across the inductor (V_L) = 10 V

### Step 2: Write the Expression for the Source Voltage (V_s)
In a series L-C-R circuit, the source voltage (V_s) can be calculated using the formula:
\[ V_s = \sqrt{V_R^2 + (V_C - V_L)^2} \]

### Step 3: Substitute the Known Values
Since V_C and V_L are both equal to 10 V, we can substitute:
\[ V_s = \sqrt{V_R^2 + (10 - 10)^2} \]
\[ V_s = \sqrt{V_R^2} \]
\[ V_s = V_R \]
Thus, we have:
\[ V_s = 10 \, \text{V} \]

### Step 4: Analyze the Effect of Short-Circuiting the Capacitor
When the capacitor is short-circuited, the voltage across the capacitor (V_C) becomes 0 V. Therefore, we can update our expression for the source voltage:
\[ V_s = \sqrt{V_R^2 + V_L^2} \]
Since V_C = 0, we can ignore it in the equation.

### Step 5: Substitute the Values Again
Now, we can write:
\[ V_s = \sqrt{V_R^2 + V_L^2} \]
Substituting V_R = 10 V:
\[ 10 = \sqrt{10^2 + V_L^2} \]
\[ 10 = \sqrt{100 + V_L^2} \]

### Step 6: Square Both Sides to Eliminate the Square Root
Squaring both sides gives us:
\[ 100 = 100 + V_L^2 \]
Subtracting 100 from both sides:
\[ 0 = V_L^2 \]
This indicates that we need to re-evaluate our approach since we have neglected the phase difference.

### Step 7: Correctly Calculate V_L
From the original formula for V_s:
\[ V_s = \sqrt{V_R^2 + V_L^2} \]
Since V_C = 0, we can say:
\[ V_L = \frac{V_s}{\sqrt{2}} \]
Substituting V_s = 10 V:
\[ V_L = \frac{10}{\sqrt{2}} \]
\[ V_L = 10 \cdot \frac{\sqrt{2}}{2} \]
\[ V_L = 5\sqrt{2} \, \text{V} \]

### Conclusion
The voltage across the inductance when the capacitance is short-circuited is \( 10\sqrt{2} \, \text{V} \).

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