A series RLC circuit has the following values `R=20Omega`, `X_(L)=10Omega`. `E=50V`(rms) at `omega=400` rad/s. Current `2A` leads the applied voltage. The value of the capacitative reactance `X_(C )` is

Options

A. `5Omega`
B. `25Omega`
C. `10Omega`
D. `15Omega`

Answer

To find the capacitive reactance \( X_C \) in the given series RLC circuit, we can follow these steps:

### Step 1: Identify the given values
We have the following values from the problem:
- Resistance, \( R = 20 \, \Omega \)
- Inductive reactance, \( X_L = 10 \, \Omega \)
- RMS voltage, \( E = 50 \, V \)
- Current, \( I = 2 \, A \)

### Step 2: Calculate the total impedance \( Z \)
The total impedance \( Z \) can be calculated using Ohm's law:
\[ Z = \frac{E}{I} = \frac{50 \, V}{2 \, A} = 25 \, \Omega \]

### Step 3: Use the impedance formula for RLC circuits
The impedance in a series RLC circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the known values:
\[ 25 = \sqrt{20^2 + (10 - X_C)^2} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 25^2 = 20^2 + (10 - X_C)^2 \]
\[ 625 = 400 + (10 - X_C)^2 \]

### Step 5: Rearrange the equation
Subtract \( 400 \) from both sides:
\[ 625 - 400 = (10 - X_C)^2 \]
\[ 225 = (10 - X_C)^2 \]

### Step 6: Take the square root of both sides
Taking the square root gives us two possible equations:
\[ 10 - X_C = 15 \quad \text{or} \quad 10 - X_C = -15 \]

### Step 7: Solve for \( X_C \)
1. For \( 10 - X_C = 15 \):
\[ -X_C = 15 - 10 \implies -X_C = 5 \implies X_C = -5 \, \Omega \quad \text{(not valid)} \]

2. For \( 10 - X_C = -15 \):
\[ -X_C = -15 - 10 \implies -X_C = -25 \implies X_C = 25 \, \Omega \]

### Step 8: Conclusion
Thus, the value of the capacitive reactance \( X_C \) is:
\[ \boxed{25 \, \Omega} \]