`In` an L-C-R series circuit R = `10Omega, X_(L) = 8Omega `and `X_(C) = 6Omega`the total impedance of the circuit is -

Options

A. `10.2 Omega`
B. `17.2 Omega`
C. `10 Omega`
D. None of the above

Answer

To find the total impedance (Z) of an L-C-R series circuit, we can use the formula:

\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

Where:
- \( R \) is the resistance,
- \( X_L \) is the inductive reactance,
- \( X_C \) is the capacitive reactance.

Given:
- \( R = 10 \, \Omega \)
- \( X_L = 8 \, \Omega \)
- \( X_C = 6 \, \Omega \)

### Step-by-Step Solution:

1. **Calculate the difference between inductive and capacitive reactance:**
\[ X_L - X_C = 8 \, \Omega - 6 \, \Omega = 2 \, \Omega \]

2. **Square the resistance and the difference of reactances:**
\[ R^2 = (10 \, \Omega)^2 = 100 \, \Omega^2 \]
\[ (X_L - X_C)^2 = (2 \, \Omega)^2 = 4 \, \Omega^2 \]

3. **Add the squared values:**
\[ R^2 + (X_L - X_C)^2 = 100 \, \Omega^2 + 4 \, \Omega^2 = 104 \, \Omega^2 \]

4. **Take the square root to find the total impedance:**
\[ Z = \sqrt{104 \, \Omega^2} \approx 10.2 \, \Omega \]

### Final Answer:
The total impedance of the circuit is approximately \( 10.2 \, \Omega \).