`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 these

Answer

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

\[ Z = R + j(X_L - X_C) \]

where:
- \( R \) is the resistance,
- \( X_L \) is the inductive reactance,
- \( X_C \) is the capacitive reactance,
- \( j \) is the imaginary unit.

### Step 1: Identify the values
From the problem statement, we have:
- \( R = 10 \, \Omega \)
- \( X_L = 8 \, \Omega \)
- \( X_C = 6 \, \Omega \)

### Step 2: Calculate the net reactance
The net reactance \( X \) is given by:
\[ X = X_L - X_C \]
Substituting the values:
\[ X = 8 \, \Omega - 6 \, \Omega = 2 \, \Omega \]

### Step 3: Write the total impedance
Now we can substitute \( R \) and \( X \) into the impedance formula:
\[ Z = R + jX = 10 \, \Omega + j(2 \, \Omega) \]

### Step 4: Calculate the magnitude of the impedance
To find the magnitude of the total impedance \( |Z| \), we use the formula:
\[ |Z| = \sqrt{R^2 + X^2} \]
Substituting the values:
\[ |Z| = \sqrt{(10 \, \Omega)^2 + (2 \, \Omega)^2} = \sqrt{100 + 4} = \sqrt{104} \]
Calculating the square root:
\[ |Z| \approx 10.2 \, \Omega \]

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

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