In series LCR circuit `3 = Omega X_(L) = 8 Omega, X_(C) = 4 Omega`, the impendance of the circuit is:

Options

A. `3Omega`
B. `7 Omega`
C. `5 Omega`
D. `25 Omega`

Explanation

To find the impedance of the given series LCR circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Resistance (R) = 3 Ω
- Inductive Reactance (X_L) = 8 Ω
- Capacitive Reactance (X_C) = 4 Ω

2. **Calculate the net reactance (X):**
In a series LCR circuit, the net reactance (X) is given by:
\[ X = X_L - X_C \]
Substituting the given values:
\[ X = 8 \, \Omega - 4 \, \Omega = 4 \, \Omega \]

3. **Use the impedance formula:**
The impedance (Z) of a series LCR circuit can be calculated using the formula:
\[ Z = \sqrt{R^2 + X^2} \]
Substituting the values of R and X:
\[ Z = \sqrt{(3 \, \Omega)^2 + (4 \, \Omega)^2} \]

4. **Calculate R² and X²:**
\[ R^2 = 3^2 = 9 \]
\[ X^2 = 4^2 = 16 \]

5. **Add R² and X²:**
\[ R^2 + X^2 = 9 + 16 = 25 \]

6. **Take the square root:**
\[ Z = \sqrt{25} = 5 \, \Omega \]

### Final Answer:
The impedance of the circuit is \( Z = 5 \, \Omega \).

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