In a series R-L-C circuit, R = 10$\Omega$, $X_{c}$= 4$\Omega$ and $X_{L}$ = 9$\Omega$. The impedance of the circuit is

Analyze the question: We have a series R-L-C circuit and need its impedance (Z) from the resistance and the two reactances.

Concept: In a series RLC circuit, the inductive and capacitive reactances act in opposite directions (180° out of phase), so they partly cancel. The net reactance is X_L − X_C. The resistance is at 90° to this net reactance, so the two combine like the sides of a right-angled triangle:

Z = √[R² + (X_L − X_C)²]

Solution:

1. Net reactance: X_L − X_C = 9 − 4 = 5 Ω
2. Substitute: Z = √(10² + 5²)
3. Z = √(100 + 25) = √125
4. Z ≈ 11.18 Ω ≈ 11.2 Ω

Answer: The impedance is about 11.2 Ω, which is option D.

Why the other options are wrong:

• A (4.1 Ω) and B (5.0 Ω) come from using only the reactances, or from a wrong combination.
• C (10.8 Ω) does not come from the correct formula.

Exam tip: Remember the steps as "subtract first, then Pythagoras." Subtract the reactances (X_L − X_C), then use the Pythagoras theorem with R. Here the numbers 10 and 5 give √125, which is close to 11.2 (since 11² = 121 and 11.2² ≈ 125.4). Estimating like this helps you check your answer quickly.