Numerical answer · +4 / −1
In an LCR circuit with R=3 Ω, X_L=6 Ω, X_C=2 Ω, the impedance (in Ω) is:
Answer and solution
Answer 5
Solution
Physics sir · 3 steps
Subtract the two reactances first, then combine with R by Pythagoras, since they are 90° apart.
- Net reactance. The voltages across L and C are exactly opposite in phase, so their reactances cancel partly. The net reactance is X_L - X_C = 6 - 2 = 4 Ω.
- Combine with R. The resistor voltage is 90° out of phase with the net reactive voltage. So R and X_L - X_C add like two perpendicular sides of a right triangle.
Z = √(R² + (X_L - X_C)²) = √(3² + 4²) = √25
- So. The impedance is Z = 5 Ω, so the answer is 5.
The common slip: 7. Adding R and X_L - X_C directly as 3 + 4 gives 7, but they are perpendicular in the phasor diagram and must be combined by Pythagoras.
The idea to keep. In a series LCR circuit, X_L and X_C subtract because they are opposite in phase, and the result adds to R only as a vector: Z = √(R² + (X_L - X_C)²).
Answer from the NTA final answer key. Solution by PrepWiser.