Single correct · +4 / −1
The impedance of a series LCR circuit is:
- (a) R + X_L + X_C
- (b) √(R² + (X_L - X_C)²)
- (c) √(R² + X_L² + X_C²)
- (d) R - X_L - X_C
Answer and solution
Answer (b) √(R² + (X_L - X_C)²)
Solution
Physics sir · 3 steps
Voltages here are out of step, so add them as phasors, not as plain numbers.
- Phase of each voltage. The same current flows through R, L and C. Across R the voltage is in phase with the current, across L it leads by 90°, and across C it lags by 90°.
- L and C oppose. V_L and V_C point in exactly opposite directions, so they partly cancel and leave V_L - V_C. This difference is at right angles to V_R, so the two add like the sides of a right triangle.
V = √(V_R² + (V_L - V_C)²) = I√(R² + (X_L - X_C)²)
- So. Dividing V by I gives the impedance Z = √(R² + (X_L - X_C)²), which is option (b).
The idea to keep. In AC, resistance and reactance sit at 90° to each other, while X_L and X_C sit at 180°. So subtract the reactances first, then combine with R using Pythagoras.
Answer from the NTA final answer key. Solution by PrepWiser.