In a series LCR circuit, the impedance is given by

The correct answer is Option (1) → \( \left[R^2+(X_L-X_C)^2\right]^{1/2} \)

In a series LCR circuit the voltages across L and C are opposite in phase, so the net reactance is their difference. The impedance diagram is a right triangle with R and the net reactance as its two sides and Z as the hypotenuse.

Option (1) \( \left[R^2+(X_L-X_C)^2\right]^{1/2} \) — Correct. By Pythagoras’ theorem on the impedance triangle, \( Z^2=R^2+(X_C-X_L)^2 \). Since the term is squared, \( (X_C-X_L)^2=(X_L-X_C)^2 \), which gives this form.

Option (2) \( \left[R^2-(X_L-X_C)^2\right]^{1/2} \) — Incorrect. R and the net reactance are perpendicular sides of the triangle, so their squares add. A minus sign would make Z smaller than R, but Z can never be less than R.

Option (3) \( \left[R^2+(X_L+X_C)^2\right]^{1/2} \) — Incorrect. The voltages across L and C are opposite in phase, so their reactances partly oppose each other. The net reactance is the difference \( X_L-X_C \). This option adds the two reactances instead.

Option (4) \( \left[R^2-(X_L+X_C)^2\right]^{1/2} \) — Incorrect. It subtracts the squares, which must add. It also adds the two reactances, which must be subtracted.

NCERT (2026-27):
Phasor-Diagram Solution – “By analogy to the resistance in a circuit, we introduce the impedance Z in an ac circuit:

    \( \displaystyle i_m=\frac{v_m}{Z} \qquad [7.25(b)] \)

where

    \( Z=\sqrt{R^2+(X_C-X_L)^2} \)\( \qquad (7.26) \)”