In a series RC circuit, the impedance is

The correct answer is Option (3) → \( \left[R^2+X_C^2\right]^{1/2} \)

In a series RC circuit the same current flows through R and C. The voltage across R is in phase with the current, while the voltage across C lags it by \( \pi/2 \). So the two voltages add as phasors at right angles.

Option (1) \( \left[R^2-X_C^2\right]^{1/2} \) — Incorrect. For two phasors at right angles the Pythagorean theorem adds the squares. A minus sign would make the impedance smaller than R. But adding a capacitor in series only increases the opposition to the current.

Option (2) \( R+X_C \) — Incorrect. The voltages across R and C are \( \pi/2 \) out of phase, so they add as phasors. The plain sum \( R+X_C \) is larger than the real impedance.

Option (3) \( \left[R^2+X_C^2\right]^{1/2} \) — Correct. With \( v_{Rm}=i_mR \) and \( v_{Cm}=i_mX_C \) at right angles, the source voltage is \( v_m=i_m\sqrt{R^2+X_C^2} \). Hence the impedance is \( \sqrt{R^2+X_C^2} \). This is the series LCR result \( Z=\sqrt{R^2+(X_C-X_L)^2} \) with \( X_L=0 \).

Option (4) \( \left[R^2+X_C^2\right]^{-1/2} \) — Incorrect. This is the reciprocal of the impedance. Its unit is \( \Omega^{-1} \), so it cannot be an impedance.

NCERT (2026-27):
Points to Ponder – “5. In an ac circuit, while adding voltages across different elements, one should take care of their phases properly. For example, if V_R and V_C are voltages across R and C, respectively in an RC circuit, then the total voltage across RC combination is

    \( V_{RC}=\sqrt{V_R^2+V_C^2} \)

and not V_R + V_C since V_C is π/2 out of phase of V_R.”

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

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