In the circuit in figure (a), R = 10 Ω, the capacitor reactance at the operating angular frequency is 1/(ωC) = 10 Ω, and the input voltage is vin = 10∠0° V (RMS). Which of the following is the internal impedance Zth of the Thevenin equivalent circuit seen from the output terminals?

Answer: B. 5 − j5 Ω

Method 1 (parallel formula): with the voltage source shorted, R and ZC = −j10 Ω appear in parallel from the output terminals, so Zth = R ZC/(R + ZC) = (10 × (−j10))/(10 − j10) = 5 − j5 Ω. Method 2 (sum of admittances): Y = 1/10 + 1/(−j10) = 0.1 + j0.1 S, so Zth = 1/Y = (0.1 − j0.1)/0.02 = 5 − j5 Ω. The equivalent voltage is the open-circuit voltage, which the voltage divider gives as Vth = 10 × (−j10)/(10 − j10) = 5 − j5 V ≈ 7.07∠−45° V. With impedances and phasors, the same Thevenin's theorem as for DC applies directly to AC.

• A. The sign of the imaginary part is reversed. A parallel circuit containing a capacitor is capacitive, so the imaginary part is negative.
• B. Correct. Zth = R in parallel with ZC = 5 − j5 Ω.
• C. 10 − j10 Ω is R and ZC in series; it misses that the two are in parallel when the voltage source is shorted.