In rectangular coordinates, what is the impedance of a network consisting of a 0.01-microfarad capacitor in parallel with a 300-ohm resistor at 50 kHz?

• A. 150 - j159
• B. 150 + j159
• C. 159 - j150
• D. 159 + j150

Answer: (C)

In a parallel network, add the admittances rather than the impedances. The capacitor at 50 kHz has Zc = 1/(jωC). With C = 0.01 μF = 1×10^-8 F and ω = 2π·50,000 ≈ 314,159 rad/s, Zc ≈ -j318 Ω. The resistor is 300 Ω real.

Compute admittance:

Y = 1/R + 1/Zc = 1/300 + 1/(-j318) = 0.003333... + j0.003144.

The total impedance is Z = 1/Y. Inverting gives Z ≈ (0.003333 − j0.003144) / (0.003333^2 + 0.003144^2) ≈ 159 − j150 Ω.

So the impedance is about 159 − j150 ohms.