What is the sign of the imaginary part of the capacitor impedance Z_C = 1/(j ω C)?

• A. Positive imaginary
• B. Negative imaginary
• C. Zero
• D. Purely real positive

Answer: (B)

The key idea is how a capacitor behaves in AC in terms of impedance. For a capacitor, the current leads the voltage by 90 degrees, and in phasor form I = jωC V. The impedance is Z = V/I, which gives Z_C = 1/(jωC) = -j/(ωC). This shows the impedance has only an imaginary part, and that imaginary part is negative.

So the imaginary component is -1/(ωC). The negative sign reflects the capacitive nature: voltage lags the current by 90 degrees. The magnitude 1/(ωC) decreases as frequency increases, which is why a capacitor acts more like a short circuit at high frequency and more like an open circuit at low frequency. This also explains why an inductor has a positive imaginary impedance, Z_L = jωL, while a pure resistor has zero imaginary part.