In phasor form, what is the impedance of a capacitor?

In phasor form, what is the impedance of a capacitor?

• A. Z_R = R
• B. Z_L = j ω L
• C. Z_C = 1/(j ω C)
• D. Z_C = j ω C

Answer: (C)

When a capacitor is in an AC circuit, its behavior is captured by the relationship i = C dv/dt. In phasor form, differentiation becomes multiplication by jω, so I = jωC V. The impedance Z is V divided by I, which gives Z_C = V/I = 1/(jωC) = -j/(ωC). This shows a magnitude of 1/(ωC) and a phase of -90°, meaning current leads voltage by 90 degrees, as expected for a capacitor. The other expressions correspond to different elements: a resistor has Z = R (purely real), an inductor has Z = jωL (positive 90° phase shift), and jωC would represent the capacitor’s admittance, not its impedance. Therefore, the impedance of a capacitor in phasor form is 1/(jωC).

When a capacitor is in an AC circuit, its behavior is captured by the relationship i = C dv/dt. In phasor form, differentiation becomes multiplication by jω, so I = jωC V. The impedance Z is V divided by I, which gives Z_C = V/I = 1/(jωC) = -j/(ωC). This shows a magnitude of 1/(ωC) and a phase of -90°, meaning current leads voltage by 90 degrees, as expected for a capacitor.

The other expressions correspond to different elements: a resistor has Z = R (purely real), an inductor has Z = jωL (positive 90° phase shift), and jωC would represent the capacitor’s admittance, not its impedance. Therefore, the impedance of a capacitor in phasor form is 1/(jωC).