In AC circuits, the impedance of a capacitor is Z_C = ?

In AC circuits, the impedance of a capacitor is Z_C = ?

• A. Z_C = 1/(i ω C)
• B. Z_C = i ω C
• C. Z_C = −i ω C
• D. Z_C = R

Answer: (A)

In AC analysis, impedance links voltage and current for a component, showing how the current responds with a phase shift. A capacitor makes current lead the voltage by 90 degrees, and in phasor form the relation is I = i ω C V. From this, the impedance is Z_C = V/I = 1/(i ω C), which is equivalently −i/(ω C). So the capacitor presents a purely imaginary, negative impedance with magnitude 1/(ω C). This means as frequency rises, the capacitor’s impedance falls, and as frequency goes to zero, the impedance becomes infinite. The other expressions don’t match this behavior: i ω C would be the impedance of an inductor-like form, not a capacitor; −i ω C would grow with frequency and still isn’t the correct reciprocal; and R represents a resistor with real (non-imaginary) impedance.

In AC analysis, impedance links voltage and current for a component, showing how the current responds with a phase shift. A capacitor makes current lead the voltage by 90 degrees, and in phasor form the relation is I = i ω C V. From this, the impedance is Z_C = V/I = 1/(i ω C), which is equivalently −i/(ω C). So the capacitor presents a purely imaginary, negative impedance with magnitude 1/(ω C). This means as frequency rises, the capacitor’s impedance falls, and as frequency goes to zero, the impedance becomes infinite.

The other expressions don’t match this behavior: i ω C would be the impedance of an inductor-like form, not a capacitor; −i ω C would grow with frequency and still isn’t the correct reciprocal; and R represents a resistor with real (non-imaginary) impedance.