Question E5B09:
What is the relationship between the AC current through a capacitor and the voltage across a capacitor?  

Choices:  
A: Voltage and current are in phase  
B: Voltage and current are 180 degrees out of phase  
C: Voltage leads current by 90 degrees  
D: Current leads voltage by 90 degrees    

Correct Answer is  D: Current leads voltage by 90 degrees  

Correct answer: D — Current leads voltage by 90 degrees.

Why D is correct:
- For a capacitor, the current is proportional to the rate of change of voltage:
  $$i(t) = C \frac{dv(t)}{dt}$$
  If $v(t) = V_m \sin(\omega t)$, then:
  $$i(t) = C \omega V_m \cos(\omega t) = C \omega V_m \sin(\omega t + 90^\circ)$$
  This shows the current peaks one-quarter cycle (90°) before the voltage.
- In phasor form, the impedance of a capacitor is:
  $$Z_C = \frac{1}{j \omega C} = -\,j \frac{1}{\omega C}$$
  The angle of $Z_C$ is $-90^\circ$, so:
  $$I = \frac{V}{Z_C} \implies \angle I = \angle V - (-90^\circ) = \angle V + 90^\circ$$
  Therefore, current leads voltage by 90°.

Why the other choices are incorrect:
- A: In phase — true only for a pure resistor, not a capacitor.
- B: 180 degrees out of phase — reactive components (ideal L or C) produce 90° phase shifts, not 180°.
- C: Voltage leads current by 90 degrees — that describes an inductor, not a capacitor.

Useful formulas and rules of thumb:
- Time domain: $$i(t) = C \frac{dv(t)}{dt}$$
- Phasor domain: $$Z_C = \frac{1}{j \omega C}, \quad I = j \omega C \, V$$
- Magnitude relationship: $$|I| = \omega C \, |V|$$
- Mnemonic: “ELI the ICE man”
  - ELI: In an inductor (L), voltage (E) leads current (I) by 90°
  - ICE: In a capacitor (C), current (I) leads voltage (E) by 90°