Question E5B10:
What is the relationship between the AC current through an inductor and the voltage across an inductor?  

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

Correct Answer is  A: Voltage leads current by 90 degrees  

To understand an inductor in AC, remember the time-derivative relationship and the phasor (impedance) form.

Why the correct answer (A) is correct:
- An inductor obeys:
  $$v_L(t) = L\,\frac{di(t)}{dt}$$
- For a sinusoidal current $i(t) = I_m \sin(\omega t)$:
  $$v_L(t) = \omega L I_m \cos(\omega t) = \omega L I_m \sin(\omega t + 90^\circ)$$
  So the voltage is a quarter cycle (90°) ahead of the current.
- In phasor form, the inductor’s impedance is:
  $$Z_L = j\omega L = \omega L \angle +90^\circ$$
  Therefore:
  $$V = I \cdot Z_L \Rightarrow \angle V = \angle I + 90^\circ$$
  Voltage leads current by 90°.

Why the other choices are incorrect:
- B: Current leads voltage by 90° — this describes a capacitor, not an inductor. For a capacitor:
  $$Z_C = \frac{1}{j\omega C} = \frac{1}{\omega C}\angle -90^\circ$$
  So current leads voltage by 90° in a capacitor.
- C: 180° out of phase — this would imply a sign inversion, not characteristic of a pure inductor.
- D: In phase — true only for a pure resistor, where:
  $$Z_R = R \angle 0^\circ$$

Useful formulas and rules of thumb:
- Inductor: $$v = L\,\frac{di}{dt},\quad Z_L = j\omega L,\quad |Z_L| = \omega L$$
- Capacitor: $$i = C\,\frac{dv}{dt},\quad Z_C = \frac{1}{j\omega C},\quad |Z_C| = \frac{1}{\omega C}$$
- Mnemonic: “ELI the ICE man”
  - E (voltage) leads I (current) in L (inductor).
  - I (current) leads E (voltage) in C (capacitor).