Question E5C03:
Which of the following represents a pure inductive reactance in polar coordinates?  

Choices:  
A: A positive 45 degree phase angle  
B: A negative 45 degree phase angle  
C: A positive 90 degree phase angle  
D: A negative 90 degree phase angle    

Correct Answer is  C: A positive 90 degree phase angle  

To represent impedance in polar form, write:
$$Z = R + jX = |Z| \angle \phi$$
where:
- $|Z| = \sqrt{R^2 + X^2}$
- $\phi = \tan^{-1}\!\left(\dfrac{X}{R}\right)$

Correct answer: C — A positive 90 degree phase angle
- A pure inductor has reactance $X_L = +\omega L$, so $Z = jX_L$ with $R=0$ and $X>0$.
- With $R=0$, $\phi = \tan^{-1}(\infty) = +90^\circ$.
- In phasor terms, an ideal inductor’s voltage leads current by $90^\circ$.

Why the others are incorrect:
- A (positive 45°): Implies both $R$ and $X$ present with $X=R>0$; not purely reactive.
- B (negative 45°): Implies both $R$ and $X$ present with $X=-R$; not purely reactive.
- D (negative 90°): Purely capacitive reactance, $Z = -jX_C$ with $X_C = 1/(\omega C)$; this is a capacitor, not an inductor.

Useful formulas and rules of thumb:
- Inductive reactance: $X_L = +\omega L$ (positive, inductive) → $+jX$
- Capacitive reactance: $X_C = -\dfrac{1}{\omega C}$ (negative, capacitive) → $-jX$
- Polar angles:
  - Pure inductor: $+90^\circ$
  - Pure capacitor: $-90^\circ$
  - Pure resistor: $0^\circ$
- Memory aid: “L is +j” (inductor positive imaginary), “C is −j.”