Question E5B03:
How is impedance in polar form converted to an equivalent admittance?  

Choices:  
A: Take the reciprocal of the angle and change the sign of the magnitude  
B: Take the reciprocal of the magnitude and change the sign of the angle  
C: Take the square root of the magnitude and add 180 degrees to the angle  
D: Square the magnitude and subtract 90 degrees from the angle    

Correct Answer is  B: Take the reciprocal of the magnitude and change the sign of the angle  

To convert impedance to admittance, use the reciprocal. In polar form:

- If $Z = |Z|\angle\theta$, then the admittance is
  $$Y = \frac{1}{Z} = \frac{1}{|Z|}\angle(-\theta)$$

Why the correct answer is correct (Choice B):
- Taking the reciprocal in polar form inverts the magnitude and negates the angle:
  $$r\angle\theta \;\to\; \frac{1}{r}\angle(-\theta)$$

Why the other choices are incorrect:
- Choice A: “Reciprocal of the angle and change the sign of the magnitude”
  - Angles are not inverted to take reciprocals; the polar rule is to negate the angle.
  - Magnitude is always nonnegative; changing its sign is not meaningful here.
- Choice C: “Square root of the magnitude and add 180 degrees”
  - Adding $180^\circ$ flips the vector direction (multiplying by $-1$), not the reciprocal.
  - Taking a square root of the magnitude has nothing to do with reciprocals.
- Choice D: “Square the magnitude and subtract 90 degrees”
  - Subtracting $90^\circ$ rotates by $-90^\circ$ (multiplying by $-j$), not inversion.
  - Squaring the magnitude is the opposite of taking a reciprocal.

Useful formulas and rules of thumb:
- Polar reciprocal:
  $$Z = |Z|\angle\theta \;\Rightarrow\; Y = \frac{1}{|Z|}\angle(-\theta)$$
- Rectangular form (sometimes handy):
  - If $Z = R + jX$, then
    $$Y = \frac{1}{Z} = \frac{R - jX}{R^2 + X^2} = G + jB$$
    where $G = \frac{R}{R^2 + X^2}$ and $B = -\frac{X}{R^2 + X^2}$.
- Quick example:
  - $Z = 100\angle 45^\circ\ \Omega \;\Rightarrow\; Y = 0.01\angle(-45^\circ)\ \text{S}$.