Question E9G04:
What are the two families of circles and arcs that make up a Smith chart?  

Choices:  
A: Inductance and capacitance  
B: Reactance and voltage  
C: Resistance and reactance  
D: Voltage and impedance    

Correct Answer is  C: Resistance and reactance  

Correct answer: C — Resistance and reactance

1) Why C is correct
- A Smith chart is a plot of normalized impedance $z = r + jx$, where:
  - $r$ = normalized resistance
  - $x$ = normalized reactance
- The chart grid is formed by:
  - Constant-resistance circles (full circles that cross the horizontal axis at the labeled $r$ values)
  - Constant-reactance arcs (curved arcs above/below the axis for $+x$ inductive and $-x$ capacitive)

2) Why the other choices are incorrect
- A: Inductance and capacitance
  - The chart does not plot $L$ and $C$ directly; it plots reactance $X$ (which depends on $L$, $C$, and frequency). The grid is by $r$ and $x$, not $L$ and $C$.
- B: Reactance and voltage
  - Voltage is not a coordinate family on the Smith chart. The chart is for impedances/admittances via reflection coefficient.
- D: Voltage and impedance
  - Again, voltage is not a chart axis or family. The grid is built from resistance and reactance components of impedance.

3) Useful formulas and rules of thumb
- Normalize impedance:  
  $$z = \frac{Z}{Z_0} = r + jx,\quad Z = R + jX$$
- Reflection coefficient mapping:  
  $$\Gamma = \frac{z - 1}{z + 1} = \frac{Z - Z_0}{Z + Z_0}$$
- Center of chart: $z = 1 + j0$ (perfect match to $Z_0$).
- Upper half: inductive ($x > 0$). Lower half: capacitive ($x < 0$).
- Outer circle: $|\Gamma| = 1$ (lossless terminations like open/short).
- Constant-resistance: circles; constant-reactance: arcs tangent to the outer circle.