Question E5C10:
Which point on Figure E5-1 best represents the impedance of a series circuit consisting of a 400-ohm resistor and a 38-picofarad capacitor at 14 MHz?  

Choices:  
A: Point 2  
B: Point 4  
C: Point 5  
D: Point 6    

Correct Answer is  B: Point 4  

To find the impedance of a series 400 Ω resistor with a 38 pF capacitor at 14 MHz:

1) Why Point 4 is correct
- Compute the capacitive reactance:
  $$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi(14\times 10^6)(38\times 10^{-12})} \approx 300\ \Omega$$
- Series RC impedance:
  $$Z = R - jX_C \approx 400 - j300\ \Omega$$
- Interpretation:
  - Positive resistance (right of origin).
  - Negative reactance (capacitive; below the real axis).
  - Magnitudes: $R=400$ is greater than $|X|=300$, so the point is in the lower-right quadrant and closer to the real axis than the imaginary axis.
- On Figure E5-1, that location corresponds to Point 4.

2) Why the other choices are incorrect
- Point 2: Lies in the inductive region ($X>0$). Our circuit is capacitive ($X<0$).
- Point 5: Also shows positive (inductive) reactance or otherwise not in the lower-right quadrant; does not match $X<0$ with $R>0$.
- Point 6: In the lower-right quadrant but with $|X|>R$ (more like $300 - j400$); our computed values have $R>|X|$ ($400$ vs $300$).

3) Useful formulas and rules of thumb
- Capacitor reactance:
  $$X_C = \frac{1}{2\pi f C}\quad\text{(take as negative in impedance, }-jX_C\text{)}$$
- Series impedance addition:
  $$Z = R - jX_C\quad\text{(for series R and C)}$$
- Quadrant rule:
  - Inductive reactance: $+jX$ (above real axis).
  - Capacitive reactance: $-jX$ (below real axis).
- Mental math at 14 MHz: 
  $$X_C \approx \frac{11.4\ \text{k}\Omega}{C\ \text{(pF)}}$$
  So with $C=38$ pF, $X_C \approx 11.4\text{k}\Omega/38 \approx 300\ \Omega$.