Question E5C12:
Which point on Figure E5-1 best represents the impedance of a series circuit consisting of a 300-ohm resistor and a 19-picofarad capacitor at 21.200 MHz?  

Choices:  
A: Point 1  
B: Point 3  
C: Point 7  
D: Point 8    

Correct Answer is  A: Point 1  

To analyze a series R–C impedance, compute the capacitive reactance and form the complex impedance.

Key formulas:
- Capacitive reactance:
  $$X_C = \frac{1}{2\pi f C}$$
- Series R–C impedance:
  $$Z = R - jX_C$$
- Phase angle:
  $$\phi = \tan^{-1}\!\left(\frac{X}{R}\right)$$
  (negative for capacitive reactance)

Compute values:
- \(f = 21.200\,\text{MHz} = 21.2\times10^6\,\text{Hz}\), \(C = 19\,\text{pF} = 19\times10^{-12}\,\text{F}\)
- $$X_C \approx \frac{1}{2\pi(21.2\times10^6)(19\times10^{-12})} \approx 395\,\Omega$$
- $$Z \approx 300 - j\,395\,\Omega$$
- $$\phi \approx \tan^{-1}\!\left(\frac{-395}{300}\right) \approx -53^\circ$$

Interpretation on the R–jX plane:
- Real part positive (to the right): resistive component is 300 Ω.
- Imaginary part negative (below axis): capacitive reactance.
- Magnitude of reactance exceeds resistance (\(|X_C| > R\)): angle more negative than −45°.

Why the correct answer (Point 1) is correct:
- Point 1 on Figure E5-1 represents the fourth quadrant (positive R, negative X) with |X| > R. That matches \(Z \approx 300 - j395\,\Omega\).

Why the other choices are incorrect:
- Point 3: Fourth quadrant but with R > |X| (angle between 0 and −45°). Our circuit has |X| > R, so not Point 3.
- Point 7: Inductive region (positive X, above the axis). Our circuit is capacitive (negative X).
- Point 8: Also in the inductive half-plane or otherwise not in the fourth-quadrant capacitive region matching |X| > R.

Rules of thumb:
- Inductive reactance: \(+jX_L\) (above axis); Capacitive reactance: \(-jX_C\) (below axis).
- Series R + reactance plots at coordinates \((R, \pm X)\).
- If \(|X| > R\), the point lies beyond the ±45° line from the origin in the corresponding quadrant.