Question E5C11:
Which point in Figure E5-1 best represents the impedance of a series circuit consisting of a 300-ohm resistor and an 18-microhenry inductor at 3.505 MHz?  

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

Correct Answer is  B: Point 3  

To analyze a series R–L circuit, use:
- Inductive reactance: $$X_L = 2\pi f L$$
- Series impedance: $$Z = R + jX_L$$

Compute the reactance at 3.505 MHz with L = 18 μH:
- Handy rule of thumb: $$X_L(\Omega) \approx 6.283 \times f_{\text{MHz}} \times L_{\mu H}$$
- Thus: $$X_L \approx 6.283 \times 3.505 \times 18 \approx 396\ \Omega$$

So the impedance is:
$$Z \approx 300 + j396\ \Omega$$
This is in the first quadrant (positive resistance and positive, inductive reactance), with reactance larger than resistance (angle ≈ arctan(396/300) ≈ 53°).

Why the correct answer (Point 3) is correct:
- Point 3 on Figure E5-1 represents an impedance with R > 0 and X > 0 (upper-right quadrant), matching a series R–L circuit. Since X > R, it would plot above the 45° line—consistent with 300 + j396 Ω.

Why the other choices are incorrect:
- Point 1: Represents a purely inductive reactance (R = 0, X > 0) or negligible R; our circuit has R = 300 Ω.
- Point 7: Left half-plane (R < 0) or otherwise not in the first quadrant; our R is positive.
- Point 8: Lower-right quadrant (R > 0, X < 0), which is capacitive; our circuit is inductive (X > 0).

Key formulas/rules:
- $$X_L = 2\pi f L$$
- $$Z_{\text{series}} = R + jX$$
- Rule of thumb: $$X_L(\Omega) \approx 6.283 \times f_{\text{MHz}} \times L_{\mu H}$$
- Inductive: X > 0 (plots above the R-axis); capacitive: X < 0 (below the R-axis).