Question E9D09:
What is the function of a loading coil in an electrically short antenna?  

Choices:  
A: To increase the SWR bandwidth by increasing net reactance  
B: To lower the losses  
C: To lower the Q  
D: To resonate the antenna by cancelling the capacitive reactance    

Correct Answer is  D: To resonate the antenna by cancelling the capacitive reactance  

To resonate the antenna by cancelling the capacitive reactance

Why this is correct:
- An electrically short antenna (shorter than about a quarter-wave) looks capacitive at the feedpoint: its reactance is $X_C < 0$.
- A loading coil adds inductive reactance $X_L > 0$. At the operating frequency, choose $L$ so:
  $$X_L + X_C = 0 \quad \Rightarrow \quad 2\pi f L = \frac{1}{2\pi f C}$$
- When the inductive and capacitive reactances cancel, the antenna is resonant (input impedance is primarily resistive), improving match potential.

Why the other choices are incorrect:
- A: “Increase SWR bandwidth by increasing net reactance” — Incorrect. At resonance the net reactance is brought to zero, not increased. Also, loading coils generally narrow the SWR bandwidth.
- B: “Lower the losses” — Incorrect. Coils add series resistance and higher circulating currents, increasing I²R loss and often reducing efficiency.
- C: “Lower the Q” — Incorrect. Loading typically increases stored reactive energy relative to radiation/ohmic resistance, raising $Q$ and narrowing bandwidth.

Useful formulas and rules of thumb:
- Inductive reactance: $$X_L = 2\pi f L$$
- Capacitive reactance: $$X_C = -\frac{1}{2\pi f C}$$
- Resonance condition: $$X_L + X_C = 0$$
- If you know the antenna’s capacitive reactance magnitude $|X_C|$ at $f$, pick:
  $$L = \frac{|X_C|}{2\pi f}$$
- Bandwidth relation: $$\frac{\Delta f}{f_0} \approx \frac{1}{Q}$$
- Short antennas have low radiation resistance and high $Q$; loading coils generally increase $Q$ (narrower bandwidth). For better efficiency and bandwidth, top-loading (capacity hats) is preferred over adding lossier inductors.