*Question G4E04:
Why should DC power for a 100-watt HF transceiver not be supplied by a vehicle’s auxiliary power socket?*  

*Correct Answer is  B: The socket’s wiring may be inadequate for the current drawn by the transceiver*  

### AI Response: (may contain errors)  

For a 100-watt HF transceiver, the correct answer is:

**B: The socket’s wiring may be inadequate for the current drawn by the transceiver**

### Explanation:

1. **Why the correct answer is correct:**
   - A 100-watt HF transceiver typically requires a significant amount of current, often around 20-25 amperes at 13.8 volts DC. Vehicle auxiliary power sockets, commonly known as cigarette lighter sockets, are usually designed for lower current loads, often around 10-15 amperes. Using a socket with inadequate wiring can lead to overheating, voltage drops, or even damage to the vehicle's electrical system.

2. **Why the other choices are incorrect:**
   - **A: The socket is not wired with an RF-shielded power cable**  
     This is not relevant to the power supply issue. RF shielding is more about preventing interference, not about handling current capacity.
   
   - **C: The DC polarity of the socket is reversed from the polarity of modern HF transceivers**  
     Most vehicle auxiliary sockets are standardized with the correct polarity for typical 12V devices. Polarity issues are rare and not the primary concern here.
   
   - **D: Drawing more than 50 watts from this socket could cause the engine to overheat**  
     The engine overheating is not directly related to the power drawn from the auxiliary socket. The concern is more about the wiring's ability to handle the current safely.

### Useful Rule of Thumb:

- **Current Calculation for Power**:  
  The current ($I$) required can be calculated using the formula:  
  $$ I = \frac{P}{V} $$  
  where:
  - $P$ = power in watts (100 watts for the transceiver)
  - $V$ = voltage in volts (typically 13.8 volts for vehicle systems)

  For a 100-watt transceiver:  
  $$ I = \frac{100}{13.8} \approx 7.25 \text{ amperes} $$

  However, this is the ideal current. In practice, due to inefficiencies and peak power demands, the actual current draw can be higher, often requiring 20-25 amperes. This exceeds the typical capacity of a vehicle's auxiliary power socket.