Question E4B06:
How much power is being absorbed by the load when a directional power meter connected between a transmitter and a terminating load reads 100 watts forward power and 25 watts reflected power?  

Choices:  
A: 100 watts  
B: 125 watts  
C: 112.5 watts  
D: 75 watts    

Correct Answer is  D: 75 watts  

To find power absorbed by the load, subtract the reflected power from the forward power measured by a directional wattmeter.

Calculation:
$$P_{\text{load}} = P_{\text{fwd}} - P_{\text{ref}} = 100\,\text{W} - 25\,\text{W} = 75\,\text{W}$$

Why D (75 watts) is correct:
- The wattmeter’s “forward” reading is the incident power on the load, and the “reflected” reading is what comes back due to mismatch. Only the difference is actually absorbed by the load.

Why the others are incorrect:
- A: 100 watts — This ignores the reflected power. You’d only get 100 W into the load if the load were perfectly matched (no reflection).
- B: 125 watts — Adding forward and reflected is wrong; reflected power is not delivered to the load, so summing double-counts energy.
- C: 112.5 watts — There’s no basis for subtracting only part of the reflected power; the correct relationship is simple subtraction of all reflected power from forward.

Useful formulas and tips:
- Net power into the load: $$P_{\text{load}} = P_{\text{fwd}} - P_{\text{ref}}$$
- Reflection coefficient magnitude: $$\rho = \sqrt{\frac{P_{\text{ref}}}{P_{\text{fwd}}}} = \sqrt{\frac{25}{100}} = 0.5$$
- Fraction of forward power absorbed: $$1 - \rho^2 = 1 - 0.25 = 0.75 \Rightarrow 75\,\text{W}$$
- Optional: SWR from ρ: $$\text{SWR} = \frac{1+\rho}{1-\rho} = \frac{1+0.5}{1-0.5} = 3:1$$