EMC Question of the Week: November 19, 2018

Which of the following receiver impedances receives the most average power from a 50-Ω source?

1. 25 Ω
2. 25 + j25 Ω
3. j50 Ω
4. 50 + j50 Ω

Answer

The correct answer is "a". Most electrical engineers would recognize that a 50-Ω receiver would maximize the power received, but 50 Ω was not one of the choices.  The average power received from a 50-Ω source can be calculated as,
$P_{REC}=\frac{1}{2}{\left|I\right|}^2{R}_{REC}=\frac{1}{2}\frac{V}{{\left|50+Z_{REC}\right|}^2}{R}_{REC}$
where V is the open circuit source voltage, Z_REC is the complex impedance of the receiver, and R_REC is the real part of the receiver impedance.

Using this formula, and assuming a 1-volt source, we can calculate the average received power for each of the receiver impedance options listed,

1. Z_REC= 2.2 mW
2. Z_REC= 2.0 mW
3. Z_REC= 0 mW
4. Z_REC= 2.0 mW

Note that the power received by a 50-Ω receiver would be 2.5 mW. It's also worth noting that option "c" could have been eliminated without any calculation, since no average power is received by a purely reactive load.

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