EMC Question of the Week: May 10, 2021
Two circuits send signals down a 300-meter cable and share the same return conductor. Circuit 1 delivers 120-volt, 60-Hz power to a 100-ohm load. 170 mV of coupled 60-Hz noise appears at the load end of Circuit 2 when Circuit 1 is active. Removing the load from Circuit 1 (leaving an open-circuit) has no effect on the coupled noise. The coupling is most likely
- conducted coupling
- capacitive coupling
- inductive coupling
- radiated coupling
Answer
The best answer is “b.” The circuit is electrically short at 60 Hz (λ = 5000 km). Both conducted coupling and inductive coupling would be proportional to the current in the Circuit 1. Removing the load would greatly reduce the coupled voltage if one of these coupling mechanisms dominated. Radiated coupling can also be ruled out because the circuits are very near each other relative to a wavelength.
Capacitive coupling is proportional to the voltage in Circuit 1. It's plausible that removing the load in Circuit 1 could have little effect on the power circuit voltage. Of the four possible EM coupling mechanisms, capacitive coupling is the only one consistent with the observations.
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