Question: Repeat the analysis that led to eq. (3.40) for a load with impedance described by a power factor cos ?.Show that the fraction of power

Repeat the analysis that led to eq. (3.40)

RPL RPL = 2- RL Piost R %3D PL v2 RM

for a load with impedance described by a power factor cos ?.Show that the fraction of power lost compared to real power consumed in the load is approximately given by eq. (38.12)

when resistive losses are much smaller than power consumed in the load.

RPL RPL = 2- RL Piost R %3D PL v2 RM

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To repeat the analysis that led to eq 340 we start with the expression for the power consumed by a load with impedance ZL PL VL2 ReZL where VL is the ... View full answer

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