In the shown source network, a 5 V source in series with 1 Ω is parallel to an upward 1 A current source. What load current IL gives maximum load power?
With load open, the upward one-ampere source drives the one-ohm branch, raising the terminal voltage one volt above the five-volt source: Vth=6 V. Deactivating independent sources leaves Rth=1 Ω. Maximum power occurs at RL=Rth, so IL=6/(1+1)=3 A, A.
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