Engineering Board Exam Reviewer · Circuit Theorems in AC Analysis · medium
A source of 60 V 0° drives a branch Z1=20-j60, in series with the parallel branch Z2=-j50, , as shown. Terminals a and b are the open terminals across Z2. What is the maximum average power that can be delivered to a load connected across a-b?
- 180 W
- 45 W
- 6.36 W
- 90 W
Answer
B. 45 W
Solution
With the source replaced by a short circuit, ZTh=Z1 Z2=Z1Z2/Z1+Z2=28.3 -81.9°, . VTh=VZ2/Z1+Z2=60 V·-j50, /20-j110, =26.8 -10.3° V. RTh=4, . With ZL=ZTh*, Pmax= |VTh|24RTh= (26.8 V)24(4, )=45 W.
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