Engineering Board Exam Reviewer · Circuit Theorems in AC Analysis · medium
A source of 240 V 0° drives a branch Z1=20-j10, in series with the parallel branch Z2=80+j50, , as shown. Terminals a and b are the open terminals across Z2. For maximum power transfer, what load impedance should be connected across a-b (rectangular form)?
- 18.8+j5.52,
- 18.8-j5.52,
- 19.6,
- 5.52+j18.8,
Answer
A. 18.8+j5.52,
Solution
With the source replaced by a short circuit, ZTh=Z1 Z2=Z1Z2/Z1+Z2=19.6 -16.4°, . Maximum power transfer needs the complex conjugate: ZL=ZTh*=18.8+j5.52, .
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