Determine the Norton equivalent of the circuit in the figure below as seen from terminalsa-b. Use the equivalent to find lo.Practice:16.10j2ll4.81-j3wwI,4/-90° A1020/0° V-j5Answer:ZN = 3.176 + j0.706 2, IN =8.396/-32.68° A, I, 1.971/-2.101° A.

Answered: Image /qna-images/answer/e0db7370-fdc0-45bb-a604-8f7311d7b7f1.jpg

Practice: Determine the Norton equivalent of the circuit in the figure below as seen from terminals 16.10 a-b. Use the equivalent to find lo. 4. j2 ww lll 8 1 -j3 ww ww 10 20/0° V 4/-90° A -j5 Answer: ZN = 3.176 + j0.706 2, IN =8.396/-32.68° A, I, 1.971/-2.101° A.

Practice: Determine the Norton equivalent of the circuit in the figure below as seen from terminals 16.10 a-b. Use the equivalent to find lo. 4. j2 ww lll 8 1 -j3 ww ww 10 20/0° V 4/-90° A -j5 Answer: ZN = 3.176 + j0.706 2, IN =8.396/-32.68° A, I, 1.971/-2.101° A.

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.61P

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Question

Determine the Norton equivalent of the circuit in the figure below as seen from terminals
a-b. Use the equivalent to find lo.
Practice:
16.10
j2
ll
4.
8
1
-j3
ww
I,
4/-90° A
10
20/0° V
-j5
Answer:
ZN = 3.176 + j0.706 2, IN =8.396/-32.68° A, I, 1.971/-2.101° A.

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