Practice:Determine the Norton equivalent of the circuit in the figure below as seen from terminalsa-b. Use the equivalent to find lo.16.104wwj2ell-138wwww1020/0° V) 4/-90° A--j5Answer:ZN = 3.176 + j0.706 2, IN = 8.396/- 32.68 A, I, = 1.971,2.101° A.

KIRCHHOFF’S LAW: KCL: The algebraic sum of all currents at a node is zero. Kirchhoff's current law…

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 ell -13 10 20/0° V 4/-90° A Answer: ZN 3.176+j0.706 2, IN =8.396/-32.68° A, L = 1.971/-2.101° A. %3D %3D %3D

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 ell -13 10 20/0° V 4/-90° A Answer: ZN 3.176+j0.706 2, IN =8.396/-32.68° A, L = 1.971/-2.101° A. %3D %3D %3D

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

Practice:
Determine the Norton equivalent of the circuit in the figure below as seen from terminals
a-b. Use the equivalent to find lo.
16.10
4
ww
j2
ell
-13
8
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.

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