Determine the Norton equivalent of the circuit in Fig. 10.30 as seen from terminals a-b. Use the equivalent to find Io Π2 Ω 10/0° V 8 Ω 4Ω 1Ω -j3Ω τ (+) 22-90° A α b Figure 10.30 Answer: Zy = 3.176 + j0.706 Ω, Ι. 1ο = 985.5/-2.101° mA. 10 Ω -j5Ω = 4.198/-32.68° A.
Determine the Norton equivalent of the circuit in Fig. 10.30 as seen from terminals a-b. Use the equivalent to find Io Π2 Ω 10/0° V 8 Ω 4Ω 1Ω -j3Ω τ (+) 22-90° A α b Figure 10.30 Answer: Zy = 3.176 + j0.706 Ω, Ι. 1ο = 985.5/-2.101° mA. 10 Ω -j5Ω = 4.198/-32.68° A.
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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
Transcribed Image Text:Determine the Norton equivalent of the circuit in Fig. 10.30 as seen
from terminals a-b. Use the equivalent to find Lo
j2 52
10/0° V
892
www
Figure 10.30
492
wwwww
=
1Ω -j3Ω
www
2/-90° A
Answer: ZN
I, 985.5/-2.101° mA.
10 Ω
-j5 52
3.176 + j0.706 2. IN = 4.198/-32.68° A.
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