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4mA ↑ Ⓒ ins lka ka 12 222mA m 4A) A lkn = I₂ Eselivered 00 -47 V₁=0.20 10 -112V/₂=16.4V 2 O O 03-2-4 V₁=16V @ -1 mA + V₂ - V₁ + L-2 -1 -2 5/2 √√√4=441 Ik V₁= V₁ = 1.6V (√₂ = @Made Abralysis ⒸY₁+ V₁ - V₁₂+ 4m = 0 TL TE ZIV₁-V₁=-4mA 2ka IMA 4 kak @-4m+V₂-V₁-2m=0 2.k 12V/₂-1k V₂ = 12 m 32m + V₂-V₁ + V₂ =0 16 2k 3kV₂-2V/4= -4m. + 1₂ 3¹ Vo ه - ولا ولا خلا ولا Ik -2kV₁- IKV₂-2V₂ +5V₂ = 2m 2k b) kop Analysis Ⓒlk (1-4) + x + lk (1₂) = 0 2k1₂-4V+x=0 ⒸIKCT+2m) + 2k (Tu)-X-0 32 14 +2V-X=6 2ks +3k 14-2V=0 2k1₂ + 6V +3kIs-2V=0 5kI3=-4V 51 5k Is=-0.2mA 14-1= Im 14 = m + 13 I4 = 1m+-0. 14 = 0.8m V.-IR. = 0.8m (2k) = 1.6V

4mA ↑ Ⓒ ins lka ka 12 222mA m 4A) A lkn = I₂ Eselivered 00 -47 V₁=0.20 10 -112V/₂=16.4V 2 O O 03-2-4 V₁=16V @ -1 mA + V₂ - V₁ + L-2 -1 -2 5/2 √√√4=441 Ik V₁= V₁ = 1.6V (√₂ = @Made Abralysis ⒸY₁+ V₁ - V₁₂+ 4m = 0 TL TE ZIV₁-V₁=-4mA 2ka IMA 4 kak @-4m+V₂-V₁-2m=0 2.k 12V/₂-1k V₂ = 12 m 32m + V₂-V₁ + V₂ =0 16 2k 3kV₂-2V/4= -4m. + 1₂ 3¹ Vo ه - ولا ولا خلا ولا Ik -2kV₁- IKV₂-2V₂ +5V₂ = 2m 2k b) kop Analysis Ⓒlk (1-4) + x + lk (1₂) = 0 2k1₂-4V+x=0 ⒸIKCT+2m) + 2k (Tu)-X-0 32 14 +2V-X=6 2ks +3k 14-2V=0 2k1₂ + 6V +3kIs-2V=0 5kI3=-4V 51 5k Is=-0.2mA 14-1= Im 14 = m + 13 I4 = 1m+-0. 14 = 0.8m V.-IR. = 0.8m (2k) = 1.6V

Introductory Circuit Analysis (13th Edition)
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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how do i verify my answer for V0 using tellegens theoerm?
1. 4 mA (↑) it a m Ika Ika {ILA NOOT 4MA lkn ² Is 2k22mA Edelivered +w 00-11-47 V₁=0.21 1 O 12 V₂=16.4V 0 3-2-4 V₁-1.6V @ 1mA + V₂ - V₁ + V₁ V₂₁ V₂ V₂ = 0 L-2 -1 -2 5 2 √√√4=441 Ik Ik 2k -2kV₁- IKV₂-2V₂ +5V4 = 2m V₁= V₁ = 1.6V OF (2A) Ⓒ 2ka a Mode Analysis S 2kak ⒸV₁ + V₁ - V₁ + 4m = 0 IL + @ -4m + V₂-V₁-2m=0 2.k 12V/₂-1k V₁ = 12 m IL 2LV₁-V₁-4mA € + 32k Vo 2m + V₂ Vy+ V₂ = 0 12 2k 3KV₂-2V4 = -4m b) keep Analysis Ⓒlk (1-4)+x+ lk (1₂) = 0 2k1₂-4V+x=0 Ⓒlk (T+ + 2m) + 2k (Tu) - X=0 32 +2V-X=6 2k13 +3k 14-2V=O 2k1s+ 6V+3k]s=2V=0 5kI3 = -4V 5k 5k 1₂=-0.2mA 14-1= Im 14 = m + 13 I4 = 1m+-0.2m I4 = 0.8m V. Luk. = 0.8m (2k) = 1.6V
Transcribed Image Text:1. 4 mA (↑) it a m Ika Ika {ILA NOOT 4MA lkn ² Is 2k22mA Edelivered +w 00-11-47 V₁=0.21 1 O 12 V₂=16.4V 0 3-2-4 V₁-1.6V @ 1mA + V₂ - V₁ + V₁ V₂₁ V₂ V₂ = 0 L-2 -1 -2 5 2 √√√4=441 Ik Ik 2k -2kV₁- IKV₂-2V₂ +5V4 = 2m V₁= V₁ = 1.6V OF (2A) Ⓒ 2ka a Mode Analysis S 2kak ⒸV₁ + V₁ - V₁ + 4m = 0 IL + @ -4m + V₂-V₁-2m=0 2.k 12V/₂-1k V₁ = 12 m IL 2LV₁-V₁-4mA € + 32k Vo 2m + V₂ Vy+ V₂ = 0 12 2k 3KV₂-2V4 = -4m b) keep Analysis Ⓒlk (1-4)+x+ lk (1₂) = 0 2k1₂-4V+x=0 Ⓒlk (T+ + 2m) + 2k (Tu) - X=0 32 +2V-X=6 2k13 +3k 14-2V=O 2k1s+ 6V+3k]s=2V=0 5kI3 = -4V 5k 5k 1₂=-0.2mA 14-1= Im 14 = m + 13 I4 = 1m+-0.2m I4 = 0.8m V. Luk. = 0.8m (2k) = 1.6V
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