For the following three problems, use resistance values R₁ 500 , R6 600 , and R7 R4 400, R5 = node-voltage analysis to solve these problems. = 18 V + R₂ www R3 VA I www 2. Using the resistance values above, solve for i in the circuit shown in Prob. 4.1 of N&R (12th ed.). However, instead of using the circuit as drawn, replace the dependent current source with a 5 V voltage source. The lower polarity is to the left and the higher polarity is to the right. Other than changing this one source, there are no other changes to the circuit from the original. = R₁ R6 www R₁ 204 R5 25iA = R₁ 100 , R₂ 700 2 (in short, Rn - = 200 N, R3 300 , = nx 100 2). Use 5 mA =

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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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100%
100 , R₂
For the following three problems, use resistance values R₁
500 , R6 600 , and R7 700 2 (in short, Rn
R4
400, R5 =
node-voltage analysis to solve these problems.
=
18 V
+
R₂
www
R3
+3
2. Using the resistance values above, solve for i in the circuit shown in Prob. 4.1 of N&R
(12th ed.). However, instead of using the circuit as drawn, replace the dependent current
source with a 5 V voltage source. The lower polarity is to the left and the higher polarity is
to the right. Other than changing this one source, there are no other changes to the circuit
from the original.
I
www
R6
=
ww
R₁
R₁
204
R5
25iA
-
R₁
=
200 N, R3
300 ,
= nx 100 2). Use
5 mA
=
Transcribed Image Text:100 , R₂ For the following three problems, use resistance values R₁ 500 , R6 600 , and R7 700 2 (in short, Rn R4 400, R5 = node-voltage analysis to solve these problems. = 18 V + R₂ www R3 +3 2. Using the resistance values above, solve for i in the circuit shown in Prob. 4.1 of N&R (12th ed.). However, instead of using the circuit as drawn, replace the dependent current source with a 5 V voltage source. The lower polarity is to the left and the higher polarity is to the right. Other than changing this one source, there are no other changes to the circuit from the original. I www R6 = ww R₁ R₁ 204 R5 25iA - R₁ = 200 N, R3 300 , = nx 100 2). Use 5 mA =
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