www V₁B + V3B + R3 W 10 Ω V2B VB 9 V + REQ V2B VEQ 7.5 Ω 9V Practice A circuit VB=7\ and V3 f FIGURE 7.6 Solution: One(c) The first step is to split the circuit into two single source circuits. These circuits (along with their series equivalents) are shown in Figures 7.6b and 7.6c. For Figure 7.6b, REQ = R2|| R3 = 15 || 10 = 6N The re Figure 7.7 ple. Note greater v . Si .Si andum REQ 6 Ω VEQ = VA REQ + R₁ = = (12 V) =2V 36 Ω Since this voltage is across the parallel combination of R2 and R3, V2A = V3A = 2 V ЗА and PTER 7 Circuit Analysis Techniques V₁ = VA - VEQ = 12 V-2 V = 10 V ✓ VA- 12 V FIGU R₁ 180 ww 20 20 V ÷ R₂ 180 www R₂ 680 VB 10 V ww www R 220

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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  1. Determine the values of V1 through V4 for the circuits shown in the following figure (Fig. 7.43b).                           I believe this Superposition Theorem. The circuit below has 4 resistors but my examples have only 3. The question seems ask for the voltage values but others have included current values. I do not know how to calculate V4. I am not sure if I use the equivalent of V3||V4 for V3 and V4 voltage. I will leave example photos out my book but it does not show how to do V4.
www
V₁B
+
V3B
+
R3
W
10 Ω
V2B
VB
9 V
+
REQ
V2B
VEQ
7.5 Ω
9V
Practice
A circuit
VB=7\
and V3 f
FIGURE 7.6
Solution:
One(c)
The first step is to split the circuit into two single source circuits. These circuits (along
with their series equivalents) are shown in Figures 7.6b and 7.6c. For Figure 7.6b,
REQ = R2|| R3 = 15 || 10
= 6N
The re
Figure 7.7
ple. Note
greater v
. Si
.Si
andum
REQ
6 Ω
VEQ = VA REQ + R₁
=
= (12 V)
=2V
36 Ω
Since this voltage is across the parallel combination of R2 and R3,
V2A = V3A = 2 V
ЗА
and
PTER 7
Circuit Analysis Techniques
V₁ = VA - VEQ = 12 V-2 V = 10 V
✓
VA-
12 V
FIGU
Transcribed Image Text:www V₁B + V3B + R3 W 10 Ω V2B VB 9 V + REQ V2B VEQ 7.5 Ω 9V Practice A circuit VB=7\ and V3 f FIGURE 7.6 Solution: One(c) The first step is to split the circuit into two single source circuits. These circuits (along with their series equivalents) are shown in Figures 7.6b and 7.6c. For Figure 7.6b, REQ = R2|| R3 = 15 || 10 = 6N The re Figure 7.7 ple. Note greater v . Si .Si andum REQ 6 Ω VEQ = VA REQ + R₁ = = (12 V) =2V 36 Ω Since this voltage is across the parallel combination of R2 and R3, V2A = V3A = 2 V ЗА and PTER 7 Circuit Analysis Techniques V₁ = VA - VEQ = 12 V-2 V = 10 V ✓ VA- 12 V FIGU
R₁
180
ww
20
20 V
÷
R₂
180
www
R₂
680
VB
10 V
ww
www
R
220
Transcribed Image Text:R₁ 180 ww 20 20 V ÷ R₂ 180 www R₂ 680 VB 10 V ww www R 220
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