Apply mesh analysis to find the mesh currents in the circuit. Use the information to determine the voltage V where Va-15, Vb=19, R1=4, R2=11, R3-15, and R4-9. Va Answer: R1 V R3 ww R2 12 R4 Vb. Hide question 2 feedback Feedback Apply mesh analysis to find the mesh currents in the circuit of Use the information to determine the voltage V. 16 V 202 w V 292 ww 3402 12 12 V Solution: Application of KVL to the two loops gives: Mesh 1: Mesh 2: -16+2+3(-12)=0, 3(2-1)+(2+4)/2+12=0, which can be simplified to 511-312-16 -311+9/2 -12. EG (1) (2) Simultaneous solution of (1) and (2) leads to Hence, 4₁ = 3 A, A. V=3(-2)=3(3+1 -3 (3+ 1/3) =1 10 - 10 V.

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.61P
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Apply mesh analysis to find the mesh currents in the circuit. Use the information to determine the voltage
V where Va-15, Vb=19, R1=4, R2=11, R3-15, and R4-9.
Va
Answer:
R1
V
R3
ww
R2
12
R4
Vb.
Hide question 2 feedback
Feedback
Apply mesh analysis to find the mesh currents in the circuit of
Use the information to determine the voltage V.
16 V
202
w
V
292
ww
3402
12
12 V
Solution: Application of KVL to the two loops gives:
Mesh 1:
Mesh 2:
-16+2+3(-12)=0,
3(2-1)+(2+4)/2+12=0,
which can be simplified to
511-312-16
-311+9/2 -12.
EG
(1)
(2)
Simultaneous solution of (1) and (2) leads to
Hence,
4₁ = 3 A,
A.
V=3(-2)=3(3+1
-3 (3+ 1/3) =1
10
- 10 V.
Transcribed Image Text:Apply mesh analysis to find the mesh currents in the circuit. Use the information to determine the voltage V where Va-15, Vb=19, R1=4, R2=11, R3-15, and R4-9. Va Answer: R1 V R3 ww R2 12 R4 Vb. Hide question 2 feedback Feedback Apply mesh analysis to find the mesh currents in the circuit of Use the information to determine the voltage V. 16 V 202 w V 292 ww 3402 12 12 V Solution: Application of KVL to the two loops gives: Mesh 1: Mesh 2: -16+2+3(-12)=0, 3(2-1)+(2+4)/2+12=0, which can be simplified to 511-312-16 -311+9/2 -12. EG (1) (2) Simultaneous solution of (1) and (2) leads to Hence, 4₁ = 3 A, A. V=3(-2)=3(3+1 -3 (3+ 1/3) =1 10 - 10 V.
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