ISC (Lapt KCL @@ Node A: I₁ = Io +1₂ @ KVL @ loop 1 A I To 1₂ 10 1₁ + 51 -15V + 10V=0 101 +5 KVL @ Loop 2 3 Vo. 61₂ +31₂ HSV-S1₂ =D 912-510 = -15V 37 32 A 37 = - 40 A 37 Use systerns of linear equation to find the 3 currents: A = Vo = 35 (1₂) =-120 las (Loop 2 = 22 (-40) 244 = 11,054 41 A 37 Vo 5V @ 1-3.24 V 352 241 a la (hopp) 24.5 f Flose Loop 2 450 th C Sol'n KVL @ Loop 1 121 + 1012-24V=D 121 + 101₂ = 24V 0 KVL @ Loop 2: 241 +450-10 5₂00 ки @ побе А To=11+1₂ 4.0 J₁-To 112-2 like cysterns of linear equation to find the 3 currents: 9A 43 T₁ = 52 A I₁ = I₂ = 42 A 43 1₁ = SA 43 -11-19 A How is this solve?
ISC (Lapt KCL @@ Node A: I₁ = Io +1₂ @ KVL @ loop 1 A I To 1₂ 10 1₁ + 51 -15V + 10V=0 101 +5 KVL @ Loop 2 3 Vo. 61₂ +31₂ HSV-S1₂ =D 912-510 = -15V 37 32 A 37 = - 40 A 37 Use systerns of linear equation to find the 3 currents: A = Vo = 35 (1₂) =-120 las (Loop 2 = 22 (-40) 244 = 11,054 41 A 37 Vo 5V @ 1-3.24 V 352 241 a la (hopp) 24.5 f Flose Loop 2 450 th C Sol'n KVL @ Loop 1 121 + 1012-24V=D 121 + 101₂ = 24V 0 KVL @ Loop 2: 241 +450-10 5₂00 ки @ побе А To=11+1₂ 4.0 J₁-To 112-2 like cysterns of linear equation to find the 3 currents: 9A 43 T₁ = 52 A I₁ = I₂ = 42 A 43 1₁ = SA 43 -11-19 A How is this solve?
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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Concept explainers
KVL and KCL
KVL stands for Kirchhoff voltage law. KVL states that the total voltage drops around the loop in any closed electric circuit is equal to the sum of total voltage drop in the same closed loop.
Sign Convention
Science and technology incorporate some ideas and techniques of their own to understand a system skilfully and easily. These techniques are called conventions. For example: Sign conventions of mirrors are used to understand the phenomenon of reflection and refraction in an easier way.
Question
1
Transcribed Image Text:LOV
ISC
(Lapt
KCL @@ Node A:
I₁ = I₁ +1₂ @
KVL @ loop 1
A
5₁
To
10 1₁ + 51 -15V + 10V=0
101 +5
= 5V Ⓒ
KVL @ Loop 2
61₂ +31₂ HSV-S1₂ =D
912-510 = -15V
1₂
Vo.
Use systems of linear equation to
find the 3 currents :
2
A
37
32 A
37
= - 40 A
37
1 las2
Vo = 35 (1₂)
=-120
(Loop 2
394 = 11,054 4 |
A
= 22 (-40)
Vo
1-324 V
352
240
In 1200
(hopp)
24.01.
f
Flose Loop 2 450
th
Coy
Sol'n
KVL @ Loop 1
121 + 1012-24V=D
121 + 101₂ = 24V 0
KVL @ Loop 2:
241
4.0
+450-10 5₂=0 @
KCL @node A
To=1₁+1₂
J₁-To 112-0
like cysterns of linear equation
to find the 3 currents:
9A
1₁ = 42²
1₁ = 52 A
43
I₂ = 42 A
43
1₁ = 51A
43
-//-19 A
How is this
solve?
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