Please show 7 detailed Solution on this Part, and how it become That answer.

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)
10th Edition
ISBN:9781337399128
Author:Russell E. Smith
Publisher:Russell E. Smith
Chapter7: Alternating Current, Power Distribution, And Voltage Systems
Section: Chapter Questions
Problem 7RQ
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B
Sol'n:
JAR
SA
20
ALL
Iza + √₂
Loop 1
KCL @node A
+
1₂622
10A + 14 = 5A + Ion
10A-SA= 16-14
SA=In In
ICCL@node B
5A = IAR + 1₂ Ⓒ
KVL@ Loop 1
JOA
• V.- 2a (²0)
400V
V.
√₂ = 62 (HA)
√√√₂ = 20V
-V₁ +43 + √₂=0
- (20) Izet (40) 142 + (61) 1₂-0
-212n + 414 + 610 @
|V₂ = 40 (-A)
√√₁₂ = -20 V
2
Use systems of linear equations to get all
three currents (Use calculator for
shortcut)
Im - 2A:
I4 = -A
Ion = 10 A
Please show
> detailed
Solution on this
Part, and
how it become
That answer.
Transcribed Image Text:B Sol'n: JAR SA 20 ALL Iza + √₂ Loop 1 KCL @node A + 1₂622 10A + 14 = 5A + Ion 10A-SA= 16-14 SA=In In ICCL@node B 5A = IAR + 1₂ Ⓒ KVL@ Loop 1 JOA • V.- 2a (²0) 400V V. √₂ = 62 (HA) √√√₂ = 20V -V₁ +43 + √₂=0 - (20) Izet (40) 142 + (61) 1₂-0 -212n + 414 + 610 @ |V₂ = 40 (-A) √√₁₂ = -20 V 2 Use systems of linear equations to get all three currents (Use calculator for shortcut) Im - 2A: I4 = -A Ion = 10 A Please show > detailed Solution on this Part, and how it become That answer.
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