Why we did not multiply by 3
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
Related questions
Question
100%
Why we did not multiply by 3 because of the three phases?? And you can see the rules in the second picture if I am mistaken so when we use them
![0-714
8:26 9
Slepi
а)
Coiven :-
Vab = 5KV (Line vo Hage)
Source
6= G0KVAR
pf.50.7
S= 100KVA
"P = ২০০k
p.f= 0-6
lead)-
Load -1
Lead - 2
Load-3
Load -1
5두 100 KVA
S, = 100 L45-573° KVA
Load - 2
१, = २०० k५
SLz = PLz
0.6
SL2 = 1000 KVA
SL2 = lo00 feos"(0. 6)
= lo00 /- 5313° kVA
Load-3
cosp = 0,7
O = 45,573°
sin iþ =0=714
Qug = '60 KUAR
Sind
Si3=
= 84.03 KVA
Step2
b)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa145743f-9cae-4226-a3fd-bd208f7f38e5%2F7d24d787-e3cb-4272-9807-8be4f0be906d%2Fq6jutv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:0-714
8:26 9
Slepi
а)
Coiven :-
Vab = 5KV (Line vo Hage)
Source
6= G0KVAR
pf.50.7
S= 100KVA
"P = ২০০k
p.f= 0-6
lead)-
Load -1
Lead - 2
Load-3
Load -1
5두 100 KVA
S, = 100 L45-573° KVA
Load - 2
१, = २०० k५
SLz = PLz
0.6
SL2 = 1000 KVA
SL2 = lo00 feos"(0. 6)
= lo00 /- 5313° kVA
Load-3
cosp = 0,7
O = 45,573°
sin iþ =0=714
Qug = '60 KUAR
Sind
Si3=
= 84.03 KVA
Step2
b)
![Zp=
Impedkince per phase
e = p.f. angle = Angle of Zph= Angle between
Tpiand ph
Then:
P= 3 Vp Ip cag O = V3 VL IL Coze Watt
Total Real Power:
Total Reactive Pawver: Q= 3 VpIp Sine = V3 VL I Sins VAR
To tal Apparant Pawer: S = 3Vp Ip = V3 VL IL
%3D
%3D
VA
%3D
Total Complex Poner: 3- P+jQ = 3p = 32pTp.
3.
Power Factor = Cgo=
=
%3D
P= Re CŠ)
Q= Im 3)
S' = IŠI
Angle of s
Also:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa145743f-9cae-4226-a3fd-bd208f7f38e5%2F7d24d787-e3cb-4272-9807-8be4f0be906d%2Fb3tvji7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Zp=
Impedkince per phase
e = p.f. angle = Angle of Zph= Angle between
Tpiand ph
Then:
P= 3 Vp Ip cag O = V3 VL IL Coze Watt
Total Real Power:
Total Reactive Pawver: Q= 3 VpIp Sine = V3 VL I Sins VAR
To tal Apparant Pawer: S = 3Vp Ip = V3 VL IL
%3D
%3D
VA
%3D
Total Complex Poner: 3- P+jQ = 3p = 32pTp.
3.
Power Factor = Cgo=
=
%3D
P= Re CŠ)
Q= Im 3)
S' = IŠI
Angle of s
Also:
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
Step 1
Condition no (01)
If the value of the load in the question is given as power [ whatever Active power (P), Reactive power (Q), Apparent Power (S) and together with the power factor ]
So we will consider it as three phase Power only (not consider as per phase power)
So there is no need to multiply it by 3
Step by step
Solved in 3 steps with 1 images
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)
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