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12th
August,2013
Third Semester
Electical Engg Dept,PEC University Of Technology,chandigarh
Course-Electric Machines1(EE214)
Lecture Notes-3
3-PHASE CIRCUITS
Course Coordinator
Dhiraj Bharat
EE Dept,PEC
Prepared by
Mehar Grewal
12104002
BE 2nd
year-electrical
-
What do you mean by three phase circuits?
3- balance circuit whether current or voltage are those which
have same
frequency and amplitude and are out of phase by 120degree.Phase
voltages- (
-
current are the same.
Delta connection-three separate inductive loads have been
connected to
form a delta connection. This connection receives its name from
the fact
that a schematic diagram of this connection resembles the Greek
letter delta
().In the delta connection, line voltage and phase voltage are
the same. The
line current of a delta connection is higher than the phase
current by a factor
of the square root of 3.
Types of connections-
1 Y Y ( Upper part represents how source is connected &lower
part
represents manner in which how node is connected)
2 Y-Y
3 Y (Y-node ,delta connected source)
-
4 Y
5
Zy=Z
-
I~CA=V~
CA/Z=3V
- V~ab =v
-
P=3(VPh IPh cos)
Cas1-connected
Hence VPh =Vl & IPh =IL/3
Real power-PSD =3 VPh IPh cos
=3VL (IL 3) cos = 3 VL
NOW node is Y connected
Hence VPh =Vl/3 & IPh =IL
P= VPN IPN cos
=3VL (IL /3) cos = 3 VL I
Reavtive power Q= 3 VPN I
What is the Q in the given circuit?
The circuit is given below in image. Theres a line
.Since the load is unbalanced so cant use =
Q
SL =(220)2/-J6(*J6)
cos
L IL cos
NOW node is Y connected
IL cos
IPN sin= 3 VL IL sin
What is the Q in the given circuit?
The circuit is given below in image. Theres a line IN FROM NODE
AT J6 to
Since the load is unbalanced so cant use = 3 VL IL sin. Only
inductor will use
FROM NODE AT J6 to AN
sin. Only inductor will use
-
=(220)2 J/6
=SC =(220)2/J3==(220)2 J/3
=SR =(220)2/6
So, P=(220)2/6+=(220)2(1/6-1/3)
Q=(220)2/6
If asked to find neutral current then=IA +IB+IC
THIS IS THE SOL WHEN NO IN LINE.
