Determination Of Equivalent Circuit Parameters Of Single Phase Induction
Motor
It is possible to find the parameters of the equivalent circuit of the single phase induction motor
experimentally as shown in Fig.1.
I1
R1
2
j
X1
2
j
Rw
2
j
Xm
2
Rw
2
j
Xm
2
X 2′
2
V1
R1
2
j
X1
2
j
X 2′
2




1 R2′  Forward
2 s 






1 R2′  Backward

2 2−s 



Fig.1: Equivalent circuit of single phase induction motor.
For this purpose, three tests should be conducted:
1- The DC Test:
The DC resistance of the stator can be measured by applying DC current to the terminals of the
main winding and taking the reading of the voltage and the current (or using ohmmeter) and
determine the DC resistance as fallows:
VDC
I DC
Then, the AC resistance is given by:
RDC =
(1)
R1 = 1.15 R DC
(2)
2- The Locked Rotor Test
When the rotor is locked (i.e. prevented from running), sb = s f = 1 . The secondary impedances
become much less than the magnetizing branches and the corresponding equivalent circuit becomes
that of Fig.2.
I BL
R1
2
j
X1
2
j
X 2′
2
R2′
2
VBL
R2′
2
R1
2
j
X1
2
j
X 2′
2
Fig.2: Approximate equivalent circuit of the single phase induction motor at standstill
The circuit in Fig.2 can be rearranged to the equivalent circuit that is shown in Fig.3.
I BL
R1
jX 1
jX 2′
R2′
VBL
Fig.3: Rearranged approximate equivalent circuit of the single phase induction motor at standstill.
The readings to be obtained from this test are:
a) Single phase power PBL
b) Phase voltage VBL
c) Phase current I BL
Then, Req , Z eq , and X eq can be obtained using the following equations:
Req =
PBL
2
I BL
(3)
Z eq =
VBL
I BL
(4)
X eq = Z eq2 − Req2
(5)
Separation of X 1 , X 2′ , R1 , and R2′ can be done as follows:
1
X eq
2
X 1 = X 2′ =
(6)
R2′ = Req − R1
(7)
3- The No Load Test
When the induction motor is allowed to run freely at no load, the forward slip s f approaches
zero and the backward slip sb approaches 2 ( s f = s , sb = 2 − s ). The secondary forward impedance
becomes very large with respect to the magnetizing branch, while the secondary backward
impedance becomes very small if compared with the magnetizing branch. Accordingly, the
equivalent circuit corresponding to these operating conditions can be approximated by that of Fig.4.
I NL
R1
2
j
X1
2
Rw
2
j
Xm
2




 Forward




E1F
VNL
R2′
4
R1
2
j
X1
2
j
X 2′
2




 Backward




Fig.4: Approximate equivalent circuit of the single phase induction motor at no load.
The circuit in Fig.4 can be rearranged to the equivalent circuit that is shown in Fig.5.
I NL
R1
jX 1
Rw
2
j
Xm
2
E1F
VNL
R2′
4
j
X 2′
2




 Forward








 Backward




Fig.5: Rearranged approximate equivalent circuit of the single phase induction motor at no load
The readings to be obtained from this test are:
d) Single phase power PNL
e) Phase voltage VNL
f) Phase current I NL
Then, Rw , and X m , can be obtained as fallows:
Note:
R′ 
2 
Pcore+ mechanical = PNL − I NL
 R1 + 2 
4 

(8)
R′  
X ′ 

E1F = VNL − I NL   R1 + 2  + j  X 1 + 2  
4  
2 

(9)
( I NL = I NL ∠ − θ , θ = cos −1
Rw = 2
Iw =
E1F
PNL
)
V NL I NL
2
Pcore+mechanical
E1F
E
= 2 1F
Rw
 Rw 


 2 
(10)
(11)
2
I m = I NL
− I w2
Xm = 2
E1F
Im
(12)
(13)