ECE 320 HW # 6
1. Use words, equations, and sketches to describe how torque is produced in a squirrel-cage induction motor (IM).

2. Draw the equivalent circuit of an induction motor. Describe each component's meaning and representation of the equivalent circuit.

3. A 3-phase 2-pole IM is rated 230 V, 60 Hz with Y-connected windings. The stator and rotor resistances are 1 and 2 ohms, magnetizing reactance is 100 ohms, stator and rotor side leakage reactance are 2 and 2 ohms, respectively. a) Calculate the starting torque. b) Calculate the noload current and power factor.

Calculate the no-load current and power factor.
ω = o
ω s
I s , 1
=
(
1
+

 j
230


2
3
+ j 100
) ( 0 .
0128
− j 1 .
302
)
= p.f.
= cos(
θ
)
= cos(
−
89 .
438 )
=
0 .
0098
(
1 .
302
∠ −
89 .
438 o
)

4.) Describe how to use motor tests to determine IM’s parameters.
The no load induction motor test is similar to the open circuit transformer test. In the no load condition, the real power delivered to the load is zero and
ω o
= ω s which means that s=0 and the equivalent circuit is as follows:
In the no load test, V s
is rated voltage, and current I s,1
is measured.
I s , 1
= and
Z eq
=
R s
V s
Z eq .
+
[eq. #1]
X ls
+
X m [eq. #2].
R s
can be found by connecting an ohm meter to the motor when it is at rest and the terminals are open. Since X m
is much larger than X ls
, we ignore X ls
to get
I s , 1
X m
=
=
R s


V s
+
V s
I s , 1
X

 m
−
R s
where R s
, V s
and I s,1
are all known

For the locked rotor test, the rotor of the induction motor is held so that it will not turn.
Vs at rated frequency is increased from 0V until I s,1
reaches rated current. Since
ω o
=
0 and therefore s=1, we have the following equivalent circuit:
In this condition with R
R
small, a V s
of small magnitude is all that is required for I s,1
to reach rated current. Therefore, with Vs being small, and X m
is large, we can ignore magnetizing current I s,m
to arrive at the following equivalent circuit
The magnitude and angle of V s
and I s,1
are measured and R s
is already known. Therefore, we are able to solve for the rotor resistance (R
R
) and the sum of the two reactance values
(X ls
+X lr
)=X leakage
.
I s , 1
=
R s
+
R
R
V s
+ jX leakage
R
R
=
V s
I s , 1 cos(
θ
)
−
R s

X leakage
=
V s
I s , 1 sin(
θ
)
Where
θ
is the difference between the voltage and current angle.
An assumption is made about the relative magnitude of X ls
and X lr
.