Three Phase Induction Machines

Question 1
A 3-phase, 415 V, 4-pole, 50 Hz induction motor draws 5 times the rated current at rated voltage at starting. It is required to bring down the starting current from the supply to 2 times of the rated current using a 3-phase autotransformer. If the magnetizing impedance of the induction motor and no load current of the autotransformer is neglected, then the transformation ratio of the autotransformer is given by _______. (round off to two decimal places).
A
0.24
B
0.48
C
0.63
D
0.97
GATE EE 2022   Electrical Machines
Question 1 Explanation: 
We have I_L=x^2I_{sc}
Here, I_L=2I_f and I_{sc}=5I_f
From the above eqiations.
2I_f=x^25I_f\Rightarrow x=\sqrt{\frac{2}{5}}=0.6324
Question 2
The frequencies of the stator and rotor currents flowing in a three-phase 8-pole induction motor are 40 Hz and 1 Hz, respectively. The motor speed, in rpm, is _______. (round off to nearest integer)
A
254
B
365
C
542
D
585
GATE EE 2022   Electrical Machines
Question 2 Explanation: 
We have, f_r=sf_s
Slip, s=\frac{1}{40}=0.025
Now, speed
N=N_s(1-s)=\frac{120 \times 40}{8}(1-0.025)=585rpm
Question 3
An 8-pole, 50 \:Hz, three-phase, slip-ring induction motor has an effective rotor resistance of 0.08\:\Omega per phase. Its speed at maximum torque is 650 \text{ RPM}. The additional resistance per phase that must be inserted in the rotor to achieve maximum torque at start is _____________ \Omega. (Round off to 2 decimal places.) Neglect magnetizing current and stator leakage impedance. Consider equivalent circuit parameters referred to stator.
A
0.25
B
0.44
C
0.52
D
0.68
GATE EE 2021   Electrical Machines
Question 3 Explanation: 
\begin{aligned} N_{\max } &=650 \mathrm{rpm}, P=8,50 \mathrm{~Hz} \\ s_{m} &=\frac{750-650}{750}=0.1333 \\ s_{m} &=\frac{R_{2}}{X_{2}} \\ \therefore\qquad \qquad X_{2} &=\frac{R_{2}}{s_{m}}=\frac{0.08}{0.133}=0.601 \Omega \\ R_{2} &=0.08 \Omega, X_{2}=0.601 \Omega \end{aligned}
Condition for maximum T_{\mathrm{s}}
\begin{aligned} \Rightarrow \qquad \qquad R_{2} & =X_{2} \\ \therefore \qquad \qquad R_{2}+ R_{\text {ext }} & =X_{2} \\ \therefore \qquad \qquad R_{\text {ext }} & =0.601-0.08=0.521 \Omega \end{aligned}
Question 4
The power input to a \text{500 V, 50 Hz, 6-pole, 3-phase} induction motor running at 975 \text{ RPM} is 40 \text{ kW} . The total stator losses are 1 \text{ kW}. If the total friction and windage losses are 2.025\text{~ kW}, then the efficiency is _____________%.
A
85
B
95
C
90
D
98
GATE EE 2021   Electrical Machines
Question 4 Explanation: 
\begin{aligned} P_{i / p} &=40 \mathrm{~kW}, \quad \text { stator loss }=1 \mathrm{~kW}, \quad \mathrm{~F} \text { and } W=2.025 \mathrm{~kW} \\ \text { Stator } \mathrm{o} / \mathrm{p} &=40-1=39 \mathrm{~kW} \\ \qquad \text { Slip } &=\frac{1000-975}{1000}=0.025 \\ \text { Rotor } \mathrm{o} / \mathrm{p} &=\text { Rotor } \mathrm{i} / \mathrm{p} \times(1-\mathrm{s}) \\ &=39(1-0.025)=38.025 \mathrm{~kW} \\ \text { Motor } \mathrm{o} / \mathrm{p} &=38.025-2.025=36 \mathrm{~kW}\\ \eta&=\frac{\text { Motor output }}{\text { Motor input }}=\frac{36}{40} \times 100=90 \% \end{aligned}
Question 5
A delta-connected, 3.7 kW, 400 V(line), three-phase, 4-pole, 50-Hz squirrel-cage induction motor has the following equivalent circuit parameters per phase referred to the stator:
R_1=5.39\Omega,R_2=5.72\Omega,X_1=X_2=8.22\Omega.
Neglect shunt branch in the equivalent circuit. The starting line current in amperes (round off to two decimal places) when it is connected to a 100 V (line), 10 Hz, three-phase AC source is ______
A
8.25
B
14.95
C
25.45
D
32.85
GATE EE 2019   Electrical Machines
Question 5 Explanation: 
\begin{aligned} I_L &=\sqrt{3}I_{ph} \\ \text{At} \;\; f &= 10Hz, \\ X_1=X_2 &=8.22 \times \frac{10}{50}=1.644\Omega \\ I_{ph} &= \frac{V_{ph}}{\sqrt{(R_1+R_2)^2+(X_1+X_2)^2}} \\ &=\frac{100}{\sqrt{(5.39+5.72)^2+(1.644+1.644)^2}} \\ &= \frac{100}{11.586}=8.63\\ I_L&= 14.95A \end{aligned}
Question 6
The parameter of an equivalent circuit of a three-phase induction motor affected by reducing the rms value of the supply voltage at the rated frequency is
A
rotor resistance
B
rotor leakage reactance
C
magnetizing reactance
D
stator resistance
GATE EE 2019   Electrical Machines
Question 7
A star-connected, 12.5 kW, 208 V (line), 3-phase, 60 Hz squirrel cage induction motor has following equivalent circuit parameters per phase referred to the stator: R_{1}=0.3\Omega ,R_{2}=0.3\Omega ,X_{1}=0.41\Omega ,X_{2}=0.41\Omega. Neglect shunt branch in the equivalent circuit. The starting current (in Ampere) for this motor when connected to an 80 V (line), 20 Hz, 3-phase AC source is __________.
A
50
B
60
C
70
D
80
GATE EE 2017-SET-2   Electrical Machines
Question 7 Explanation: 
Equivalent circuit representation during starting
-according to question

Total resistance to stator =0.6\Omega
Total reactance reference to stator =0.82\Omega
Reactance for 60Hz \;\rightarrow \;0.82\Omega
Reactance for 60Hz \;\rightarrow ?
\because \;\; X_e\propto F
Reactance for 20Hz\;\Rightarrow \; \frac{20}{60} \times 0.82
\therefore \;\; X for 20 Hz frequency =0.273\Omega
\therefore \;\; Starting current
\Rightarrow \;\; I_{st}=\frac{80\sqrt{3}}{\sqrt{(0.6)^2+(0.273)^2}}=70.05A
Question 8
A 3-phase, 4-pole, 400 V, 50 Hz squirrel-cage induction motor is operating at a slip of 0.02. The speed of the rotor flux in mechanical rad/sec, sensed by a stationary observer, is closest to
A
1500
B
1470
C
157
D
154
GATE EE 2017-SET-2   Electrical Machines
Question 8 Explanation: 
\begin{aligned} N_s &= \frac{120 \times 50}{4}=1500\; \text{rpm}\\ s &=0.02 \\ N&=N_s(1-s) \\ &=1500(1-0.02) \\ &=1470 \; \text{rpm} \end{aligned}
As it was specified rotor rmf with respect to stationary part/sector.
Rotor rmf w.r.t. stator is always at N_s.
\therefore \; Speed of rotor flux in mechanical radian/sec.
=\frac{2 \times 3.14 \times 1500}{60}=157 \; rad/sec
Question 9
A 4 pole induction machine is working as an induction generator. The generator supply frequency is 60 Hz. The rotor current frequency is 5 Hz. The mechanical speed of the rotor in RPM is
A
1350
B
1650
C
1950
D
2250
GATE EE 2017-SET-1   Electrical Machines
Question 9 Explanation: 
Induction Generator Mode:
\begin{aligned} N & \gt N_s \;\;\text{i.e.(s is negative)} \\ \because \;\; P &=4,\; f=60Hz \\ N_s&=1800 \; \text{rpm} \end{aligned}
Rotor frequency given = 5Hz \Rightarrow (s \times f)
\begin{aligned} \because \;\; f &=60Hz \\ \therefore \;\; s &=\frac{5}{60}=0.0833 \end{aligned}
\because It is induction generator slip will be -0.0833.
\begin{aligned} s &=\frac{1800-N}{1800}=-0.0833 \\ N &=1950 \; \text{rpm} \end{aligned}
Question 10
The starting line current of a 415 V, 3-phase, delta connected induction motor is 120 A, when the rated voltage is applied to its stator winding. The starting line current at a reduced voltage of 110 V, in ampere, is _________.
A
11.4
B
22.5
C
31.8
D
44.2
GATE EE 2016-SET-2   Electrical Machines
Question 10 Explanation: 
415V, 3-phase, \Delta connected induction motor
(I_{st})_{line}=120A at rated voltage.
at, V=110V, i.e. reduced voltage
\begin{aligned} I_{st} &=x(I_{st})_{rated} \\ \text{where}\;\; x&= \frac{V_{reduced}}{V_{rated}}\\ &=\frac{110}{415} \\ (I_{st})_{at \; 110V} &= \left ( \frac{110}{415} \right )\times 120\\ &=31.807A \end{aligned}
There are 10 questions to complete.