# Three Phase Induction Machines

 Question 1
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 1 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 2
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 3
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 3 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 4
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 4 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 5
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 5 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 6
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 6 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}
 Question 7
In a constant V/f induction motor drive, the slip at the maximum torque
 A is directly proportional to the synchronous speed. B remains constant with respect to the synchronous speed. C has an inverse relation with the synchronous speed. D has no relation with the synchronous speed.
GATE EE 2016-SET-1   Electrical Machines
Question 7 Explanation:
\begin{aligned} f_0 &=\text{nominal frequency} \\ \omega _s &=\left ( \frac{f}{f_0} \right )\omega _{s0}\;\;...(i) \\ S_{max,T}&=\left ( \frac{f}{f_0} \right ) \left ( \frac{R_2}{X_{20}^1} \right )\;\;...(ii) \\ &\text{From (i) and (ii)} \\ S_{max,T} &= \left ( \frac{\omega _{s0}}{\omega _s} \right )\left ( \frac{R_2}{X_{20}^1} \right )\\ S_{max,T} &\propto \frac{1}{\omega _s} \end{aligned}
 Question 8
The figure shows the per-phase equivalent circuit of a two-pole three-phase induction motor operating at 50 Hz. The "air-gap" voltage, $V_{g}$ across the magnetizing inductance, is 210 V rms, and the slip, s, is 0.05. The torque (in Nm) produced by the motor is ________.
 A 200.45 B 305.25 C 401.64 D 455.65
GATE EE 2015-SET-2   Electrical Machines
Question 8 Explanation:

\begin{aligned} i &=\frac{210}{1+j0.22} \\ &=205.09\angle -12.40^{\circ} \\ \text{Active power} &= (205.09)^2 \times 1=P_g\\ P_m &= (1-0.05) \times P_g \\ P_m &= 39958.81 \; \text{Watt}\\ \text{Torque} &= \frac{39958.81 \times 60}{2 \pi \times 2850} \times 3\\ &= 133.88 \times 3=401.64 N-m \end{aligned}
 Question 9
A 3-phase 50 Hz square wave (6-step) VSI feeds a 3-phase, 4 pole induction motor. The VSI line voltage has a dominant $5^{th}$ harmonic component. If the operating slip of the motor with respect to fundamental component voltage is 0.04, the slip of the motor with respect to $5^{th}$ harmonic component of voltage is _________.
 A 1.2 B 1.82 C 2.14 D 6.23
GATE EE 2015-SET-1   Electrical Machines
Question 9 Explanation:
$N_S=\frac{120f}{P}=1500$
Slip due to 5th harmmonic
$S_5=\frac{N_S+\frac{N_s}{5}}{N_S}=\frac{6}{5}=1.2$
 Question 10
The torque speed characteristics of motor ($T_M$) and load ($T_L$) for two cases are shown in the figures (a) and (b). The load torque is equal to motor torque at points P, Q, R and S.

The stable operating points are
 A P and R B P and S C Q and R D Q and S
GATE EE 2014-SET-3   Electrical Machines
Question 10 Explanation:

$S=\frac{N_s-N_r}{N_s}$
For given system,
slip=(2-s)(speed =negative)
Below point P, $T_M-T_l \lt 0$
$\Rightarrow$ speed decreases
$\therefore$ S increases
$\therefore$ slip=(2-s) decreases
Above point P, $T_m-T_l \gt 0$
$\Rightarrow$ speed increases
$\therefore$ S decreases and (2-s) increasaes.
Thus, system is stable near point P.

POINT R:
Below point R: $T_M-T_l \lt 0\;\Rightarrow \;$ speed decreases
Above point R: $T_M-T_l \gt 0\;\Rightarrow \;$ speed increases
Thus, system can't regain stability at point R.
Therefore, R is unstable point.
POINT S:
Below point S: $T_M-T_l \gt 0\;\Rightarrow \;$ speed increases
Above point S: $T_M-T_l \lt 0\;\Rightarrow \;$ speed decreases
Thus, system can regain stability near point S.
There are 10 questions to complete.