# Single Phase Induction Motors, Special Purpose Machines and Electromechanical Energy Conversion System

 Question 1
The four stator conductor $\left(A, A^{\prime}, B\right.$ and $\left.B^{\prime}\right)$ of a rotating machine are carrying DC currents of the same value, the directions of which are shown in figure (i). The rotor coils $a-a^{\prime}$ and $b-b^{\prime}$ are formed by connecting the back ends of conductor '$a$' and '$a^{\prime}$' and '$b$' and '$b^{\prime}$', respectively, as shown in figure (ii). The e.m.f. induced in coil $a-a^{\prime}$ and coil $b-b^{\prime}$ are denoted by $E_{a-a^{\prime}}$ and $E_{b-b^{\prime}}$; respectively. If the rotor is rotated at uniform angular speed $\omega \mathrm{rad} / \mathrm{s}$ in the clockwise direction then which of the following correctly describes the $\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}$ and $\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}$ ? A $E_{a-a^{\prime}}$ and $E_{b-b^{\prime}}$ have finite magnitudes and are in the same phase B $E_{a-a^{\prime}}$ and $E_{b-b^{\prime}}$ have finite magnitudes with $\mathrm{E}_{\mathrm{b}-b^{\prime}}$ leading $\mathrm{E}_{\mathrm{a-a}}$ C $\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}$ and $\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}$ have finite magnitudes with $E_{a-a^{\prime}}$ leading $E_{b-b^{\prime}}$ D $\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}=\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}=0$
GATE EE 2023   Electrical Machines
Question 1 Explanation:
Since, a-a' and b-b' are placed at GNA (Geometric Neutral Axis) therefore,
$\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}=\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}=0$
 Question 2
A 10-pole, $50 \mathrm{~Hz}, 240 \mathrm{~V}$, single phase induction motor runs at 540 RPM while driving rated load. The frequency of induced rotor currents due to backward field is
 A $100 \mathrm{~Hz}$ B $95 \mathrm{~Hz}$ C $10 \mathrm{~Hz}$ D $5 \mathrm{~Hz}$
GATE EE 2023   Electrical Machines
Question 2 Explanation:
Synchronous speed,
\begin{aligned} N_{s} & =\frac{120 f}{P} \\ & =\frac{120 \times 50}{10}=600 \mathrm{rpm} \end{aligned}

Now, forward slip,
\begin{aligned} S & =\frac{N_{\mathrm{s}}-N}{N_{\mathrm{s}}} \\ & =\frac{600-540}{600}=0.1 \end{aligned}

$\therefore$ Backward slip $=2-\mathrm{S}=1.9$
Thus, required frequency $=1.9 \times 50=95 \mathrm{~Hz}$

 Question 3
The following columns present various modes of induction machine operation and the ranges of slip

$\begin{array}{ll} \textbf{A (Mode of operation)}& \textbf{B (Range of slip)}\\\\ \text{a. Running in generator mode}&\text{p) From 0.0 to 1.0}\\\\ \text{b. Running in motor mode} & \text{q) From 1.0 to 2.0}\\\\ \text{c. Plugging in motor mode} & \text{r) From - 1.0 to 0.0} \end{array}$
The correct matching between the elements in column A with those of column B is
 A a-r, b-p, and c-q B a-r, b-q, and c-p C a-p, b-r, and c-q D a-q, b-p, and c-r
GATE EE 2023   Electrical Machines
Question 3 Explanation:
Torque speed characteristic of $3-\phi$ IM : $\mathrm{S} \gt 1 \Rightarrow$ Plugging mode
$0 \lt \mathrm{S} \lt 1 \Rightarrow$ Motoring mode
$\mathrm{S} \lt 0 \Rightarrow$ Generating Mode
 Question 4
A 4-pole induction motor with inertia of 0.1 $kg-m^2$ drives a constant load torque of 2 Nm. The speed of the motor is increased linearly from 1000 rpm to 1500 rpm in 4 seconds as shown in the figure below. Neglect losses in the motor. The energy, in joules, consumed by the motor during the speed change is ____________. (round off to nearest integer) A 1732 B 2534 C 1245 D 3251
GATE EE 2022   Electrical Machines
Question 4 Explanation:
We have
\begin{aligned} J\frac{d\omega }{dt}&=\tau _e-\tau _L\\ J\omega \frac{d\omega }{dt}&=P_e-P_L \;\;(\because P=\tau \omega )\\ \therefore P_e&=J\omega \frac{d\omega }{dt}+P_L\\ E&=\int P_edt\\ R&=J\int_{1000}^{1500}\omega d\omega +\int_{1000}^{1500}P_Ldt\\ &=\frac{J}{2}\left ( \frac{2 \pi}{60} \right )^2[1500^2-1000^2]+\frac{2 \pi \tau }{60}\int_{1000}^{1500}Ndt\\ &=\frac{0.1}{2}\left ( \frac{2 \pi}{60} \right )^2\times 125 \times 10^4+\frac{2 \pi}{60} \times 2\times [1000 \times 4+\frac{1}{2} \times 500 \times 4]\\ &=1732.586J \end{aligned}
 Question 5
The type of single-phase induction motor, expected to have the maximum power factor during steady state running condition, is
 A split phase (resistance start). B shaded pole. C capacitor start. D capacitor start, capacitor run.
GATE EE 2022   Electrical Machines

There are 5 questions to complete.