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

 Question 1
A three-phase, 50 Hz, 4-pole induction motor runs at no-load with a slip of 1%. With full load, the slip increases to 5 %. The % speed regulation of the motor (rounded off to 2 decimal places) is _________ .
 A 4.21 B 16.45 C 2.21 D 8.44
GATE EE 2020   Electrical Machines
Question 1 Explanation:
\begin{aligned} \text{Synchronous speed, }\\ N_s&=\frac{120f}{P}\\ N_s&=\frac{120 \times 50}{4}rpm\\ N_s&=1500 \; rpm\\ \text{Speed at no-load,}\\ &=N_s(1-S_{nL})\\ &=1500(1-0.01)\\ &=1485 \; rpm\\ \text{Speed at full-load,}\\ &=N_s(1-S_{fL})\\ &=1500(1-0.05)\\ &=1425 \; rpm\\ \% \text{ speed regulation}&=\frac{N_{NL}-N_{FL}}{N_{FL}} \times 100\\ &=\frac{1485-1425}{1425} \times 100\\ &=4.21\% \end{aligned}
 Question 2
The magnetic circuit shown below has uniform cross-sectional area and air gap of 0.2 cm. The mean path length of the core is 40 cm. Assume that leakage and fringing fluxes are negligible. When the core relative permeability is assumed to be infinite, the magnetic flux density computed in the air gap is 1 tesla. With same Ampere-turns, if the core relative permeability is assumed to be 1000 (linear), the flux density in tesla(round off to three decimal places) calculated in the air gap is __________
 A 0.257 B 0.456 C 0.833 D 0.658
GATE EE 2019   Electrical Machines
Question 2 Explanation:
\begin{aligned} l_{ag} &=0.2 cm \\ l_m&= 40cm\\ B_0 &= 1 \text{Tesla}\\ \phi &= \frac{mmf}{\Re }\\ &\text{For the same mmf,} \\ \phi &\propto \frac{1}{\Re } \\ \text{In case-1:} &\\ \Re _1&=\frac{l_m}{\mu _0A} \\ &= \frac{l_{ag}}{\mu _0 A}+\frac{l_m}{\mu _0 \mu _r A}\\ \text{as} \;\;\mu _0=0& \\ \therefore \;\;\Re _1&= \frac{0.2 \times 10^{-2}}{\mu _0 A}\\ \text{In case-2:} &\\ \Re _2&= \frac{l_{ag}}{\mu _0 A}+\frac{l_m}{\mu _0 \mu _r A}\\ &=\frac{0.2 \times 10^{-2}}{\mu _0 A}+\frac{40 \times 10^{-2}}{1000\mu _0 A}\\ &=\frac{0.24\times 10^{-2}}{\mu _0 A} \end{aligned}
As flux $\phi \propto B$ for uniform cross section area.
\begin{aligned} \therefore \; B_2&=\frac{B_1 \times \Re _1}{\Re _2} \\ B_2&= \frac{1 \times 0.2 \times 10^{-2}/\mu _0 A}{0.24 \times 10^{-2}/\mu _0A}\\ &= 0.833T \end{aligned}
 Question 3
The equivalent circuit of a single phase induction motor is shown in the figure, where the parameters are $R_{1}=R'_{2}=X_{i1}=X'_{i2}=12\Omega$, $X_{M}=240\Omega$ and s is the slip. At no-load, the motor speed can be approximated to be the synchronous speed. The no-load lagging power factor of the motor is___________ (up to 3 decimal places).
 A 0.182 B 0.412 C 0.214 D 0.106
GATE EE 2018   Electrical Machines
Question 3 Explanation:

Simplifying the above circuit into a simple R-L circuit,

\begin{aligned} Z_{eq} &=\frac{(3+j6)(j120)}{(3+j126)}+(12+j32) \\ &=138.563\angle \\ &\text{Current drawn by motor is } \\ I&= \frac{V\angle 0}{Z\angle \theta } \end{aligned}
($\therefore \; \theta :$ impedance angle will be p.f. angle)
$\therefore$ No laod lagging p.f. of motor is $(\cos\theta ),$
$\cos(83.9)=0.106$ lagging power factor
 Question 4
A transformer with toroidal core of permeability $\mu$ is shown in the figure. Assuming uniform flux density across the circular core cross-section of radius $r\lt \lt R$, and neglecting any leakage flux, the best estimate for the mean radius R is
 A $\frac{\mu Vr^{2}{N^{2}}_{P}\omega }{I}$ B $\frac{\mu VIr^{2}N_{P}N_{S}\omega }{V}$ C $\frac{\mu Vr^{2}{N^{2}}_{P}\omega }{2I}$ D $\frac{\mu Ir^{2}{N^{2}}_{P}\omega }{2V}$
GATE EE 2018   Electrical Machines
Question 4 Explanation:

Since we know reluctance of core:
$\Re =\frac{l_c}{\mu a}$
Here, $l_c=$ mean core length, a= area of core reluctance,
\begin{aligned} \Re &=\frac{2 \pi R}{\mu \times \pi r^2}\\ &=\frac{2R}{\mu r^2}\\ [\text{where} R &= \text{mean radius of core}] \\ \text{or}\;\; R&=\frac{\Re \mu r^2}{2} \\ \text{and}\;\; \Re &= \frac{mmf}{flux}=\frac{N_pI_p}{\phi }\\ \text{Here,}\;\; \phi &=-\frac{1}{N_p}\int_{0}^{t}e\;dt \\ [\text{where}\; e &\text{is primary generated voltage}] \\ \text{or}\;\; \phi &=-\frac{1}{N_p}\int_{0}^{t}-(V_p)\;dt \\ \phi &= \frac{1}{N_p}\int_{0}^{t}V \cos \omega t \;dt\\ &= \frac{V}{\omega N_p}\sin \omega t\\ &\text{Now reluctance} \\ \Re &= \frac{N_p I \sin \omega t}{\frac{V}{\omega N_p} \sin \omega t}\\ &= \frac{N_p^2\omega I}{V}\\ & \text{So mean radius,}\\ R&=\frac{N_p^2 \times \omega I \times \mu r^2}{2V} \end{aligned}
 Question 5
A 375W, 230 V, 50 Hz capacitor start single-phase induction motor has the following constants for the main and auxiliary windings (at starting): $Z_{m}=(24.50+j12.75)\Omega$ (main winding), $Z_{n}=(24.50+j12.75)\Omega$ (auxiliary winding). Neglecting the magnetizing branch the value of the capacitance (in $\mu$F ) to be added in series with the auxiliary winding to obtain maximum torque at starting is _______.
 A 38.43 B 200 C 150 D 50
GATE EE 2017-SET-1   Electrical Machines
Question 5 Explanation:
Auxiliary winding impedance $=(24.5+j12.75)\Omega$
Let $X_c$ be the reactance of the capacitor connected in auxiliary winding
\begin{aligned} Z_a &= (24.5+j12.75-jX_c)\Omega \\ X_a &=(X_c-12.75),\; R_a=24.5\Omega \\ \therefore \;\; \phi _a &=\tan^{-1}\frac{X_c-12.75}{24.5}\;...(i) \\ Z_m&= 12.5+j15.75\\ &= 20.10\angle 51.56^{\circ}\Omega \\ \phi _m&= 51.56^{\circ} \end{aligned}
For maximum torque at starting, condition on $\phi _a$ and $\phi _m$ is given by $\left [ \phi _a=\frac{90^{\circ}-\phi _m}{2} \right ]$
$\phi _a=\frac{90-51.56}{2}=19.22^{\circ}}\;...(ii)$
Comparing equation (i) and (ii),
\begin{aligned} \tan(19.22^{\circ}) &=\frac{X_c-12.75}{24.5} \\ 0.3486 &= \frac{X_c-12.75}{24.5}\\ X_c &=8.541+12.75 \\ \text{Capacitive reactance,}& \\ X_c &= 21.291\Omega \\ \frac{1}{2 \pi fc}&= 21.291\\ \Rightarrow \;C&= \frac{1}{2\pi \times 50 \times 21.291}\\ &= 149.51\mu F \end{aligned}
 Question 6
The flux linkage ($\lambda$) and current (i) relation for an electromagnetic system is $\lambda = (\sqrt{i})/g$. When i=2A and g (air-gap length)=10cm, the magnitude of mechanical force on the moving part, in N, is ________.
 A 128.52 B 165.65 C 188.56 D 214.25
GATE EE 2016-SET-2   Electrical machines
Question 6 Explanation:
Energy in magnetic system,
\begin{aligned} W_f &= \int_{0}^{\lambda } i(\lambda )\; d\lambda \\ \because \;\; i &=\lambda ^2g^2 \\ W_f &= \int_{0}^{\lambda } \lambda ^2g^2 \; d\lambda \\ &=g^2\frac{\lambda ^3}{2} \end{aligned}
Now, mechanical force,
\begin{aligned} F_F &=-\frac{\partial W_f (\lambda ,g)}{\partial g} \\ &=-\frac{\partial }{\partial g} \left ( \frac{g^2\lambda ^3}{3} \right )\\ &= -\frac{2}{3}\lambda ^3 g\\ \because \;\; i=2, \;g&=10cm ,\; \lambda =10\sqrt{2} \\ \text{Hence,}\;\;|F_F|&=\frac{2}{3} \times (10\sqrt{2})^3 \times 0.1\\ &=188.56N \end{aligned}
 Question 7
The direction of rotation of a single-phase capacitor run induction motor is reversed by
 A interchanging the terminals of the AC supply B interchanging the terminals of the capacitor C interchanging the terminals of the auxiliary winding D interchanging the terminals of both the windings
GATE EE 2016-SET-2   Electrical Machines
Question 7 Explanation:
Inter changing the terminals of the auxiliary winding.
 Question 8
A 220 V, 3-phase, 4-pole, 50 Hz inductor motor of wound rotor type is supplied at rated voltage and frequency. The stator resistance, magnetizing reactance, and core loss are negligible. The maximum torque produced by the rotor is 225 % of full load torque and it occurs at 15% slip. The actual rotor resistance is 0.03 $\Omega$/phase. The value of external resistance (in Ohm) which must be inserted in a rotor phase if the maximum torque is to occur at start is ______.
 A 0 B 0.18 C 0.24 D 0.28
GATE EE 2015-SET-2   Electrical Machines
 Question 9
A 200 V, 50 Hz, single-phase induction motor has the following connection diagram and winding orientations as shown. MM' is the axis of the main stator winding($M_1M_2$) and AA' is that of the auxiliary winding($A_1A_2$). Directions of the winding axis indicate direction of flux when currents in the windings are in the directions shown. Parameters of each winding are indicated. When switch S is closed, the motor
 A rotates clockwise B rotates anti-clockwise C does not rotate D rotates momentarily and comes to a halt
GATE EE 2009   Electrical Machines
Question 9 Explanation:
$f=50Hz$
Impedance of main winding
\begin{aligned} Z_m &= r_m+j2 \pi f L_m\\ &=0.1+ j 2 \pi \times 20 \times \frac{0.1}{\pi} \\ Z_m&=0.1+j10\Omega \end{aligned}
Impedance of auxiliary winding,
\begin{aligned} Z_a &=r_a+j2 \pi fL_a \\ &= 1+j 2 \pi \times 50 \times \frac{10}{\pi}\\ Z_a&= 1+j1000\Omega \end{aligned}
Current through main winding,
$I_m=\frac{V_s}{Z_m}=\frac{220}{0.1+j10}$
$I_m=200\angle -89.427^{\circ}A$
Current through auxiliary winding
$I_a=\frac{V_s}{Z_a}$
$\;\;\;=\frac{220}{1+j1000}=0.22\angle -89.942$
Thaing $V_s$ as the reference

$I_m$ leads $I_a$ the field crreated by the two current also have same difference thereby constituting an unbalanced field system, The result is the production of the starting torque.
Space orientation of the field

The motor rotates in the direction of leading phase to lagging phase.In this case, the motor rotates anti-clockwise.
 Question 10
A 230 V, 50 Hz, 4-pole, single-phase induction motor is rotating in the clockwise (forward) direction at a speed of 1425 rpm. If the rotor resistance at standstill is 7.8 $\Omega$, then the effective rotor resistance in the backward branch of the equivalent circuit will be
 A 2$\Omega$ B 4$\Omega$ C 78$\Omega$ D 156$\Omega$
GATE EE 2008   Electrical Machines
Question 10 Explanation:
Rotor resistance at stand still, $R=7.8\Omega$
Synchronous speed,
$N_s=\frac{120f}{P}$
$\;\;=\frac{120 \times 50}{4}=1500 \text{rpm}$
The slip(s) of rotor with respect to forward field
$s=\frac{N_s-N_r}{N_s}$
$\;\;=\frac{1500-1425}{1500}=0.05$
The slip of rotor with respect to backward field
$=2-s=2-0.05=1.95$
Effective rotor resistance
$\;\;=\frac{R_2}{2(2-s)}=7.8$
$\;\;=\frac{7.8}{2 \times 1.95}=2\Omega$
There are 10 questions to complete.