# Electrical 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      Three Phase Induction 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
A star-connected 3-phase, 400 V, 50 kVA, 50 Hz synchronous motor has a synchronous reactance of 1 ohm per phase with negligible armature resistance. The shaft load on the motor is 10 kW while the power factor is 0.8 leading. The loss in the motor is 2 kW. The magnitude of the per phase excitation emf of the motor, in volts, is __________. (round off to nearest integer).
 A 258 B 324 C 128 D 245
GATE EE 2022      Synchronous Machines
Question 2 Explanation:
Power input$= 10 + 2 = 12 kW$
\begin{aligned} P&=\sqrt{3}VI \cos\phi \\ I_a&=\frac{10 \times 10^3}{\sqrt{3} \times 400 \times 0.8}\\ &=21.65\angle 36.86^{\circ}A \end{aligned}
We have, Emf equation for motor
\begin{aligned} \bar{E_{ph}}&=\bar{V_{ph}}-\bar{I_{a}}\bar{Z_{s}}\\ &=\frac{400}{\sqrt{3}}-21.65\angle 36.83^{\circ}\times 1\angle 90^{\circ}\\ &=244.542\angle -4.062^{\circ}V \end{aligned}
 Question 3
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      Singal Phase Induction Motors, Special Purpose Machines and Electromechanical Energy Conversion System
Question 3 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 4
A 280 V, separately excited DC motor with armature resistance of $1\Omega$ and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value $10\Omega$ is connected in series with the armature, is _______. (round off to nearest integer)
 A 482 B 254 C 365 D 625
GATE EE 2022      DC Machines
Question 4 Explanation:
Back emf,
$E_{b1}=280-30 \times 1 =250V$
When additional $10\Omega$ resistance connected in series with armature.
$E_{b2}=280-(10+1) I_{a_2} =280-11I_{a_2}$
Given: $\tau \propto N$
We know, $\tau \propto \phi I_a \text{ (Here, }\phi \rightarrow constant)$
So,
\begin{aligned} \tau &\propto I_a \\ \therefore \; N &\propto I_a \\ or \; \frac{N_2}{N_1}&=\frac{I_{a_2}}{I_{a_1}}\\ I_{a_2}&=N_2 \times \frac{30}{1000}\\ \therefore \; E_{b_2}&=280-\left ( 11 \times \frac{30}{1000} \right )N_2\\ &\text{We also have,}\\ E_b &\propto\phi \omega \\ \therefore \frac{E_{b_2}}{E_{b_1}}&=\frac{N_2}{N_1}\\ &\frac{280-\left ( 11 \times \frac{30}{1000} \right )N_2}{250}=\frac{N_2}{1000}\\ \Rightarrow N_2&=482.758rpm \end{aligned}
 Question 5
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      Three Phase Induction Machines
Question 5 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 6
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      Singal Phase Induction Motors, Special Purpose Machines and Electromechanical Energy Conversion System
 Question 7
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      Three Phase Induction Machines
Question 7 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 8
A belt-driven $\text{DC}$ shunt generator running at $300\: \text{RPM}$ delivers $100 \:\text{kW}$ to a $200 \:V\: \text{DC}$ grid. It continues to run as a motor when the belt breaks, taking $10 \:\text{kW}$ from the $\text{DC}$ grid. The armature resistance is $0.025 \Omega$, field resistance is $50\:\Omega$, and brush drop is $2 \:V$. Ignoring armature reaction, the speed of the motor is _____ $\text{RPM}$. (Round off to 2 decimal places.)
 A 275.18 B 254.12 C 362.24 D 148.44
GATE EE 2021      DC Machines
Question 8 Explanation:
\begin{aligned} I_{L} &=\frac{100 \times 10^{3}}{200}=500 \mathrm{~A} \\ I_{S h} &=\frac{200}{50}=4 \mathrm{~A} \\ I_{a} &=504 \mathrm{~A} \\ E_{q} &=V+I_{a} R_{a}+\text { Brush drop } \\ &=200+504(0.025)+2 \mathrm{~V} \\ &=214.6 \mathrm{~V}\\ \text{In motoring case:} V I&=10 \mathrm{~kW}, V=200 \mathrm{~V}\\ \therefore \qquad \qquad I &=\frac{10000}{200}=50 A \\ I_{f} &=4 \mathrm{~A}, I_{a}=I_{L}-I_{f} \\ &=46 \mathrm{~A} \\ E_{b} &=V-I_{a} R_{a}-\text { Brush Drop } \\ &=200-46(0.025)-2=196.85 \mathrm{~V} \\ \frac{N_{m}}{N_{g}} &=\frac{E_{b}}{E_{g}} \\ N_{m} &=\frac{E_{b}}{E_{g}} \times N_{g} \\ \therefore \qquad \qquad N_{m} &=\frac{196.85}{214.6} \times 300=275.18 \mathrm{rpm} \end{aligned}
 Question 9
In a single-phase transformer, the total iron loss is $2500\:W$ at nominal voltage of $440\:V$ and frequency $50\:Hz$. The total iron loss is $850\:W$ at $220\:V$ and $25\:Hz$. Then, at nominal voltage and frequency, the hysteresis loss and eddy current loss respectively are
 A $1600\:W$ and $900\:W$ B $900\:W$ and $1600\:W$ C $250\:W$ and $600\:W$ D $600\:W$ and $250\:W$
GATE EE 2021      Transformers
Question 9 Explanation:
\begin{aligned} W_{i 1} &=2500 \mathrm{~W} \text { at } 440 \mathrm{~V}, 50 \mathrm{~Hz} \\ W_{i 2} &=850 \mathrm{~W} \text { at } 220 \mathrm{~V}, 25 \mathrm{~Hz} \\ W_{i 3} &=R_{e_{3}}+P_{H_{3}} \text { at } 440 \mathrm{~V}, 50 \mathrm{~Hz} \\ R_{\theta_{3}} &=?, \quad P_{h_{3}}=?, \quad \frac{v}{f}=\mathrm{constant} \\ \Rightarrow \qquad \qquad\quad 2500 &=A f+B f^{2} &\left\{\frac{400}{50}=\frac{220}{25}=\text { Constant }\right\} \\ \text { or, }\qquad \qquad \quad \frac{2500}{f} &=A+B f \\ \text{or},\qquad \qquad \frac{2500}{50}&=A+B(50) &\ldots(i)\\ \text{and}\qquad \qquad \frac{850}{25}&=\mathrm{A}+\mathrm{B}(25)&\ldots(ii) \end{aligned}
Solving (i) and (ii), we get
\begin{aligned} 25 B &=\frac{2500}{50}-\frac{850}{25}=\frac{2500-1700}{50} \\ &=\frac{800}{50}=16 \\ B &=\frac{16}{25} \\ \text{and from (i)},\qquad A &=50-\frac{16}{25} \times 50=50-32=18\\ \text{So, at} 50 \mathrm{~Hz}\\ P_{h}&=A f=18 \times 50=900 \mathrm{~W} \\ P_{e}&=B f=\left(\frac{16}{25}\right) \times(50)^{2}=1600 \mathrm{~W} \end{aligned}
 Question 10
An alternator with internal voltage of $1\angle \delta _{1}\text{p.u}$ and synchronous reactance of $\text{0.4 p.u}$ is connected by a transmission line of reactance $\text{0.1 p.u}$ to a synchronous motor having synchronous reactance $\text{0.35 p.u}$ and internal voltage of $0.85\angle \delta _{2}\text{p.u}$.
If the real power supplied by the alternator is $\text{0.866 p.u}$, then $\left( \delta _{1} -\delta _{2}\right)$ is _________ degrees. (Round off to 2 decimal places.)
(Machines are of non-salient type. Neglect resistances.)
 A 60 B 25.35 C 68.6 D 88
GATE EE 2021      Synchronous Machines
Question 10 Explanation:

Given real power in (p.u.), $P=0.866$
\begin{aligned} P &=\frac{E V \sin \left(\delta_{1}-\delta_{2}\right)}{X_{\mathrm{eq}}}=\left|\frac{1 \times 0.85}{0.4+0.1+0.35}\right| \sin \left(\delta_{1}-\delta_{2}\right) \\ \left(\delta_{1}-\delta_{2}\right) &=60^{\circ} \end{aligned}

There are 10 questions to complete.