# Electrical Machines

 Question 1
When the winding $c-d$ of the singlephase, 50 $\mathrm{Hz}$, two winding transformer is supplied from an AC current source of frequency $50 \mathrm{~Hz}$, the rated voltage of $200 \mathrm{~V}$ (rms), $50 \mathrm{~Hz}$. is obtained at the open-circuited terminals a-b. The cross sectional area of the core $5000 \mathrm{~mm}^{2}$ and the average core length traversed by the mutual flux is $500 \mathrm{~mm}$. The maximum allowable flux density in the core is $B_{\max }=1 \mathrm{~Wb} / \mathrm{m}^{2}$ and the relative permeability of the core material is $5000$. The leakage impedance of the winding $a-b$ and winding c-d at $50 \mathrm{~Hz}$ are $(5+j 100 \pi \times 0.16) \Omega$ and $(11.5+j 100 \pi \times 0.36) \Omega$, respectively. Considering the magnetizing characteristics to be linear and neglecting core loss, the self-inductance of the winding $a-b$ in millihenry is ____ (Round off to 1 decimal place).

 A 1256.3 B 4152.4 C 2218.4 D 6523.8
GATE EE 2023      Transformers
Question 1 Explanation:
\begin{aligned} \mathrm{A} & =5000 \mathrm{~mm}^{2} \\ \mathrm{I} & =500 \mathrm{~mm} \\ \mathrm{~B}_{\max } & =1 \mathrm{~Wb} / \mathrm{m}^{2} \\ \mathrm{~V}_{\mathrm{ab}} & =200 \mathrm{~V} \\ \sqrt{2} \pi \mathrm{fN}_{1} \phi_{\max } & =200 \\ \sqrt{2 \pi} \times 50 \times \mathrm{N}_{1} \times 1 & \times 5000 \times 10^{-6}=200 \quad[\because \phi=\beta \mathrm{A}] \\ \Rightarrow \quad \mathrm{N}_{1} & =181 \end{aligned}

Now, mutual inductance,
\begin{aligned} M&=\frac{N^{2}}{I / \mu A} \\ & =\frac{4 \pi \times 10^{-7} \times 5000 \times 5000 \times 10^{-6} \times(181)^{2}}{0.5} \\ & =2.05843 \mathrm{H} \end{aligned}

Therefore, self inductance of winding.
\begin{aligned} L & =\text { Mutual + leakage } \\ & =2.05843+0.16 \\ & =2.21843 \mathrm{H} \text { or } 2218.43 \mathrm{mH} \end{aligned}
 Question 2
A three phase $415 \mathrm{~V}, 50 \mathrm{~Hz}, 6-pole, 960 RPM, 4 HP$ squirrel cage induction motor drives a constant torque load at rated speed operating from rated supply and delivering rated output. If the supply voltage and frequency are reduced by $20 \%$ the resultant speed of the motor in RPM (neglecting the stator leakage impednace and rotational losses) is ____ (Round off to the nearest integer)
 A 542 B 632 C 760 D 852
GATE EE 2023      Three Phase Induction Machines
Question 2 Explanation:
Given : Torque is constant.
We have, $\quad \tau \propto \frac{s V^{2}}{f} \quad ...(1)$
Synchronous speed, $N_{S}=\frac{120 f}{P}$
$=\frac{120 \times 50}{6}=1000 \mathrm{rpm}$

$\text { Slip, } S=\frac{1000-960}{1000}=0.04$

Now, from eqn. (1), we get
\begin{aligned} \frac{0.04 \times \mathrm{V}^{2}}{\mathrm{~F}} & =\frac{\mathrm{S}_{\text {new }} \times(0.8)^{2} \mathrm{~V}}{0.8 \mathrm{~F}} \\ \Rightarrow \quad \mathrm{S}_{\text {new }} & =0.05 \end{aligned}

Synchronous speed at reduced frequency i.e. $50 \times 0.8=40 \mathrm{~Hz}$.
$\mathrm{N}_{\mathrm{s}(\mathrm{new})}=\frac{120 \times 40}{6}=800 \mathrm{rpm}$
$\therefore$ Motor speed,

\begin{aligned} \mathrm{N}&=\left(1-\mathrm{S}_{\text {new }}\right) \mathrm{N}_{\mathrm{s} \text { (new) }} & =(1-0.05) \times 800 \\ & =760 \mathrm{rpm} \end{aligned}

 Question 3
A three-phase synchronous motor with synchronous impedance of $0.1+j0.3$ per unit per phase has a static stability limit of 2.5 per unit. The corresponding excitation voltage in per unit is ____ (Round off to 2 decimal places).
 A 0.82 B 1.59 C 2.25 D 3.64
GATE EE 2023      Synchronous Machines
Question 3 Explanation:
Developed power for motor,
$P_{d e v}=\frac{V_{f}}{Z_{s}} \cos \left(\theta_{s}-\delta\right)-\frac{E_{F}^{2}}{2} \cos \theta_{s}$

At max. power developed,
$\delta=\theta_{\mathrm{S}}$

and this is decide steady state stability of motor.
$\therefore \mathrm{SSSL}, \mathrm{P}_{\text {dev }(max)}=\frac{V E_{F}}{\mathrm{Z}_{\mathrm{S}}}-\frac{\mathrm{E}_{\mathrm{F}}^{2}}{\mathrm{Z}_{\mathrm{S}}} \cos \theta_{\mathrm{s}}$
where, $V=1.0 \mathrm{pu}$

$\mathrm{Z}_{\mathrm{S}}=0.1+\mathrm{j} 0.3=0.316 \angle 71.6^{\circ} \mathrm{pu}$
and $\mathrm{SSSL}=2.5$

$\therefore \quad 2.5=\frac{1 \times E_{F}}{0.316}-\frac{E_{F}^{2}}{0.316} \cos 71.56^{\circ}$
$1.001 \mathrm{E}_{F}^{2}-3.16 \mathrm{E}_{F}+2.5=0$
$\Rightarrow \quad \mathrm{E}_{\mathrm{F}}=1.595 \mathrm{pu}$
 Question 4
A separately excited DC motor rated $400 \mathrm{~V}, 15A, 1500 RPM$ drives a constant torque load at rated speed operating from $400 \mathrm{~V}$ DC supply drawing rated current. The armature resistance is $1.2 \Omega$. If the supply voltage drops by $10 \%$ with field current unaltered then the resultant speed of the motor in RPM is ___ (Round off to the nearest integer).
 A 1343 B 2541 C 1145 D 1852
GATE EE 2023      DC Machines
Question 4 Explanation:
Seperately excited motor :

Back emf, $\mathrm{E}_{\mathrm{b}_{1}}=400-15 \times 1.2=382 \mathrm{~V}$
Given: $\quad \tau=$ constant
$\because \quad \tau \propto \phi I_{a}$
where $\phi \rightarrow$ constant
$\therefore \mathrm{I}_{\mathrm{a}} \rightarrow$ consant
If supply voltage is drops by $10 \%$, then new value of supply voltage $=0.9 \times 400=360 \mathrm{~V}$
Now, back emf, $\mathrm{E}_{\mathrm{b}_{2}}=360-15 \times 1.2=342 \mathrm{~V}$

We have,
\begin{aligned} \mathrm{E}_{\mathrm{b}} & =\phi \omega_{\mathrm{m}} \\ \therefore \quad \frac{\mathrm{E}_{\mathrm{b}_{2}}}{\mathrm{E}_{\mathrm{b}_{1}}} & =\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \\ \Rightarrow \quad \frac{342}{382} & =\frac{\mathrm{N}_{2}}{1500} \\ \Rightarrow \quad \mathrm{N}_{2} & =342.93 \mathrm{rpm} \\ &\approx 1343 \mathrm{rpm} \end{aligned}
 Question 5
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      Single Phase Induction Motors, Special Purpose Machines and Electromechanical Energy Conversion System
Question 5 Explanation:
Since, a-a' and b-b' are placed at GNA (Geometric Neutral Axis) therefore,
$\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}=\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}=0$

There are 5 questions to complete.