# Transformers

 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   Electrical Machines
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
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   Electrical Machines
Question 2 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 3
A conducting square loop of side length 1 m is placed at a distance of 1 m from a long straight wire carrying a current l=2 A as shown below. The mutual inductance, in nH (rounded off to 2 decimal places), between the conducting loop and the long wire is __________. A 138.63 B 122.44 C 156.82 D 186.12
GATE EE 2020   Electrical Machines
Question 3 Explanation:
\begin{aligned} \phi &\propto I \\ \phi &=MI \\ \vec{B}&=\frac{\mu _{o}I}{2\pi \rho }\hat{a_{\phi }} \;\; (\vec{B} \text{ due to infinite long line}) \\ \text{Magnetic flux}& \text{ crossing square loop is}\\ \phi &=\int \int \vec{B}\cdot \vec{ds} \\ &=\int \int \frac{\mu _{o}I}{2\pi \rho }\hat{a_{\phi }}\cdot (d\rho dz)\hat{a_{\phi }}\\ &=\frac{\mu _{0}I}{2\pi }\int_{\rho =1}^{2}\frac{d\rho }{\rho }\int_{z=0}^{1}dz \\ \phi &=\frac{\mu _{o} I}{2\pi }(\ln \rho )_{\rho =1}^{2}(z)_{z=0}^{1} \\ \phi &=\frac{\mu _{o} I}{2\pi }(\ln 2)\\ M&=\frac{\phi }{I} \\ M&=\frac{\phi_{o}(\ln2)}{2\pi }=\frac{4\pi \times 10^{-7}(\ln _{2})}{2\pi } \\ M&=1.386\times 10^{-7}\, Henry \simeq 138.63\, nH \end{aligned}
 Question 4
Windings 'A', 'B' and 'C' have 20 turns each and are wound on the same iron core as shown, along with winding 'X' which has 2 turns. The figure shows the sense (clockwise/ anti-clockwise) of each of the windings only and does not reflect the exact number of turns, If windings 'A', 'B' and 'C' are supplied with balanced 3-phase voltages at 50 Hz and there is no core saturation, the no-load RMS voltage (in V, rounded off to 2 decimal places) across winding 'X' is _________ . A 36 B 46 C 12 D 58
GATE EE 2020   Electrical Machines
Question 4 Explanation:
As per GATE official answer key MTA (Marks to ALL)
$V_{X}=\frac{2}{20}(230\angle 0^{\circ}-230\angle 120^{\circ}-230\angle -120^{\circ})=46\angle 0^{\circ}\: V$
 Question 5
The figure below shows the per-phase Open Circuit Characteristics (measured in V) and Short Circuit Characteristics (measured in A) of a 14 kVA, 400 V, 50 Hz, 4-pole, 3-phase, delta connected alternator, driven at 1500 rpm. The field current, $I_f$ is measured in A. Readings taken are marked as respective (x,y) coordinates in the figure. Ratio of the unsaturated and saturated synchronous impedances ($Z_{s(unsat)}/Z_{s(sat)}$) of the alternator is closest to A 2.1 B 2.025 C 2 D 1
GATE EE 2020   Electrical Machines
Question 5 Explanation:
For unsaturated synchronous impedance \begin{aligned} \text{At } I_f&=2A, \; V_{OC}=210V \\ \text{From SCC}\\ \text{at } I_f&=4A, \; I_{SC}=20A \\ \therefore \; I_f&=2A, \; I_{SC}=10A \\ Z_s, \text{unsaturated }&=\left.\begin{matrix} \frac{V_{OC}}{I_{SC}} \end{matrix}\right|_{I_f=const.} \\ &= \frac{210}{10}=21 \Omega \\ \text{For } Z_s, \text{ saurated}:\\ \text{when, } V_{OC}&=V_{rated}=400V\\ I_f&=8A \\ \text{From SCC, } I_f&=8A, \; I_{SC}=40A \\ Z_s, \text{saturated }&=\frac{400}{40}=10\Omega \\ \text{Hence,}\\ \frac{Z_s, \text{unsaturated }}{Z_s, \text{ saturated }}&=\frac{21}{10}=2.1\Omega \end{aligned}

There are 5 questions to complete.