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).
1256.3 | |
4152.4 | |
2218.4 | |
6523.8 |
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}
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
1600\:W and 900\:W
| |
900\:W and 1600\:W
| |
250\:W and 600\:W
| |
600\:W and 250\:W
|
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}
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 __________.
138.63 | |
122.44 | |
156.82 | |
186.12 |
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 _________ .
36 | |
46 | |
12 | |
58 |
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
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
2.1 | |
2.025 | |
2 | |
1 |
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.