Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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}
Question 4 |
A single-phase, 4 kVA, 200 V/100 V, 50 Hz transformer with laminated CRGO steel core
has rated no-load loss of 450 W. When the high-voltage winding is excited with 160 V,
40 Hz sinusoidal ac supply, the no-load losses are found to be 320 W. When the highvoltage winding of the same transformer is supplied from a 100 V, 25 Hz sinusoidal ac
source, the no-load losses will be _________W (rounded off to 2 decimal places).
162.5 | |
12.45 | |
188.66 | |
212.46 |
Question 4 Explanation:
\begin{aligned}200 V,50 Hz, P_{c}&=450 Watt\\ 160 V,40 Hz,
P_{c}&=320 Watt \\ 100 V,25 Hz,]P_{c}&=? Watt \\ \frac{v}{f}=\text{constant}&=\frac{200}{50}=\frac{160}{40}=\frac{100}{25}\\ \text{So, } P_{c}&=Af+Bf^{2} \\ 450&=A\times (50)+B\times (50)^{2} \;\; ...(i)\\ 320&=A\times (40)+B\times (40)^{2} \;\; ...(ii)\\ \text{From (i) and (ii),}\\ \frac{450}{50}&=A+B(50) \;\;...(iii)
\\ \frac{320}{40}&=A+B(40) \;\;...(iv)
\\ \text{Equation (iii)-(iv),} \\ (9-8)&=B(10)\\ B&=\frac{1}{10} \\ A&=9-\frac{1}{10}\times 50=4 \\ \text{Now at 100V,25 Hz,} \\P_{c}&=4\times 25+\frac{1}{10}\times (25)^{2}\\
&=100+62.5=162.50 Watt
\end{aligned}
Question 5 |
A single-phase transformer of rating 25 kVA, supplies a 12 kW load at power factor of 0.6 lagging. The additional load at unity power factor in kW (round off to two decimal places) that may be added before this transformer exceeds its rated kVA is __________.
7.21 | |
6.24 | |
5.69 | |
8.78 |
Question 5 Explanation:
For a 12 kW, 0.6 pf lagging load,
\begin{aligned} P_L &= 12kW\\ Q_L&= \frac{12}{0.6} \times \sin (cos^{-1} 0.6)\\ &= 16kVAR \end{aligned}
Trasformer rating , S=25kVA
Let us assume load thet can be added is P kW then,
\begin{aligned} S^2 &=(P+P_L)^2+Q_L^2 \\ 25^2&= (P+12)^2+16^2\\ P &= 7.21kW \end{aligned}
\begin{aligned} P_L &= 12kW\\ Q_L&= \frac{12}{0.6} \times \sin (cos^{-1} 0.6)\\ &= 16kVAR \end{aligned}
Trasformer rating , S=25kVA
Let us assume load thet can be added is P kW then,
\begin{aligned} S^2 &=(P+P_L)^2+Q_L^2 \\ 25^2&= (P+12)^2+16^2\\ P &= 7.21kW \end{aligned}
Question 6 |
A 5 kVA, 50 V/100 V, single-phase transformer has a secondary terminal voltage of 95 V when loaded. The regulation of the transformer is
4.50% | |
9% | |
5% | |
1% |
Question 6 Explanation:
\begin{aligned} \text{Voltage regulation } &=\frac{V_{NL}-V_{FL}}{V_{NL}} \times 100 \\ &= \frac{100-95}{100} \times 100=5\% \end{aligned}
Question 7 |
A 3-phase 900kVA, 3kV/\sqrt{3}kV(\Delta /Y), 50Hz transformer has primary (high voltage side) resistance per phase of 0.3 \Omega and secondary (low voltage side) resistance per phase of 0.02 \Omega. Iron loss of the transformer is 10 kW. The full load % efficiency of the transformer operated at unity power factor is _______ (up to 2 decimal places).
56.25 | |
47.25 | |
97.36 | |
112.25 |
Question 7 Explanation:
900kVA, \Delta /Y, 3-phase transformer
\begin{aligned} \text{Given,}\;\; R_1 &=300W/ph \\ R_2&= 0.02 W/ph\\ \text{Iron loss}&=10kW \end{aligned}
Primary line current,
I_{l_1}=\frac{900 \times 1000}{\sqrt{3} \times 3000}=173.2A
Primary phase current,
I_{ph_1}=100A
Secondary line current,
I_{l_2}=\frac{900 \times 1000}{\sqrt{3} \times 1732}=300A
Secondary phase current
I_{ph_2}=300A
Cu loss of H.V. side
=3 \times I_1^2 R_1=3(100)^2 (0.3) =9000 Watts
Cu loss of L.V. side
= 3 \times _2^2 \times R_2 =3(300)^2(0.02)=5400Watts
Full load efficiency at u.p.f.
\begin{aligned} \eta &=\frac{1 \times 900 \times 1000 \times 1}{1 \times 900 \times 1000+9000+5400+10000} \\ &= \frac{900000}{924400}\\ \therefore \;\;\eta &=\eta 0.9736 \; \text{or} \; 97.36\% \end{aligned}
\begin{aligned} \text{Given,}\;\; R_1 &=300W/ph \\ R_2&= 0.02 W/ph\\ \text{Iron loss}&=10kW \end{aligned}
Primary line current,
I_{l_1}=\frac{900 \times 1000}{\sqrt{3} \times 3000}=173.2A
Primary phase current,
I_{ph_1}=100A
Secondary line current,
I_{l_2}=\frac{900 \times 1000}{\sqrt{3} \times 1732}=300A
Secondary phase current
I_{ph_2}=300A
Cu loss of H.V. side
=3 \times I_1^2 R_1=3(100)^2 (0.3) =9000 Watts
Cu loss of L.V. side
= 3 \times _2^2 \times R_2 =3(300)^2(0.02)=5400Watts
Full load efficiency at u.p.f.
\begin{aligned} \eta &=\frac{1 \times 900 \times 1000 \times 1}{1 \times 900 \times 1000+9000+5400+10000} \\ &= \frac{900000}{924400}\\ \therefore \;\;\eta &=\eta 0.9736 \; \text{or} \; 97.36\% \end{aligned}
Question 8 |
A single-phase 100 kVA, 1000 V / 100 V, 50 Hz transformer has a voltage drop of 5% across its series impedance at full load. Of this, 3% is due to resistance. The percentage regulation of the transformer at full load with 0.8 lagging power factor is
4.8 | |
6.8 | |
8.8 | |
10.8 |
Question 8 Explanation:
Percent voltage regulation
=\left ( \frac{I_2R_{02}}{E_2} \times 100 \right )\cos \phi _2\pm \left ( \frac{I_2X_{02}}{E_2} \times 100 \right ) \sin \phi _2
\% \; VR.=V_r \cos \phi _2\pm V_x \sin \phi _2
(where, '+' lag p.f. and '-' lead p.f.)
At full load:
Given, V_r=3\%
Impedance drop, V_z=5\%
\therefore \; Reluctance drop, V_x=\sqrt{5^2-3^2}=4\%
Voltage regulation at full load at 0.8 p.f. lagging
V.R.=3(0.8)+4(0.6)=4.8\%
=\left ( \frac{I_2R_{02}}{E_2} \times 100 \right )\cos \phi _2\pm \left ( \frac{I_2X_{02}}{E_2} \times 100 \right ) \sin \phi _2
\% \; VR.=V_r \cos \phi _2\pm V_x \sin \phi _2
(where, '+' lag p.f. and '-' lead p.f.)
At full load:
Given, V_r=3\%
Impedance drop, V_z=5\%
\therefore \; Reluctance drop, V_x=\sqrt{5^2-3^2}=4\%
Voltage regulation at full load at 0.8 p.f. lagging
V.R.=3(0.8)+4(0.6)=4.8\%
Question 9 |
If the primary line voltage rating is 3.3 kV (Y side) of a 25 kVA. Y-\Delta transformer (the per phase turns ratio is 5:1), then the line current rating of the secondary side (in Ampere) is_____.
38 | |
44 | |
22 | |
82 |
Question 9 Explanation:
From turn ratio,
\begin{aligned} V_2 &=381 V/phase \\ V_{2 \; line}&= 381 V\\ \sqrt{3} V_2I_2&= 25000\\ \therefore \;\; I_2&=\frac{25000}{\sqrt{3} \times 381}=37.88A \end{aligned}
Question 10 |
A three-phase, three winding \Delta / \Delta / Y (1.1kV/6.6kV/400 V) transformer is energized from AC mains at the 1.1 kV side. It supplies 900 kVA load at 0.8 power factor lag from the 6.6 kV winding and 300 kVA load at 0.6 power factor lag from the 400 V winding. The RMS line current in ampere drawn by the 1.1 kV winding from the mains is _______.
456 | |
286 | |
625 | |
789 |
Question 10 Explanation:
Load on 6.6 kV winding = 900 kVA @0.8 pf lag
\therefore \;\; I_2=\frac{900 \times 10^3}{\sqrt{3} \times 6.6 \times 10^3}
\;\;\;\;=78.73\angle -36.86^{\circ} A
Transforamtion Ratio
\begin{aligned} k &=\frac{1.1}{6.6}=\frac{1}{6} \\ \therefore \;\; I'_2&=\frac{I_2}{k}=472.4\angle -36.86^{\circ} A \end{aligned}
Load on 400 V winding = 300 kVA at 0.6 p.f.
\therefore \;\; I_3=\frac{300 \times 10^3}{\sqrt{3} \times 400} =433.025\angle -53.13^{\circ}A
Transformation ratio = \frac{1.1 \times 1000}{\frac{400}{\sqrt{3}}}=4.76
\begin{aligned} I'_3 &=\frac{I_3}{k}=\frac{4333.025}{4.76} \\ &= 90.97\angle -53.13^{\circ}A \end{aligned}
So, line value of current
\begin{aligned} &=\sqrt{3} \times 90.97\angle -53.13^{\circ} \\ &= 157.56 \angle -53.13^{\circ} \end{aligned}
Current drawn by 1.1 kV winding will be
\begin{aligned} I_1&=I'_2+I'_3 \\ &=472.4\angle -36.86 +157.56\angle -53.13^{\circ} \\ &=377.96 -j283.37 +94.53 -126.04 \\ &= 472.49 -j409.41\\ &= 625.19\angle -40.9^{\circ}A \end{aligned}
\therefore \;\; I_2=\frac{900 \times 10^3}{\sqrt{3} \times 6.6 \times 10^3}
\;\;\;\;=78.73\angle -36.86^{\circ} A
Transforamtion Ratio
\begin{aligned} k &=\frac{1.1}{6.6}=\frac{1}{6} \\ \therefore \;\; I'_2&=\frac{I_2}{k}=472.4\angle -36.86^{\circ} A \end{aligned}
Load on 400 V winding = 300 kVA at 0.6 p.f.
\therefore \;\; I_3=\frac{300 \times 10^3}{\sqrt{3} \times 400} =433.025\angle -53.13^{\circ}A
Transformation ratio = \frac{1.1 \times 1000}{\frac{400}{\sqrt{3}}}=4.76
\begin{aligned} I'_3 &=\frac{I_3}{k}=\frac{4333.025}{4.76} \\ &= 90.97\angle -53.13^{\circ}A \end{aligned}
So, line value of current
\begin{aligned} &=\sqrt{3} \times 90.97\angle -53.13^{\circ} \\ &= 157.56 \angle -53.13^{\circ} \end{aligned}
Current drawn by 1.1 kV winding will be
\begin{aligned} I_1&=I'_2+I'_3 \\ &=472.4\angle -36.86 +157.56\angle -53.13^{\circ} \\ &=377.96 -j283.37 +94.53 -126.04 \\ &= 472.49 -j409.41\\ &= 625.19\angle -40.9^{\circ}A \end{aligned}
There are 10 questions to complete.