Electrical Machines

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 __________.
A
138.63
B
122.44
C
156.82
D
186.12
GATE EE 2020      Transformers
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
A cylindrical rotor synchronous generator has steady state synchronous reactance of 0.7 pu and subtransient reactance of 0.2 pu. It is operating at (1+j0) pu terminal voltage with an internal emf of (1+j0.7) pu. Following a three-phase solid short circuit fault at the terminal of the generator, the magnitude of the subtransient internal emf (rounded off to 2 decimal places) is ________ pu.
A
0.42
B
0.82
C
1.02
D
1.26
GATE EE 2020      Synchronous Machines
Question 2 Explanation: 
\begin{aligned} &\text{Prefault current,}\\ I_{0}&=\frac{E_{f}-V_{t}}{jX_{d}} \\ &=\frac{1+j0.7-1}{j0.7}=1 \\ &\text{Subtransient induced emf,} \\ {E_{f}}''&=V_{0}+j{X_{d}}''I_{0} \\ &=1+j0.2\times 1=1+j0.2 \\ | {E_{f}''}|&=\sqrt{1^{2}+0.2^{2}}=1.02 \end{aligned}
Question 3
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      Transformers
Question 3 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 4
A cylindrical rotor synchronous generator with constant real power output and constant terminal voltage is supplying 100 A current to a 0.9 lagging power factor load. An ideal reactor is now connected in parallel with the load, as a result of which the total lagging reactive power requirement of the load is twice the previous value while the real power remains unchanged. The armature current is now _______ A (rounded off to 2 decimal places).
A
125.29
B
35.45
C
85.12
D
158.36
GATE EE 2020      Synchronous Machines
Question 4 Explanation: 
At P_{constant},
I_{a1}\cos \phi _{1}=I_{a2}\cos \phi _{2}
\cos \phi _{1}=0.9
\tan \phi _{1}=0.484=\frac{Q}{P}
\Rightarrow \, \, \frac{2Q}{P}=0.9686=\tan \phi _{2}
\cos \phi _{2}=0.7182
\therefore \, 100\times 0.9=I_{a2}\times 0.7182
\Rightarrow \, \, I_{a2}=125.29\: A
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      Transformers
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}
Question 6
Consider a permanent magnet dc (PMDC) motor which is initially at rest. At t=0, a dc voltage of 5 V is applied to the motor. Its speed monotonically increases from 0 rad/s to 6.32 rad/s in 0.5 s and finally settles to 10 rad/s. Assuming that the armature inductance of the motor is negligible, the transfer function for the motor is
A
\frac{10}{0.5s+1}
B
\frac{2}{0.5s+1}
C
\frac{2}{s+0.5}
D
\frac{10}{s+0.5}
GATE EE 2020      DC Machines
Question 6 Explanation: 
\begin{aligned} \text{Input} &= 5 V \\ \Rightarrow \; R(s)&=\frac{5}{s} \\ \frac{C(s)}{R(s)}&=T.F.=\frac{K}{1+Ts} \\ C(s)&=\frac{5K}{s(1+Ts)} \\ \lim_{s\rightarrow 0}&=5K=10\\ \Rightarrow \; K&=2\\ \text{Steady state speed}&=10 \text{ rad/sec.(Given)} \\ C(s)&=\frac{2}{1+0.5s}\times \frac{5}{s} \\ \Rightarrow \; \lim_{s\rightarrow 0}sC(s)&=10 \end{aligned}
Question 7
A 250 V dc shunt motor has an armature resistance of 0.2\Omega and a field resistance of 100\Omega. When the motor is operated on no-load at rated voltage. It draws an armature current of 5 A and runs at 1200 rpm. When a load is coupled to the motor, it draws total line current of 50 A at rated voltage, with a 5% reduction in the air-gap flux due to armature reaction. Voltage drop across the brushes can be taken as 1 V per brush under all operating conditions. The speed of the motor, in rpm, under this loaded condition, is closest to:
A
1200
B
1000
C
1220
D
900
GATE EE 2020      DC Machines
Question 7 Explanation: 
no load current 5 A

\begin{aligned} B.R.D&=\text{1 V per brush}\\ \text{loaded, }I_{L}&=50 A \\R_{sh}&=100 \Omega\\ I_{sh}&=\frac{250}{100}=2.5\: A \\I_{a0}&=2.5\: A\\I_{aL}&=47.5\: A \\V&=E_{b}+I_{a}R_{a}+B.R.D\\ E_{b\, no\, load}&=V-I_{a0}R_{a}-B.R.D \\&=250-2.5(0.2)-1\times 2 \\&=247.5\, Volts \\E_{b\, load}&=250-47.5(0.2)-1\times 2 \\&=238.5\, volts\\ \frac{N_{2}}{N_{1}}&=\frac{E_{b_{2}}}{E_{b_{1}}}\times \frac{\phi_{1} }{\phi _{2}} \\ \frac{N_{2}}{1200}&=\frac{238.5}{247.5}\times \frac{\phi _{1}}{0.95\phi_{1}}\\ N_{2}&=1217.22\, rpm \end{aligned}
Question 8
A single 50 Hz synchronous generator on droop control was delivering 100 MW power to a system. Due to increase in load, generator power had to be increased by 10 MW. as a result of which, system frequency dropped to 49.75 Hz. Further increase in load in the system resulted in a frequency of 49.25 Hz. At this condition, the power in MW supplied by the generator is ________ (rounded off to 2 decimal places)
A
110
B
150
C
130
D
90
GATE EE 2020      Synchronous Machines
Question 8 Explanation: 
Assumed full load frequency is 50 Hz
\begin{aligned}\tan \theta &=\frac{50-49.75}{110-100} \\ &=\frac{49.75-49.25}{(x-(110))} \\ \frac{0.25}{10}&=\frac{0.5}{(x-150)} \\ x-110&=\frac{0.5\times 10}{0.25} \\ x&=110+20=130\, MW\end{aligned}

Question 9
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).
A
162.5
B
12.45
C
188.66
D
212.46
GATE EE 2020      Transformers
Question 9 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 10
A three-phase, 50 Hz, 4-pole induction motor runs at no-load with a slip of 1%. With full load, the slip increases to 5 %. The % speed regulation of the motor (rounded off to 2 decimal places) is _________ .
A
4.21
B
16.45
C
2.21
D
8.44
GATE EE 2020      Single Phase Induction Motors, Special Purpose Machines and Electromechanical Energy Conversion System
Question 10 Explanation: 
\begin{aligned} \text{Synchronous speed, }\\ N_s&=\frac{120f}{P}\\ N_s&=\frac{120 \times 50}{4}rpm\\ N_s&=1500 \; rpm\\ \text{Speed at no-load,}\\ &=N_s(1-S_{nL})\\ &=1500(1-0.01)\\ &=1485 \; rpm\\ \text{Speed at full-load,}\\ &=N_s(1-S_{fL})\\ &=1500(1-0.05)\\ &=1425 \; rpm\\ \% \text{ speed regulation}&=\frac{N_{NL}-N_{FL}}{N_{FL}} \times 100\\ &=\frac{1485-1425}{1425} \times 100\\ &=4.21\% \end{aligned}


There are 10 questions to complete.
Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.