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 |

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.

0.42 | |

0.82 | |

1.02 | |

1.26 |

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 _________ .

36 | |

46 | |

12 | |

58 |

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

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).

125.29 | |

35.45 | |

85.12 | |

158.36 |

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

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

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}

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

\frac{10}{0.5s+1} | |

\frac{2}{0.5s+1} | |

\frac{2}{s+0.5} | |

\frac{10}{s+0.5} |

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:

1200 | |

1000 | |

1220 | |

900 |

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}

\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)

110 | |

150 | |

130 | |

90 |

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}

\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).

162.5 | |

12.45 | |

188.66 | |

212.46 |

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 _________ .

4.21 | |

16.45 | |

2.21 | |

8.44 |

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.