GATE EE 2018

 Question 1
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
 A 4.8 B 6.8 C 8.8 D 10.8
Electrical Machines   Transformers
Question 1 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.)
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 2
In a salient pole synchronous motor, the developed reluctance torque attains the maximum value when the load angle in electrical degrees is
 A 0 B 45 C 60 D 90
Electrical Machines   Synchronous Machines
Question 2 Explanation:
Salient pole synchronous motor power and torque relations per phase:
$P=\frac{E_bV}{X_t}\sin \delta +\frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \;Watts$
$T=\frac{60}{2 \pi N_S}\left [ \frac{E_bV}{X_d} \sin \delta + \frac{V^2}{2}\left ( \frac{X_d-X_q}{X_dX_q} \right ) \sin (2\delta ) \right ]\; N-m$
The second term is reluctance power or reluctance torque, which is directly proportional to $\sin 2\delta$.
Therefore, reluctance torque will be maximum.
When, $\delta =45^{\circ}$
$\because \;\;\sin2 (45^{\circ})\Rightarrow \sin 90^{\circ}=1$ (Maximum)
 Question 3
A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load voltage and current?
 A $v_{0}\geq 0 \text{ and } i_{0} \lt 0$ B $v_{0} \lt 0 \text{ and } i_{0}\lt 0$ C $v_{0}\geq 0 \text{ and } i_{0}\geq 0$ D $v_{0} \lt 0 \text{ and } i_{0}\geq 0$
Power Electronics   Phase Controlled Rectifiers
 Question 4
Four power semiconductor devices are shown in the figure along with their relevant terminals. The device(s) that can carry dc current continuously in the direction shown when gated appropriately is (are)
 A Triac only B Triac and MOSFET C Triac and GTO D Thyristor and Triac
Power Electronics   Power Semiconductor Devices and Commutation Techniques
 Question 5
Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. If one of the wattmeters reads half of the other (both positive), then the power factor of the load is
 A 0.532 B 0.632 C 0.707 D 0.866
Electrical and Electronic Measurements   Measurement of Energy and Power
Question 5 Explanation:
In two wattmeter method,
$\tan \phi =\frac{\sqrt{3}(W_1-W_2)}{W_1+W_2}$
\begin{aligned} \text{Given}, \; W_2 &=\frac{W_1}{2} \\ \tan \phi &=\frac{\sqrt{3}\left (W_1-\frac{W_1}{2} \right ) }{\left (W_1+\frac{W_1}{2} \right )} \\ \phi &= 30^{\circ}\\ \cos \phi &=\cos 30^{\circ}=0.866 \end{aligned}
 Question 6
Consider a lossy transmission line with $V_{1} \; and \; V_{2}$ as the sending and receiving end voltages, respectively. Z and X are the series impedance and reactance of the line, respectively. The steady-state stability limit for the transmission line will be
 A greter than $|\frac{V_{1}V_{2}}{X}|$ B less than $|\frac{V_{1}V_{2}}{X}|$ C equal to $|\frac{V_{1}V_{2}}{X}|$ D equal to $|\frac{V_{1}V_{2}}{Z}|$
Power Systems   Power System Stability
Question 6 Explanation:
With only x:
$P_{max}=\left | \frac{V_1V_2}{x} \right |$

With Lossy Tr, Line
$P=\left | \frac{V_1V_2}{z} \right | \cos (\beta -\delta )-\left | \frac{AV_2^2}{z} \right |\cos (\beta -\delta )$

Therefore, with Lossy Line $P_{max} \lt \left | \frac{V_1V_2}{x} \right |$
 Question 7
The graph of a network has 8 nodes and 5 independent loops. The number of branches of the graph is
 A 11 B 12 C 13 D 14
Electric Circuits   Magnetically Coupled Circuits, Network Topology and Filters
Question 7 Explanation:
Loops =b-(N-1)
5=b-(8-1)
b=12
 Question 8
In the figure, the voltages are
$v_{1}(t)=100cos(\omega t)$,
$v_{2} (t) = 100cos(\omega t + \pi /18)$ and
$v_{3}(t) = 100cos(\omega t + \pi /36)$.
The circuit is in sinusoidal steady state, and $R \lt \lt \omega L$. $P_{1}$,$P_{2}$ and $P_{3}$ are the average power outputs. Which one of the following statements is true?
 A $P_{1}=P_{2}=P_{3}=0$ B $P_{1} \lt 0,P_{2} \gt 0,P_{3} \gt 0$ C $P_{1} \lt 0,P_{2} \gt 0,P_{3} \lt 0$ D $P_{1} \gt 0,P_{2} \lt 0,P_{3} \gt 0$
Electric Circuits   Steady State AC Analysis
Question 8 Explanation:

$V_2:\frac{\pi}{18}=\frac{180^{\circ}}{18}=10^{\circ}$
$V_3:\frac{\pi}{36}=\frac{180^{\circ}}{36}=5^{\circ}$
$V_2$ leads $V_1$ and $V_3$,
So, $V_2$is a source, $V_1$ and $V_3$ are absorbing.
Hence, $P_2 \gt 0$ and $P_1,P_3 \lt 0$
 Question 9
Match the transfer functions of the second-order systems with the nature of the systems given below.
 A P-I, Q-II, R-III B P-II, Q-I, R-III C P-III, Q-II, R-I D P-III, Q-I, R-II
Control Systems   Time Response Analysis
Question 9 Explanation:
$P=\frac{15}{s^2+5s+15}$
$\omega _n=\sqrt{15}=3.872$ rad/sec
$2 \xi \times 3.872=5$
$\xi=\frac{5}{2 \times 3.872}=0.64 \;\;\;\;(Underdamped)$
$Q=\frac{25}{s^2+10s+25}$
$\omega _n=\sqrt{25}=5$ rad/sec
$2 \xi \times 5=10$
$\xi=1 \;\;\;\;\;(Critically \; damped)$
Observing all the options, option (C) is correct.
 Question 10
A positive charge of 1 nC is placed at (0,0,0.2) where all dimensions are in metres. Consider the x-y plane to be a conducting ground plane. Take $\epsilon _{0}=8.85 \times 10^{-12}$ F/m. The z component of the E field at (0,0,0.1) is closest to
 A 899.18 V/m B $-899.18 V/m$ C 999.09 V/m D $-999.09 V/m$
Electromagnetic Fields   Electrostatic Fields
Question 10 Explanation:

Net electric field at point P due to charge Q is,
\begin{aligned} \vec{E_{12}}&=\frac{Q\vec{R_{12}}}{4 \pi \varepsilon _0|\vec{R_{12}}|^3} \\ &= \frac{1 \times 10^{-9}(-0.1\hat{a_z})}{4 \pi (8.854 \times 10^{-12})(0.1)^3}\\ &+ \frac{-1 \times 10^{-9}(-0.3\hat{a_z})}{4 \pi (8.854 \times 10^{-12})(0.3)^3}\\ &= \left [ \frac{-10^5}{4 \pi (8.854)}- \frac{10^5}{4 \pi (8.854)9} \right ]\hat{a_z}\\ &=(-898.774-99.863) \hat{a_z}\\ &= -999.09\hat{a_z}\; V/m \end{aligned}
There are 10 questions to complete.