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}

There are 5 questions to complete.