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
4.8 | |
6.8 | |
8.8 | |
10.8 |
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.)
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 2 |
In a salient pole synchronous motor, the developed reluctance torque attains the maximum value when the load angle in electrical degrees is
0 | |
45 | |
60 | |
90 |
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)
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?
v_{0}\geq 0 \text{ and } i_{0} \lt 0 | |
v_{0} \lt 0 \text{ and } i_{0}\lt 0 | |
v_{0}\geq 0 \text{ and } i_{0}\geq 0 | |
v_{0} \lt 0 \text{ and } i_{0}\geq 0 |
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)
Triac only | |
Triac and MOSFET | |
Triac and GTO | |
Thyristor and Triac |
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
0.532 | |
0.632 | |
0.707 | |
0.866 |
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}
\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
greter than |\frac{V_{1}V_{2}}{X}| | |
less than |\frac{V_{1}V_{2}}{X}| | |
equal to |\frac{V_{1}V_{2}}{X}| | |
equal to |\frac{V_{1}V_{2}}{Z}| |
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 |
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
11 | |
12 | |
13 | |
14 |
Question 7 Explanation:
Loops =b-(N-1)
5=b-(8-1)
b=12
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?
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?
P_{1}=P_{2}=P_{3}=0 | |
P_{1} \lt 0,P_{2} \gt 0,P_{3} \gt 0 | |
P_{1} \lt 0,P_{2} \gt 0,P_{3} \lt 0 | |
P_{1} \gt 0,P_{2} \lt 0,P_{3} \gt 0 |
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_2is 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.
P-I, Q-II, R-III | |
P-II, Q-I, R-III | |
P-III, Q-II, R-I | |
P-III, Q-I, R-II |
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.
\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
899.18 V/m | |
-899.18 V/m | |
999.09 V/m | |
-999.09 V/m |
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.