1
GATE EE 2003
MCQ (Single Correct Answer)
+2
-0.6
An inverter has a periodic output voltage with the output waveform as shown in figure GATE EE 2003 Power Electronics - Inverters Question 18 English

With reference to the output waveform given in figure, the output of the converter will be free form $${5^{th}}$$ harmonic when

A
$$\alpha = {72^ \circ }$$
B
$$\alpha = {36^ \circ }$$
C
$$\alpha = {150^ \circ }$$
D
$$\alpha = {120^ \circ }$$
2
GATE EE 2002
MCQ (Single Correct Answer)
+2
-0.6
Fig. $$(a)$$ shows an inverter circuit with a $$dc$$ source voltage $${V_{s}}$$. The semiconductor switches of the inverter are operated in such a manner that the pole voltages $${V_{10}}$$ and $${V_{20}}$$ are as shown in fig.
Fig. $$(b).$$ What is the rms value of the pole-to-pole voltage $${V_{12}}$$: GATE EE 2002 Power Electronics - Inverters Question 20 English 1 GATE EE 2002 Power Electronics - Inverters Question 20 English 2
A
$${{{V_s}\phi } \over {\pi \sqrt 2 }}$$
B
$${V_s}\sqrt {{\phi \over \pi }} $$
C
$${V_s}\sqrt {{\phi \over {2\pi }}} $$
D
$${{{V_s}} \over \pi }$$
3
GATE EE 2001
MCQ (Single Correct Answer)
+2
-0.6
A single - phase full-bridge voltage source inverter feeds a purely inductive load, as shown in figure. when $${T_1},{T_2},\,\,\,\,\,{T_3},{T_4}$$ are power transistors and $${D_1},{D_2},\,{D_3},{D_4}$$ are feedback diodes. The inverter is operated in square-wave mode with a frequency of $$50$$ $$Hz.$$ If the average load current is zero, what is the time duration of conduction of each feedback diode in a cycle? GATE EE 2001 Power Electronics - Inverters Question 21 English
A
$$5\,\,m\,\,\sec $$
B
$$10\,\,m\,\,\sec $$
C
$$20\,\,m\,\,\sec $$
D
$$2.5\,\,m\,\,\sec $$
4
GATE EE 2000
MCQ (Single Correct Answer)
+2
-0.6
A three phase voltage source inverter supplies a purely inductive three phase load. Upon Fourier analysis, the output voltage waveform is found to have an $${h^{th}}$$ order harmonic of magnitude α h times that of the fundamental frequency component $$\left( {{\alpha _h} < 1} \right),$$ the load current would then have an $${h^{th}}$$ order harmonic of magnitude
A
zero
B
$${{\alpha _h}}$$ times the fundamental frequency component
C
$${h{\alpha _h}}$$ times the fundamental frequency component
D
$${\raise0.5ex\hbox{$\scriptstyle {{\alpha _h}}$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle h$}}$$ times the fundamental frequency component
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