Question 1 |
The single phase rectifier consisting of three thyristors T_{1}, T_{2}, T_{3} and a diode D_{1} feed power to a 10 \mathrm{~A} constant current load. T_{1} and T_{3} are fired at \alpha=60^{\circ} and T_{2} is fired at \alpha=240^{\circ}. The reference for \alpha is the positive zero crossing of \mathrm{V}_{\text {in }}. The average voltage \mathrm{V}_{o} across the load in volts is _____ (Round off to 2 decimal places).
25.35 | |
36.47 | |
39.79 | |
48.64 |
Question 1 Explanation:
\therefore Average output voltage,
\begin{gathered} \mathrm{V}_{o}=\frac{1}{2 \pi}\left[\int_{\alpha}^{\pi} \mathrm{V}_{\mathrm{m}} \sin \omega t d \omega t+\int_{\pi+\alpha}^{2 \pi+\alpha}-\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t} d \omega t\right] \\ =\frac{V_{m}}{2 \pi}[1+3 \cos \alpha] \end{gathered}
Put the values,
\begin{aligned} V_{o} & =\frac{100}{2 \pi}\left[1+3 \cos 60^{\circ}\right] \\ & =39.79 \mathrm{~V} \end{aligned}
Question 2 |
The circuit shown in the figure has reached steady state with thyristor 'T' in OFF condition. Assume that the latching and holding currents of the thyristor are zero.
The thyristor is turned ON at t=0 sec. The duration in microseconds for which the thyristor would conduct, before it turns off, is ___ (Round off to 2 decimal places).
The thyristor is turned ON at t=0 sec. The duration in microseconds for which the thyristor would conduct, before it turns off, is ___ (Round off to 2 decimal places).
7.33 | |
5.36 | |
4.25 | |
8.23 |
Question 2 Explanation:
Case (i) t\lt 0 sec :
Steady state circuit :
Case (ii) \mathrm{t} \gt 0 sec:
Thyristor is ON
Redraw the circuit :
\mathrm{I}_{4 \Omega}=\frac{100}{4}=25 \mathrm{~A}
and
\begin{aligned} \mathrm{i}_{\mathrm{C}} & =\mathrm{I}_{\mathrm{P}} \sin \omega_{0} \mathrm{t} \\ & =\mathrm{V}_{\mathrm{S}} \sqrt{\frac{C}{L}} \sin \omega_{0} \mathrm{t} \\ & =100 \sqrt{\frac{1}{4}} \sin \omega_{0} \mathrm{t}=50 \sin \omega_{0} \mathrm{t} \end{aligned}
\therefore Current through thyristor :
\mathrm{I}_{\mathrm{T}}=25+50 \sin \omega_{0} \mathrm{t}
When \mathrm{I}_{\mathrm{T}}=0 \Rightarrow Thyristor is turn off.
0=25+50 \sin \omega_{0} t
\Rightarrow \quad \omega_{0} \mathrm{t}=210^{\circ} \times \frac{\pi}{180^{\circ}}
\Rightarrow \quad \mathrm{t}=\sqrt{\mathrm{LC}} \times \frac{7 \pi}{6}
=\sqrt{10^{-6} \times 4 \times 10^{-6}} \times \frac{7 \pi}{6}
=7.33 \mu \mathrm{sec}
Steady state circuit :
Case (ii) \mathrm{t} \gt 0 sec:
Thyristor is ON
Redraw the circuit :
\mathrm{I}_{4 \Omega}=\frac{100}{4}=25 \mathrm{~A}
and
\begin{aligned} \mathrm{i}_{\mathrm{C}} & =\mathrm{I}_{\mathrm{P}} \sin \omega_{0} \mathrm{t} \\ & =\mathrm{V}_{\mathrm{S}} \sqrt{\frac{C}{L}} \sin \omega_{0} \mathrm{t} \\ & =100 \sqrt{\frac{1}{4}} \sin \omega_{0} \mathrm{t}=50 \sin \omega_{0} \mathrm{t} \end{aligned}
\therefore Current through thyristor :
\mathrm{I}_{\mathrm{T}}=25+50 \sin \omega_{0} \mathrm{t}
When \mathrm{I}_{\mathrm{T}}=0 \Rightarrow Thyristor is turn off.
0=25+50 \sin \omega_{0} t
\Rightarrow \quad \omega_{0} \mathrm{t}=210^{\circ} \times \frac{\pi}{180^{\circ}}
\Rightarrow \quad \mathrm{t}=\sqrt{\mathrm{LC}} \times \frac{7 \pi}{6}
=\sqrt{10^{-6} \times 4 \times 10^{-6}} \times \frac{7 \pi}{6}
=7.33 \mu \mathrm{sec}
Question 3 |
For the ideal AC-DC rectifier circuit shown in the figure below, the load current
magnitude is I_{dc} = 15 A and is ripple free. The thyristors are fired with a delay angle
of 45^{\circ}. The amplitude of the fundamental component of the source current, in
amperes, is __________. (round off to two decimal places)
12.45 | |
25.32 | |
14.25 | |
17.64 |
Question 3 Explanation:
Given rectifier circuit is a 1-\phi semiconverter Waveform of source current:
\therefore \; i_{sn}=\sum_{n=1,3,5...}^{\infty }\frac{4I_o}{n\pi} \cos \left ( \frac{n\alpha }{2} \right ) \sin (n\omega t+\phi _n)
Now, fundamental component,
I_{s1}=\frac{4I_o}{\pi} \cos \left ( \frac{\alpha }{2} \right )=\frac{4 \times 15}{\pi} \cos \left ( \frac{45^{\circ} }{2} \right )=17.64A
\therefore \; i_{sn}=\sum_{n=1,3,5...}^{\infty }\frac{4I_o}{n\pi} \cos \left ( \frac{n\alpha }{2} \right ) \sin (n\omega t+\phi _n)
Now, fundamental component,
I_{s1}=\frac{4I_o}{\pi} \cos \left ( \frac{\alpha }{2} \right )=\frac{4 \times 15}{\pi} \cos \left ( \frac{45^{\circ} }{2} \right )=17.64A
Question 4 |
The voltage at the input of an AC-DC rectifier is given by v(t)=230\sqrt{2}\sin \omega t where \omega = 2 \pi \times 50 rad/s. The input current drawn by the rectifier is given by
i(t)=10\sin \left (\omega t -\frac{\pi}{3} \right )+4\sin \left (3\omega t -\frac{\pi}{6} \right )+3\sin \left (5\omega t -\frac{\pi}{3} \right )
The input power factor, (rounded off to two decimal places), is, _________ lag.
i(t)=10\sin \left (\omega t -\frac{\pi}{3} \right )+4\sin \left (3\omega t -\frac{\pi}{6} \right )+3\sin \left (5\omega t -\frac{\pi}{3} \right )
The input power factor, (rounded off to two decimal places), is, _________ lag.
0.44 | |
0.22 | |
0.66 | |
0.88 |
Question 4 Explanation:
We have,
\begin{aligned} PF&=g.FDF \\ g&=\frac{I_{{s1}}}{I_{sr}} \\ &= \frac{10/\sqrt{2}}{\sqrt{\left ( \frac{10}{\sqrt{2}} \right )^2+\left ( \frac{4}{\sqrt{2}} \right )^2+\left ( \frac{3}{\sqrt{2}} \right )^2}}\\ &=0.894 \\ FDF&= \cos \frac{\pi}{3}=0.5 \\ PF&=0.894 \times 0.5=0.447 \end{aligned}
\begin{aligned} PF&=g.FDF \\ g&=\frac{I_{{s1}}}{I_{sr}} \\ &= \frac{10/\sqrt{2}}{\sqrt{\left ( \frac{10}{\sqrt{2}} \right )^2+\left ( \frac{4}{\sqrt{2}} \right )^2+\left ( \frac{3}{\sqrt{2}} \right )^2}}\\ &=0.894 \\ FDF&= \cos \frac{\pi}{3}=0.5 \\ PF&=0.894 \times 0.5=0.447 \end{aligned}
Question 5 |
A single-phase full-bridge diode rectifier feeds a resistive load of 50 \Omega from a 200 V,
50 Hz single phase AC supply. If the diodes are ideal, then the active power, in watts,
drawn by the load is _____________. (round off to nearest integer).
200 | |
1600 | |
600 | |
800 |
Question 5 Explanation:
For 1-\phi full bridge diode rectifier:
V_{or}=\frac{V_m}{\sqrt{2}} =\frac{\sqrt{2} \times 200}{\sqrt{2}}=200V
Active power drawn by the load
P_{o}=\frac{V_{or}^2}{R} =\frac{200^2}{50}=800W
V_{or}=\frac{V_m}{\sqrt{2}} =\frac{\sqrt{2} \times 200}{\sqrt{2}}=200V
Active power drawn by the load
P_{o}=\frac{V_{or}^2}{R} =\frac{200^2}{50}=800W
There are 5 questions to complete.