Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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
Question 4 |
The waveform shown in solid line is obtained by clipping a full-wave rectified sinusoid (shown dashed). The ratio of the \text{RMS} value of the full-wave rectified waveform to the \text{RMS} value of the clipped waveform is _______________. (Round off to 2 decimal places.)
1.21 | |
2.24 | |
1.82 | |
0.65 |
Question 4 Explanation:
We know,
RMS value of full wave rectified sine =\frac{V_{m}}{\sqrt{2}}=0.707 V_{m}
For clipped wave form
As 0 \rightarrow \frac{\pi}{4} and \frac{3 \pi}{4} \rightarrow \pi wave form are identical
\begin{aligned} &\text { rms value of clipped wave } \\ &\left.=\sqrt{\frac{1}{\pi}\left[2 \int_{0}^{\pi / 4} V_{m}^{2} \sin ^{2} \omega t+\int_{\pi / 4}^{3 \pi / 4}\left(0.707 V_{m}\right)^{2} d \omega t\right.}\right] \\ &= \sqrt{\frac{1}{\pi}\left(\int_{0}^{\pi / 4} \frac{2 V_{m}^{2}}{2}[1-\cos 2 \omega t] d \omega t+\frac{1}{2} V_{m}^{2} \times \frac{\pi}{2}\right)^{1 / 2}}\\ &=\left[\frac{1}{\pi}\left[V_{m}^{2}\left(\frac{\pi}{4}-\frac{1}{2} \sin \frac{\pi}{2}\right)+\frac{V_{m}^{2}}{4} \pi\right]\right]^{1 / 2} \\ &=\left[\frac{V_{m}^{2}}{\pi}\left[\frac{\pi}{4}-\frac{1}{2}\right]+\frac{V_{m}^{2}}{4}\right]^{1 / 2}=0.5838 V_{m} \\ \text { Ratio } &=\frac{0.707}{0.5838}=1.21 \end{aligned}
Question 5 |
In the circuit shown, the input V_{i} is a sinusoidal \text{AC} voltage having an \text{RMS} value of 230\;V\pm 20\%. The worst-case peak-inverse voltage seen across any diode is ____________V. (Round off to 2 decimal places.)
390.32 | |
124.25 | |
258.36 | |
471.56 |
Question 5 Explanation:
\begin{aligned} \left(V_{D}\right)_{\text {peak }} \text { for worst case } &=(230+20 \%) \times \sqrt{2} \\ &=\left(230+230 \times \frac{20}{100}\right) \times \sqrt{2}=390.32 \mathrm{~V} \end{aligned}
Question 6 |
A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a
10\Omega resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance
and a total inductance of 2.28 mH. If the thyristors are triggered at an angle \alpha = 45^{\circ},
the commutation overlap angle in degree (rounded off to 2 decimal places) is_____.
2.4 | |
4.8 | |
6.4 | |
8.2 |
Question 6 Explanation:
\begin{aligned}&1-\phi\text{ SCR bridge rectifier} \\ \alpha &=45^{\circ},\; \; R=10\, \Omega\\ &\text{supply 230 V, 50 Hz}\\ L_{s}&=2.28\, mH \\ \mu &=? \\ \Delta V_{d}&=\frac{V_{m}}{\pi }[\cos \alpha -\cos (\alpha +\mu )]\\ &= 4fL_{s}I_{0} \\ V_{0}&=\frac{2V_{m}}{\pi }\cos \alpha -4fL_{s}I_{0} \\ I_{0}R&=\frac{2V_{m}}{\pi}\cos \alpha -4fL_{s}I_{0} \\ &\text{Find }I_{0} \\ I_{0}\times 10&=\frac{2\times 230\sqrt{2}}{\pi } \cos 45 \\&-4\times 50\times 2.28\times 10^{-3}I_{0} \\ I_{0}(10+0.456)&=146.42 \\ I_{0}&=\frac{146.49}{10.456}=14.0036\: A \\ \Delta V_{d0 }&=\frac{230\sqrt{2}}{\pi }[\cos 45-\cos (45+\mu)] \\ &=4\times 50\times 2.28\times 10^{-3}\times 14 \\ &=6.384 \\ \cos 45^{\circ}-\cos (45^{\circ}+\mu )&=0.061659 \\ 45+\mu &=49.80 \\ \therefore \, \, \mu =4.80^{\circ}\end{aligned}
Question 7 |
Consider the diode circuit shown below. The diode, D, obeys the current-voltage
characteristic I_D=I_S\left ( exp\left ( \frac{V_D}{nV_T} \right )-1 \right ), where n \gt 1, V_T \gt 0, V_D is the voltage across
the diode and I_D is the current through it. The circuit is biased so that voltage, V \gt 0 and current, I \lt 0. If you had to design this circuit to transfer maximum power from
the current source (I_1) to a resistive load (not shown) at the output, what values R_1 \; and \; R_2 would you choose?
Large R_1 and large R_2. | |
Small R_1 and small R_2. | |
Large R_1 and small R_2. | |
Small R_1 and large R_2. |
Question 7 Explanation:
R_{1}-low, R_{2}- high
V_{D}=V\times \frac{R_{2}}{R_{1}+R_{2}}
If R_{2} is large V_{D} becomes high
If R_{1} is less V_{D}=V
So for maximum power, R_{1} is small and R_{2} is large.
V_{D}=V\times \frac{R_{2}}{R_{1}+R_{2}}
If R_{2} is large V_{D} becomes high
If R_{1} is less V_{D}=V
So for maximum power, R_{1} is small and R_{2} is large.
Question 8 |
Thyristor T_1 is triggered at an angle \alpha (in degree), and T_2 at angle 180^{\circ} + \alpha, in each
cycle of the sinusoidal input voltage. Assume both thyristors to be ideal. To control the
load power over the range 0 to 2 kW, the minimum range of variation in \alpha is:
0^{\circ} \; to \; 60^{\circ} | |
0^{\circ} \; to \; 120^{\circ} | |
60^{\circ} \; to \; 120^{\circ} | |
60^{\circ} \; to \; 180^{\circ} |
Question 8 Explanation:
As per GATE official answer key, MTA (Marks to All)
As load is capacitive, for any \alpha :
\begin{aligned} V_{or}&=\frac{V_m}{\sqrt{2\pi}}\left [ \pi - \alpha +\frac{1}{2} \sin 2\alpha \right ]^{1/2}\\ P_0&=\left ( \frac{V_{or}}{z} \right )^2 \times R\\ &=\frac{V_{or}^2}{100} \times 5=\frac{V_{or}^2}{20}\\ \text{For }\alpha &=0^{\circ}\\ V_{or}&=\frac{V_m}{\sqrt{2}}=200\\ P_0&=\frac{200}{20}=2KW \end{aligned}
For \alpha =120^{\circ}, as current distinguishes, so P=0.
So, \alpha should be 0^{\circ} \text{ to } 120^{\circ}.
As load is capacitive, for any \alpha :
\begin{aligned} V_{or}&=\frac{V_m}{\sqrt{2\pi}}\left [ \pi - \alpha +\frac{1}{2} \sin 2\alpha \right ]^{1/2}\\ P_0&=\left ( \frac{V_{or}}{z} \right )^2 \times R\\ &=\frac{V_{or}^2}{100} \times 5=\frac{V_{or}^2}{20}\\ \text{For }\alpha &=0^{\circ}\\ V_{or}&=\frac{V_m}{\sqrt{2}}=200\\ P_0&=\frac{200}{20}=2KW \end{aligned}
For \alpha =120^{\circ}, as current distinguishes, so P=0.
So, \alpha should be 0^{\circ} \text{ to } 120^{\circ}.
Question 9 |
A single-phase, full-bridge diode rectifier fed from a 230 V, 50 Hz sinusoidal source
supplies a series combination of finite resistance. R, and a very large inductance, L,
The two most dominant frequency components in the source current are:
50 Hz, 0 Hz | |
50 Hz, 100 Hz | |
50 Hz, 150 Hz | |
50 Hz, 250 Hz |
Question 9 Explanation:
For full bridge rectifier, High inductive load
\begin{aligned} I_{source}&=\sum_{n=1,3,5}^{\infty }\frac{4I_0}{n \pi} \sin nd \sin \frac{n \pi}{2} \sin (n\omega t+\phi _n) \\ 2d &=\pi , \;\; d=\frac{\pi}{2}\\ I_{source}&= \frac{4I_0}{n \pi} \sum_{n=1,3,5}^{\infty } \sin ^2 n\left ( \frac{\pi}{2} \right ) \sin (n\omega t+\phi _n)\\ I_{source}&\neq 0 \end{aligned} For n=1,3,5,...
Since the most dominant frequency component will be f, 3f, f = 50
So dominant frequency = 50 Hz, 150 Hz.
\begin{aligned} I_{source}&=\sum_{n=1,3,5}^{\infty }\frac{4I_0}{n \pi} \sin nd \sin \frac{n \pi}{2} \sin (n\omega t+\phi _n) \\ 2d &=\pi , \;\; d=\frac{\pi}{2}\\ I_{source}&= \frac{4I_0}{n \pi} \sum_{n=1,3,5}^{\infty } \sin ^2 n\left ( \frac{\pi}{2} \right ) \sin (n\omega t+\phi _n)\\ I_{source}&\neq 0 \end{aligned} For n=1,3,5,...
Since the most dominant frequency component will be f, 3f, f = 50
So dominant frequency = 50 Hz, 150 Hz.
Question 10 |
A single-phase fully-controlled thyristor converter is used to obtain an average voltage of 180 V with 10 A constant current to feed a DC load. It is fed from single-phase AC supply of 230 V, 50 Hz. Neglect the source impedance. The power factor (round off to two decimal places) of AC mains is________
0.25 | |
0.55 | |
0.78 | |
0.95 |
Question 10 Explanation:
V_{sr}I_{sr} \cos \phi =V_0I_0
For single-phase fully controlled converter,
I_0=I_{sr}=10A
\cos \phi =\frac{V_0}{V_{sr}}=\frac{180}{230}=0.78
For single-phase fully controlled converter,
I_0=I_{sr}=10A
\cos \phi =\frac{V_0}{V_{sr}}=\frac{180}{230}=0.78
There are 10 questions to complete.