Q21GATE 2018MCQ

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?

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Q22GATE 2017MCQ

A phase-controlled, single-phase, full-bridge converter is supplying a highly inductive DC load. The converter is fed from a 230 V, 50 Hz, AC source. The fundamental frequency in Hz of the voltage ripple on the DC side is

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📖 Explanation

All even harmonics are present. Fundamental frequency $=2f_s=2 \times 50=100 Hz$

Q23GATE 2017NAT

In the circuit shown in the figure, the diode used is ideal. The input power factor is _______. (Give the answer up to two decimal places.)

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📖 Explanation

\begin{aligned} PF&=\frac{V_{or}}{V_{sr}}=\frac{V_m/2}{V_m/\sqrt{2}} \\ &=\frac{1}{\sqrt{2}}=0.707 \end{aligned}

Q24GATE 2017MCQ

In the circuit shown, the diodes are ideal, the inductance is small, and $I_o \neq 0$ . Which one of the following statements is true?

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📖 Explanation

Both diodes will conduct for more than $180^{\circ}$ .

Q25GATE 2017NAT

The figure below shows the circuit diagram of a controlled rectifier supplied from a 230 V, 50 Hz, 1-phase voltage source and a 10:1 ideal transformer. Assume that all devices are ideal. The firing angles of the thyristors $T_1 \; and\; T_2$ are $90^{\circ} \; and\; 270^{\circ}$ , respectively. The RMS value of the current through diode $D_3$ in amperes is ____

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📖 Explanation

Diode current, $I_{D3(rms)}=0$ FD will not conduct for resistive load.

Q26GATE 2017NAT

The figure below shows an uncontrolled diode bridge rectifier supplied form a 220 V, 50 Hz 1-phase ac source. The load draws a constant current $I_o=14A$ . The conduction angle of the diode $D_1$ in degrees is___________.

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📖 Explanation

Average reduction in output voltage due to source inductance $=\Delta V_{d0}$

\begin{aligned} \Delta V_{d0}&=4fL_s I_0\\ &=4 \times 50 \times (10.10^{-3})\times 14\\ &=28V\\ \Delta V_{d0}&=\frac{V_m}{\pi}[( \cos \alpha - \cos (\alpha +\mu ))]\\ \alpha =0,\\ 28&=\frac{V_m}{\pi}[1-\cos \alpha ]\\ 0.282&=1- \cos \alpha \\ \cos \alpha &=0.717\\ \mu &=44.17^{\circ}\\ \therefore \; & \text{Conduction angle of diode}\\ &=180^{\circ}+44.17^{\circ}\\ &=224.17^{\circ} \end{aligned}

Q27GATE 2016NAT

A full-bridge converter supplying an RLE load is shown in figure. The firing angle of the bridge converter is 120 $^{\circ}$ . The supply voltage $v_{m}(t) = 200\pi sin (100 \pi t) V$ , $R=20 \Omega$ , E=800 V. The inductor L is large enough to make the output current $I_L$ a smooth dc current. Switches are lossless. The real power fed back to the source, in kW, is __________.

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📖 Explanation

\begin{aligned} V_0&=2\frac{V_m}{\pi}\cos \alpha \\ &=2\frac{200 \pi}{\pi} \cos 120^{\circ}\\ V_0&=-200V\\ |V_0|&=200V\\ &\text{Power balance equation}\\ EI_0&=I_0^2 R+V_0I_0\\ 800I_0&=I_0^2(20)+200I_0\\ I+0&=30A\\ &\text{Power fed to source}\\ &=V_0I_0\\ &=200 \times 30\\ &=6kW \end{aligned}

Q28GATE 2016NAT

A single-phase bi-directional voltage source converter (VSC) is shown in the figure below. All devices are ideal. It is used to charge a battery at 400 V with power of 5 kW from a source $V_s$ =220 V (rms), 50 Hz sinusoidal AC mains at unity p.f. If its AC side interfacing inductor is 5 mH and the switches are operated at 20 kHz, then the phase shift ( $\delta$ ) between AC mains voltage ( $V_s$ ) and fundamental AC rms VSC voltage ( $V_{c1}$ ), in degree, is _________.

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📖 Explanation

\begin{aligned} P&=V_sI_s \;p.f.\\ 5 \times 10^3 &= 220 \times I_s \times 1\\ I_s&=22.72A\\ \tan \delta &=\frac{I_s X_s}{V_s}\\ \delta &=\tan^{-1}\left ( \frac{I_sX_s}{V_s} \right )\\ \delta &=\tan^{-1}\left ( \frac{22.72 \times 2 \pi \times 50 \times 5 \times 10^{-3}}{220} \right )\\ \delta &=9.21^{\circ} \end{aligned}

Q29GATE 2016NAT

A three-phase diode bridge rectifier is feeding a constant DC current of 100 A to a highly inductive load. If three-phase, 415 V, 50 Hz AC source is supplying to this bridge rectifier then the rms value of the current in each diode, in ampere, is _____________.

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📖 Explanation

In the $3-\phi$ diode bridge rectifier each diode conducts for $120^{\circ}$ for one complete cycle.

\begin{aligned} I_{Drms}&=\sqrt{\frac{1}{2 \pi}\int_{0}^{2\pi/3} I_0^2 d\omega t} \\ &= I_0\sqrt{\frac{2 \pi}{2 \pi \times 3}}\\ &= \frac{I_0}{\sqrt{3}}=\frac{100}{\sqrt{3}}=57.7A \end{aligned}

Q30GATE 2016MCQ

A single-phase thyristor-bridge rectifier is fed from a 230 V, 50 Hz, single-phase AC mains. If it is delivering a constant DC current of 10 A, at firing angle of 30 $^{\circ}$ , then value of the power factor at AC mains is

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📖 Explanation

Input power factor = power factor at ac mains = $C.D.F. \times D.F.$ $=\frac{2\sqrt{2}}{\pi}\cos \alpha =\frac{2\sqrt{2}}{\pi}\cos 30^{\circ}=0.78$

Q31GATE 2015MCQ

In the following circuit, the input voltage $V_{\in}$ is 100sin(100 $\pi$ t). For 100 $\pi$ RC=50, the average voltage across R (in Volts) under steady-state is nearest to

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📖 Explanation

Given,

\begin{aligned} 100 \pi RC&=50\\ 100 \pi\cdot \frac{RC}{2}&=25\\ \omega \tau &=25 \end{aligned}

For one cycle $\omega \tau =2 \pi \; rad=6.28$ $[\omega \tau =25] \gt \gt [\omega \tau =2 \pi \; rad=6.28]$ Each capacitor voltage is approximately equal to

\begin{aligned} V_m&=100V \\ V_0&=2V_m\simeq 200V \end{aligned}

Q32GATE 2015MCQ

In the given rectifier, the delay angle of the thyristor $T_1$ measured from the positive going zero crossing of $V_s$ is 30 $^{\circ}$ . If the input voltage $V_s$ is 100sin(100 $\pi$ t) V, the average voltage across R (in Volt) under steady-state is _______.

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📖 Explanation

\begin{aligned} V_0&=\frac{1}{2 \pi}\left[ \int_{\alpha }^{\pi}V_m \sin \omega t\cdot d(\omega t)-\int_{\pi }^{2\pi}V_m \sin \omega t\cdot d(\omega t) \right]\$ \\ &=\frac{V_m}{2 \pi}[3+\cos \alpha ]\\ &=\frac{100}{2 \pi}\$[3+ \cos30^{\circ}]\$=61.53V \end{aligned}

Q33GATE 2014NAT

The figure shows the circuit of a rectifier fed from a 230 V (rms), 50-Hz sinusoidal voltage source. If we want to replace the current source with a resistor so that the rms value of the current supplied by the voltage source remains unchanged, the value of the resistance (in ohms) is _____. (Assume diodes to be ideal.)

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📖 Explanation

\begin{aligned} I_{0,rms} &=\frac{V_s}{R}\\ R&=\frac{230}{10}=23\Omega \end{aligned}

Q34GATE 2014NAT

The figure shows the circuit diagram of a rectifier. The load consists of a resistance 10 $\Omega$ and an inductance 0.05 H connected in series. Assuming ideal thyristor and ideal diode, the thyristor firing angle (in degree) needed to obtain an average load voltage of 70 V is ____.

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📖 Explanation

\begin{aligned} V_0&=\frac{V_m}{2 \pi}(1+ \cos \alpha )\\ \frac{70 \times 2 \pi}{325}&= \cos \alpha \\ \alpha &=\cos ^{-1}\left ( \frac{70 \times 2 \pi}{325}-1 \right )\\ \alpha &=69^{\circ}31' \end{aligned}

Q35GATE 2014NAT

A fully controlled converter bridge feeds a highly inductive load with ripple free load current. The input supply ( $V_s$ ) to the bridge is a sinusoidal source. Triggering angle of the bridge converter is $\alpha =30^{\circ}$ . The input power factor of the bridge is

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📖 Explanation

The input power factor (PF) is defined as

\begin{aligned} P.F.&=\frac{V_sI_{s1}}{V_sI_s}\cos \phi \\ &=\frac{I_{s1}}{I_s} \cos \phi \\ P.F.&=\frac{2\sqrt{2}}{\pi} \cos \alpha \\ &=\frac{2\sqrt{2}}{\pi} \cos 30^{\circ}\\ P.F.&=0.78 \text{( Lagging)} \end{aligned}

Q36GATE 2014NAT

A diode circuit feeds an ideal inductor as shown in the figure. Given $v_s=100 \sin (\omega t)$ V, where $\omega = 100\pi$ rad/s, and L = 31.83mH. The initial value of inductor current is zero. Switch S is closed at t = 2.5ms. The peak value of inductor current $i_L$ (in A) in the first cycle is _____.

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📖 Explanation

At t=2.5 ms the circuit is as shown below

\begin{aligned} -V_s+V_L&=0\\ V_S&=L\frac{di}{dt}\\ \frac{V_S}{L}dt&=di \end{aligned}

Integrating on both sides

\begin{aligned} \int \frac{V_m \sin \omega t d(\omega t)}{\omega L}&=\int di\\ i(t)&=\frac{V_m}{\omega L}[-\cos (\omega t)]+K \end{aligned}

when t=0.0025 sec, at this instant i(t)=0

\begin{aligned} 0&=\frac{100}{100 \pi \times 31.83 \times 10^{-3}}\\ &[-\cos (100 \pi \times 0.0025)]+K\\ &=-7.07+K\\ K&=7.07 \end{aligned}

The peak value of the inductor current will be at $(\omega t=\pi^c)$

\begin{aligned} i(t)_{max}&=\frac{V_m}{\omega L}[-\cos (\pi)]+K\\ I_{max}&=\frac{100}{100 \pi \times 31.83 \times 10^{-3}}+7.07\\ I_{max}&=17.07 \end{aligned}

Q37GATE 2014MCQ

A three-phase fully controlled bridge converter is fed through star-delta transformer as shown in the figure. The converter is operated at a firing angle of 30 $^{\circ}$ . Assuming the load current ( $I_0$ ) to be virtually constant at 1 p.u. and transformer to be an ideal one, the input phase current waveform is

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📖 Explanation

Let us assume all the impedance are Z and to flow $I_0$ only closed path is required. So, in three phases one phase is open circuited. According to correct division,

\begin{aligned} I_R&=\frac{I_0 2Z}{3Z}=\frac{2I_0}{3} \\ &= \frac{2}{3}p.u. \;\;\;[\because \; I_0=1\;p.u.]\\ I_{YB}&=\frac{I_0Z}{3Z}=\frac{I_0}{3}=\frac{1}{3}p.u. \end{aligned}

The transformer ratio is given as 1:K

\begin{aligned} \therefore \;\; I_{R(primary)}&=\frac{2K}{3}p.u. \\ I_{YB}&=\frac{K}{3}p.u. \end{aligned}

According to 3 phase source waveform,

Q38GATE 2013MCQ

Thyristor T in the figure below is initially off and is triggered with a single pulse of width 10 $\mu$ s. It is given that $L=(\frac{100}{\pi })\mu H$ and $C=(\frac{100}{\pi})\mu F$ Assuming latching and holding currents of the thyristor are both zero and the initial charge on C is zero, T conducts for

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📖 Explanation

Conduction time,

\begin{aligned} &=\pi \sqrt{LC} \\ &=\pi \sqrt{\left ( \frac{100}{\pi}\times 10^{-6} \right )\left ( \frac{100}{\pi}\times 10^{-6} \right )} \\ &= 100\mu s \end{aligned}

Q39GATE 2012MCQ

A half-controlled single-phase bridge rectifier is supplying an R-L load. It is operated at a firing angle $\alpha$ and the load current is continuous. The fraction of cycle that the freewheeling diode conducts is

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📖 Explanation

Freewheeling diode conducts for $2\alpha$ over complete cycle $\therefore \; \text{ Fraction}=\frac{2\alpha}{2 \pi} =\frac{\alpha}{\pi}$

Q40GATE 2010MCQ

The fully controlled thyristor converter in the figure is fed from a single-phase source. When the firing angle is 0 $^{\circ}$ , the dc output voltage of the converter is 300 V. What will be the output voltage for a firing angle of 60 $^{\circ}$ , assuming continuous conduction

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📖 Explanation

For firing angle, $\alpha$ , average output voltage of the converter is given by $V_0=\frac{2V_m}{\pi}\cos \alpha$ when, $\alpha =0^{\circ}$

\begin{aligned} V_0=\frac{2V_m}{\pi}\cos 0^{\circ}&=300 \\ \frac{2V_m}{\pi}&=300 \\ \text{when, } \alpha &= 60^{\circ}\\ V_0&=\frac{2V_m}{\pi} \cos 60^{\circ} \\ &= 300 \cos 60^{\circ}\\ &=150V \end{aligned}