Q2GATE 2017MCQ

For the circuit shown in the figure below, assume that diodes $D_1,D_2$ and $D_3$ are ideal. The DC components of voltages $v_1 \; and \; v_2$ , respectively are

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📖 Explanation

\begin{aligned} V_{2\; avg}&=\frac{V_m}{\pi}=\frac{\pi/2}{\pi}=\frac{1}{2}=0.5V \\ V_{1\; avg} &=\frac{1}{2 \pi}\$[ \int_{0}^{\pi}\frac{\pi}{2}\sin 100 \pi t\cdot d(\omega t)\\ &+\int_{\pi}^{2\pi} \pi \sin 100 \pi t\cdot d(\omega t) ]\$\\ &= -0.5V \end{aligned}

Q3GATE 2015MCQ

Consider a HVDC link which uses thyristor based line-commutated converters as shown in the figure. For a power flow of 750 MW from System 1 to System 2, the voltages at the two ends, and the current, are given by: $V_{1}$ =500 kV, $V_{2}$ =485 kV and $V_{3}$ =1.5 kA. If the direction of power flow is to be reversed (that is, from System 2 to System 1) without changing the electrical connections, then which one of the following combinations is feasible?

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📖 Explanation

To maintain the direction of power flow from system 2 to system 1, the voltage $V_1=-485kV$ and voltage $V_2=500kV$ and $I=1.5kA$ . Since, current cannot flow in reverse direction. Option (B) is correct answer.

Q4GATE 2014NAT

The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 $\mu$ s is applied to the SCR. The maximum value of R in $\Omega$ to ensure successful firing of the SCR is ______.

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📖 Explanation

Let us assume the SCR is conducting,

\begin{aligned} &I_{ss}=\frac{100}{500}=0.2A\\ &[\because \text{inductor will be dhort circuited \in DC}]\\ &i(t)=I_{ss}(1-e^{-t/\tau })\\ &\tau =\frac{L}{R}=\frac{200 \times 10^{-3}}{500}\\ &\;\;=4 \times 10^{-4}\; sec\\ &\text{Given }t=50 \times 10^{-6}\; sec\\ &\therefore \; i(t)=0.2\left ( 1-e^{\frac{50 \times 10^{-6}}{4 \times 10^{-4}}} \right )=23.5A \end{aligned}

$V=I \times R$ $R=\frac{V}{I}=\frac{100}{16.5 \times 10^{-3}} =6060 \Omega$

Q5GATE 2014MCQ

A single-phase SCR based ac regulator is feeding power to a load consisting of 5 $\Omega$ resistance and 16 mH inductance. The input supply is 230 V, 50 Hz ac. The maximum firing angle at which the voltage across the device becomes zero all throughout and the rms value of current through SCR, under this operating condition, are

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📖 Explanation

The maximum firing angle at which the voltage across the device becomes $'\phi '$ = load angle.

\begin{aligned} \phi &= \tan ^{-1} \left ( \frac{\omega L}{R} \right )\\ &= \tan^{-1}\left ( \frac{2 \pi \times 50 \times 16 \times 10^{-3}}{5} \right ) \\ \phi &=45.15\simeq 45^{\circ} \end{aligned}

Rms value of current through SCR is

\begin{aligned} I_{T_{rms}} &=\sqrt{\left[ \frac{1}{2 \pi}\int_{\alpha }^{\pi+\alpha }\left ( \frac{V_m}{z}\sin (\omega t-\phi ) \right )^2 d(\omega t) \right]\$}\\ &=\sqrt{\frac{V_m^2}{2 \pi z^2} \int_{\alpha }^{\pi+\alpha } \left[ \frac{1-\cos 2(\omega t-\phi )}{2} \right] d(\omega t)}\\ &=\sqrt{\frac{V_m^2}{2 \times 2 \pi z^2} \left[\pi- \left.\begin{matrix} \frac{\sin 2(\omega t-\phi )}{2} \end{matrix}\right|_\alpha ^{(\pi+\alpha )} \right]\$ }\\ &=\frac{V_m}{2z \sqrt{\pi} } \sqrt{\left[ \pi+\frac{-\sin 2(\pi+\alpha -\alpha )+\sin 2(\alpha -\alpha )}{2} \right] }\\ &=\frac{V_m}{2z \sqrt{\pi} }\sqrt{\pi}=\frac{V_m}{2z}\\ I_{T_{rms}}&=\frac{230\sqrt{2}}{2 \times \sqrt{5^2(2 \pi \times 50 \times 16 \times 10^{-3})^2}}\\ &=22.93\simeq 23A\\ I_{T_{rms}}&=23A \end{aligned}

Q6GATE 2011MCQ

A solar energy installation utilize a three-phase bridge converter to feed energy into power system through a transformer of 400V/400 V, as shown below. The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance10 $\Omega$ . The kVA rating of the input transformer is

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📖 Explanation

RMS value of supply current in case of $3-\phi$ bridge converter $I_s=I_0\sqrt{\frac{2}{3}}=40\sqrt{\frac{2}{3}}=32.66A$ KVA rating of the input transformrer

\begin{aligned} &=\sqrt{3}V_sI_s \\ &= \sqrt{3} \times 400 \times 32.66 \times 10^{-3}\; kVA\\ &=22.62\; kVA \end{aligned}

Q7GATE 2011MCQ

The input voltage given to a converter is $v_i=100\sqrt{2}\sin (100 \pi t)$ V The current drawn by the converter is $i_i=10\sqrt{2}\sin (100\pi t-\pi/3)$ + $5\sqrt{2}\sin (300\pi t+\pi/4)$ + $2\sqrt{2}\sin (500\pi t-\pi/6)$ A The active power drawn by the converter is

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📖 Explanation

Rms value of input voltag, $V_{rms}=\frac{100\sqrt{2}}{\sqrt{2}}=100V$ Rms value of current, $I_{rms}=\sqrt{\left (\frac{10\sqrt{2}}{\sqrt{2}} \right )^{2+}\left (\frac{5\sqrt{2}}{\sqrt{2}} \right )^{2+}\left (\frac{2\sqrt{2}}{\sqrt{2}} \right )^2}=11.358A$ Let input power factor $\cos \phi$ $V_{rms}I_{rms} \cos \phi =$ active power drawn by the coverter $\Rightarrow \;100 \times 11.358 \times \cos \phi =500W$ $\Rightarrow \; \cos \phi =0.44$

Q8GATE 2010MCQ

The power electronic converter shown in the figure has a single-pole double-throw switch. The pole P of the switch is connected alternately to throws A and B. The converter shown is a

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📖 Explanation

When switch is connected to A for time duration $T_1$ $V_{out}=V_{\in}$ When switch is connected to B for time duration $T_2$ Average output voltage $=\frac{V_{\in}T_1}{T_1+T_2}=\alpha V_{\in}$ where, $\alpha =\text{duty cycle}=\frac{T_1}{T_1+T_2}$ Therefore, the converter shown is a step down chooper.

Q9GATE 2008MCQ

In the single phase voltage controller circuit shown in the figure, for what range of triggering angle ( $\alpha$ ), the input voltage ( $V_0$ ) is not controllable ?

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📖 Explanation

Impedance of the load

\begin{aligned} Z_L&= R_L+jX_L\\ &=50+j50 \Omega \\ &=50\sqrt{2}\angle 45^{\circ} \\ \phi _L&=45^{\circ} \end{aligned}

The output voltage ( $V_0$ ) is controllable when $\alpha \geq \phi _L$ , So for $0^{\circ} \lt \alpha \lt 45^{\circ}$ , output voltage is not controllable.

Q10GATE 2004MCQ

The triac circuit shown in figure controls the ac output power to the resistive load. The peak power dissipation in the load is

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📖 Explanation

Let $V_m$ is the peak value of input voltage $V_m=230\sqrt{2}V$ As firing angle $=\alpha =\frac{\pi}{4}$ , so peak voltage across resistance load is also $V_m$ . Peak power dissipation in the field $P=\frac{V_m^2}{R}=\frac{(230\sqrt{2})^2}{10}=10580W$

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