# Power Electronics Miscellaneous

 Question 1
A double pulse measurement for an inductively loaded circuit controlled by the IGBT switch is carried out to evaluate the reverse recovery characteristics of the diode. D, represented approximately as a piecewise linear plot of current vs time at diode turn-off. $L_{par}$ is a parasitic inductance due to the wiring of the circuit, and is in series with the diode. The point on the plot (indicate your choice by entering 1. 2, 3 or 4) at which the IGBT experiences the highest current stress is ______.
 A 1 B 2 C 3 D 4
GATE EE 2020   Power Electronics
Question 1 Explanation:

Using KCL, $I_s=I_L-I_D$
$\therefore \; I_s$ is maximum when, $I_D$ is minimum, i.e. at point 3.
Therefore, IGBT experiences highest current stress at point 3.
 Question 2
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
 A 0 V and 1 V B -0.5 V and 0.5 V C 1 V and 0.5 V D 1 V and 1 V
GATE EE 2017-SET-1   Power Electronics
Question 2 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}
 Question 3
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?
 A $V_{1}=-500 kV,V_{2}=-485 kV$ and latex]I=1.5kA[/latex] B $V_{1}=-485 kV,V_{2}=500 kV$ and latex]I=1.5kA[/latex] C $V_{1}=500 kV,V_{2}=-485 kV$ and latex]I=-1.5kA[/latex] D $V_{1}=-500 kV,V_{2}=-485 kV$ and latex]I=-1.5kA[/latex]
GATE EE 2015-SET-1   Power Electronics
Question 3 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.
 Question 4
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 ______.
 A 4050 B 5560 C 6060 D 8015
GATE EE 2014-SET-2   Power Electronics
Question 4 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}

$V=I \times R$
$R=\frac{V}{I}=\frac{100}{16.5 \times 10^{-3}} =6060 \Omega$
 Question 5
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
 A 30$^{\circ}$ and 46 A B 30$^{\circ}$ and 23 A C 45$^{\circ}$ and 23 A D 45$^{\circ}$ and 32 A
GATE EE 2014-SET-2   Power Electronics
Question 5 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}
 Question 6
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
 A 53.2 kVA B 46.0 kVA C 22.6 kVA D 7.5 kVA
GATE EE 2011   Power Electronics
Question 6 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}
 Question 7
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 maximum current through the battery will be
 A 14A B 40A C 80A D 94A
GATE EE 2011   Power Electronics
Question 7 Explanation:

Average output voltage of the converter
$V_0=\frac{3 V_{ml}}{\pi} \cos \alpha$
The converter acts as line commutated inverter and for such mode $\alpha \gt 90^{\circ}$ and $V_0$ is negative.THerefore, battery supplies energy to AC system. So, current through battery
$I_0=\frac{400-V_0}{R}$
For $V_0=0 \text{ or }\alpha =90^{\circ}$,
Maximum current flow through battery
$(I_0)_{max}=\frac{400}{10}=40A$
 Question 8
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
 A 181W B 500W C 707W D 887W
GATE EE 2011   Power Electronics
Question 8 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
$\Rightarrow \;100 \times 11.358 \times \cos \phi =500W$
$\Rightarrow \; \cos \phi =0.44$
 Question 9
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 input power factor of the converter is
 A 0.31 B 0.44 C 0.5 D 0.71
GATE EE 2011   Power Electronics
Question 9 Explanation:
\begin{aligned} V_i&=100\sqrt{2}\sin (100 \pi t)\\ i_i&=10\sqrt{2} \sin \left ( 100 \pi t -\frac{\pi}{3} \right )\\ &+5\sqrt{2} \sin \left ( 300 \pi t +\frac{\pi}{4} \right )\\ &+2\sqrt{2} \sin \left ( 500 \pi t -\frac{\pi}{4} \right )A \end{aligned}
Fundamental component of input voltage
\begin{aligned} (V_i)_1&=100\sqrt{2} \sin (100 \pi t)\\ (V_i)_{1,rms}&=\frac{100\sqrt{2}}{\sqrt{2}}=100V \end{aligned}
Fundamental component of current
\begin{aligned} (i_L)_1&=10\sqrt{2} \sin (100 \pi t-\frac{\pi}{3})\\ (i_L)_{1,rms}&=\frac{10\sqrt{2}}{\sqrt{2}}=10 \end{aligned}
Phase difference between these two components
$\phi _1=\frac{\pi}{3}, \cos \phi _1= \cos \frac{\pi}{3}=0.5$
Active power due to fundamental components
$P_1=(V_i)_{1,rms} \times (i_i)_{1,rms} \cos \phi =100 \times 10 \times 0.5=500W$
Since $3^{rd} \text{ and } 5^{th}$ harmonic are absent in input voltage, there is no active power due to the these components.
Hence, active power drawn by the converter
$P_0=$ Active power due to fundamental components =500 W
 Question 10
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
 A step down chopper (buck converter) B half-wave rectifier C step-up chopper (boost converter) D full-wave rectifier
GATE EE 2010   Power Electronics
Question 10 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.
There are 10 questions to complete.