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Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE) PDF Download

Q1: 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. Lpar 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 ______.          (2020)
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)(a) 1
(b) 2
(c) 3
(d) 4
Ans:
(c)
Sol: Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Using KCL, Is = I− ID
For inductively loaded circuits, load can be assumed to be constant.
∴ Is is maximum when, ID is minimum, i.e. at point 3.
Therefore, IGBT experiences highest current stress at point 3.

Q2: For the circuit shown in the figure below, assume that diodes  D1, D2 and D3 are ideal.
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)The DC components of voltages v1 and v2, respectively are      (SET-1 (2017))
(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
Ans: 
(b)
Sol: Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)
Q3: 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: V1 = 500 kV, V2 = 485 kV and V3 = 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?     (SET-1 (2015))
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)(a) V= −500kV, V= −485kV and latex]I = 1.5kA[/latex]
(b) V= −485kV, V2 = 500kV and latex]I = 1.5kA[/latex]
(c) V1=500kV,V2=485kVV= 500kV, V2 = −485kV and latex]I = -1.5kA[/latex]
(d) V1 = −500kV, V2 = −485kV and latex]I = -1.5kA[/latex]
Ans:
(b)
Sol: To maintain the direction of power flow from system 2 to system 1, the voltage V1 = −485kV and voltage V2 = 500kV and  I = 1.5kA.
Since, current cannot flow in reverse direction. Option (B) is correct answer.

Q4: The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 μs is applied to the SCR. The maximum value of R in Ω to ensure successful firing of the SCR is ______.      (SET-2  (2014))
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)(a) 4050
(b) 5560
(c) 6060
(d) 8015
Ans:
(c)
Sol: Let us assume the SCR is conducting,
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)
Q5: A single-phase SCR based ac regulator is feeding power to a load consisting of 5 Ω 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    (SET-2 (2014))
(a) 30° and 46 A
(b) 30° and 23 A
(c) 45° and 23 A
(d) 45° and 32 A
Ans:
(c)
Sol: The maximum firing angle at which the voltage across the device becomes ′ϕ′ = load angle.
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Rms value of current through SCR is
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)
Q6: 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.
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)

The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance 10Ω.
The kVA rating of the input transformer is      (2011)
(a) 53.2 kVA
(b) 46.0 kVA
(c) 22.6 kVA
(d) 7.5 kVA
Ans:
(c)
Sol: RMS value of supply current in case of 3 − ϕ bridge converter
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)KVA rating of the input transformrer
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)
Q7: 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.

Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance10 Ω.
The maximum current through the battery will be        (2011)
(a) 14 A
(b) 40 A
(c) 80 A
(d) 94 A
Ans: 
(b)
Sol: Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Average output voltage of the converter
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)The converter acts as line commutated inverter and for such mode α > 90° and  V0 is negative.THerefore, battery supplies energy to AC system. So, current through battery
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Maximum current flow through battery
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)
Q8: The input voltage given to a converter is Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE) The current drawn by the converter is
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)The active power drawn by the converter is       (2011)
(a) 181 W
(b) 500 W
(c) 707 W
(d) 887 W
Ans:
(b)
Sol: Rms value of input voltag,
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Rms value of current,
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Let input power factor cosϕ
Vrms Irmscosϕ = active power drawn by the coverter  
⇒ 100 × 11.358 × cosϕ = 500W
⇒ cosϕ = 0.44

Q9: The input voltage given to a converter is Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE) The current drawn by the converter is
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE) The input power factor of the converter is       (2011)
(a) 0.31
(b) 0.44
(c) 0.5
(d) 0.71
Ans: 
(c)
Sol: Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Fundamental component of input voltage
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Fundamental component of current
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Phase difference between these two components
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Active power due to fundamental components
P1 = (Vi)1,rms × (ii)1,rms cosϕ = 100 × 10 × 0.5 = 500W
Since 3rd and 5th harmonic are absent in input voltage, there is no active power due to the these components.
Hence, active power drawn by the converter
P0 = Active power due to fundamental components = 500 W  

Q10: 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      (2010)
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)(a) step down chopper (buck converter)
(b) half-wave rectifier
(c) step-up chopper (boost converter)
(d) full-wave rectifier
Ans:
(a)
Sol: When switch is connected to A for time duration T1
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Vout = Vin
When switch is connected to B for time duration T2
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Average output voltage Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)
where, α = duty cycle Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)
Therefore, the converter shown is a step down chooper.

Q11: In the single phase voltage controller circuit shown in the figure, for what range of triggering angle (α), the input voltage (V0) is not controllable ?      (2008)
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)(a) 0° < α < 45°
(b) 45° < α < 135°
(c) 90° < α < 180°
(d) 135° < α < 180°
Ans:
(a)
Sol: Impedance of the load
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)The output voltage (V0) is controllable when α ≥ ϕL, So for 0° < α < 45°, output voltage is not controllable.

Q12: The triac circuit shown in figure controls the ac output power to the resistive load. The peak power dissipation in the load is     (2004)
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)(a) 3968 W
(b) 5290 W
(c) 7935 W
(d) 10580 W
Ans:
(d)
Sol: Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)Let Vm is the peak value of input voltage
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)As firing angle = α = (π/4), so peak voltage across resistance load is also Vm.  
Peak power dissipation in the field
Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE)

The document Previous Year Questions- Power Electronics Miscellaneous | Power Electronics - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Power Electronics.
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FAQs on Previous Year Questions- Power Electronics Miscellaneous - Power Electronics - Electrical Engineering (EE)

1. What are the basic components of a power electronics system?
Ans. A power electronics system typically consists of the following basic components: power semiconductor devices (such as diodes, transistors, and thyristors), passive components (like inductors and capacitors), control circuits, and a load. These components work together to convert and control electrical power efficiently.
2. How does a DC-DC converter work in power electronics?
Ans. A DC-DC converter is used to convert one level of direct current (DC) voltage to another. It operates by switching the input voltage on and off rapidly, using an inductor or transformer to store energy during the "on" phase and release it during the "off" phase. This process allows for step-up (boost) or step-down (buck) voltage conversion.
3. What is the role of control techniques in power electronics?
Ans. Control techniques in power electronics are crucial for regulating the output voltage and current, improving system performance, and ensuring stability. Common control methods include PWM (Pulse Width Modulation), P-I-D (Proportional-Integral-Derivative) control, and digital controllers, which help manage the operation of converters and inverters efficiently.
4. What are the different types of inverters used in power electronics?
Ans. The main types of inverters used in power electronics include: 1. Voltage Source Inverters (VSI) 2. Current Source Inverters (CSI) 3. Grid-tied Inverters 4. Standalone Inverters 5. Multilevel Inverters Each type has its specific applications and advantages depending on the requirements of the electrical system.
5. What are the applications of power electronics in renewable energy systems?
Ans. Power electronics play a vital role in renewable energy systems, such as solar and wind power. They are used in applications like maximum power point tracking (MPPT) for solar inverters, grid integration of renewable sources, energy storage systems (like battery inverters), and improving the efficiency of energy conversion and management in these systems.
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