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Test: Power Electronics- 3 - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test GATE Electrical Engineering (EE) Mock Test Series 2025 - Test: Power Electronics- 3

Test: Power Electronics- 3 for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) Mock Test Series 2025 preparation. The Test: Power Electronics- 3 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Power Electronics- 3 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Power Electronics- 3 below.
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Test: Power Electronics- 3 - Question 1

In a type A chopper, the maximum value of Ripple current is

Detailed Solution for Test: Power Electronics- 3 - Question 1

Concept:

Steady state ripple

The peak to peak ripple current has maximum value when duty cycle ∝ = 0.5

In case 4FL ≫ R, then tan 
Calculation

∴ maximum value of Ripple current is inversely proportional to chopping frequency and the circuit Inductance.

*Answer can only contain numeric values
Test: Power Electronics- 3 - Question 2

A boost – regulator has an Input voltage 8V and the average output voltage of 24 V the duty cycle is _


Detailed Solution for Test: Power Electronics- 3 - Question 2

We known Boost regulator is a step – up chopper

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*Answer can only contain numeric values
Test: Power Electronics- 3 - Question 3

A DC chopper operates on 220 V dc and frequency of 500 Hz, feeds on RL load of the output voltage of chopper is 180 V the ON time of the chopper is –(in msec)


Detailed Solution for Test: Power Electronics- 3 - Question 3

Vdc = 220V, f = 500 Hz, Vo = 180 V

(2×10−3) = 1.636 msec

Test: Power Electronics- 3 - Question 4

DC chopper feed an RLE load. If the value of E increased by 20%, the current ripple

Detailed Solution for Test: Power Electronics- 3 - Question 4

Current ripple does not depend on the value of E
ΔI= Vdc ∝ (1−∝)T

Test: Power Electronics- 3 - Question 5

A class C chopper is operated from 230 V battery the load is a dc motor with R = 0.15 Ω, L = 12 mH and Eb = 120 V. The duty cycle of the chopper to achieve regenerative braking at the rated current of 12 ampere would be equal to

Detailed Solution for Test: Power Electronics- 3 - Question 5

First find rated load current

For Regeneration braking load current should be negative.

-1.8 = ∝230 – 120

230∝ = 118.2

∝ = 0.5139

∴ Duty cycle of chopper = 51.39 %

*Answer can only contain numeric values
Test: Power Electronics- 3 - Question 6

A single - quadrant chopper is operating with the following specification: Ideal battery of 240 V; on time (Ton) = 1.5 ms, OFF time = 2 ms. The Form factor factor will be –


Detailed Solution for Test: Power Electronics- 3 - Question 6

Concept:

For Single – quadrant (type A) Chopper

Average value Vo = ∝ Vs
R.M.S value Vrms
where ∝ = Ton/T
We known the form factor F.F = 
 

Calculation:

Form factor = 

∴ form factor = 1.528

*Answer can only contain numeric values
Test: Power Electronics- 3 - Question 7

A class C chopper is operated from a 220 V battery. The load is a dc motor with R = 0.1 Ω, L = 10 mH and E= 100 V. The rated current is 10 A. The duty cycle in motoring mode is


Detailed Solution for Test: Power Electronics- 3 - Question 7

Test: Power Electronics- 3 - Question 8

A buck converter is to be designed where Vs = 10V, Vo = 5V,R = 500Ω. The switching frequency is decided to be 20 kHz. Allowable current ripple is 5 mA. The optimum value of inductor is

Detailed Solution for Test: Power Electronics- 3 - Question 8

 

*Answer can only contain numeric values
Test: Power Electronics- 3 - Question 9

A step-up chopper shown in figure is to deliver 3A
In to the 10 Ω load. The battery voltage is 12 V, L = 20 μH, C = 100 μF and chopper frequency is 50 kHz. Determine the battery current variation in ampere


Detailed Solution for Test: Power Electronics- 3 - Question 9

Calculation:

Vo = IoRL = 3 × 10 = 30 V

Vb = Vo (1 – D)
1−D = 12/30 = 0.4
D = 0.6

Periodic time 

On time = DT = 0.6 × 20 × 10-6 = 12 μs 

over an on-time the change in battery current will be



Concept:


When the transistor is switched on, current ramps up in the inductor and energy is stored.

When the transistor is switched off, the inductor voltage reverses and acts together with the battery voltage to forward bias the diode, transferring energy to the capacitor.

When the transistor is switched on again, load current is maintained by the capacitor, energy is stored in the inductor and the cycle can start again.

The value of the load voltage is increased by increasing the duty cycle or the on-time of the transistor.

During the transistor on-period, assuming an ideal inductor and transistor

Vb = VL = Ldi/dt

Vb/s = (sL) i(s)

i(s) = Vb/s2L

i(t) = (Vb/L)t = kt

Over an on-time of TON, the change of battery current will be

Δi = k TON

During the off-period

V= Vb + V= Vb + L(di/dt) = Vb + L(Δi/Δt)

= Vb + L(Vb/L)TON /Toff = Vb(1 + TON /Toff)

Let the duty cycle

D = TON T, and Toff = T – TON = T(1 - D). Now

Vo = Vb (1 – DT/T(1 – D))

which simplifies to

Vo = Vb/(1 - D)

In an ideal circuit, VbI= V0I0. Therefore, from eq.

Ib = (Vo/Vb)Ib = Io/(1 – D)

Ib = Ib + Δi/2

Ibo = Ib – Δi/2

When the steady variation of battery current has been reached, the variation will be between lo and I1, as shown in.

Test: Power Electronics- 3 - Question 10

In the circuit shown below, if load R = 500 Ω, switching frequency is 25 kHz and peak to peak ripple current of inductor is limited to 0.9 A then the filter inductance L is"

Detailed Solution for Test: Power Electronics- 3 - Question 10

For Buck Converter
V= αVdc ⇒ α = 5/12
The peak to peak ripple current is

L = 129.62 μH

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