# Power Electronics Questions and Answers – Type E Choppers-2

This set of tough Power Electronics questions and answers focuses on “Type E Choppers-2”.

1. The below given circuit can

a) operate in all the four quadrants
b) operate in only the first and second quadrant
c) operate in only the fourth and third quadrant
d) operate in only the first and fourth quadrant

Explanation: For operation in the third and the fourth quadrant, the polarity of the load emf E should be reversed.

2. In the below shown chopper circuit, when CH3 is on

a) both load voltage and current are zero
b) both load voltage and current are positive
c) both load voltage and current are negative
d) none of the mentioned

Explanation: When CH3 is switched on, IIIrd quadrant operation is obtained in which both load current and load voltage are negative. D2-CH3-Vs.

3. In the below given type E chopper circuit, when CH3 and CH2 are switched on and then CH3 is switched off after some time, then

a) load current falls to zero
b) negative current freewheels through CH2, D4
c) positive current freewheels through CH2, D4
d) none of the mentioned

Explanation: When CH3 is first on then, load gets connected to Vs and charges the inductor L. When it CH3 switched off, current has to flows through through the same direction (negative) because of the L. Current starts to freewheel from CH2 and D4.

4. For the type E chopper to be operated in the fourth quadrant
a) only one switch is operated
b) two switches are operated
c) three switches are operated
d) all the switches are operated

Explanation: For IVth quadrant, only CH4 is operated.
The difference between “operated” and “on” should be noted. Operated means it can is literally operated i.e. it is switched on and off and on and off, whereas on means it is on continuously.

5. For the fourth quadrant operation, the current flows through

a) CH1, D2, L and D1
b) CH4, D3, L and D4
c) CH4, D2, L and E
d) CH1, D2, L and D3

Explanation: With CH4 on in the fourth quadrant, positive current flows through CH4, D2, L and E.
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6. When CH4 is turned on and then off

a) load current falls to zero
b) negative current freewheels through D2, D3
c) positive current freewheels through D2, D3
d) none of the mentioned

Explanation: When CH4 is on, positive current flows through CH4, D2, L and E. When CH4 is opened, current is fed back to the source from D2, D3.

7. Find the expression for the ripple factor in terms of duty cycle α for a type A step down chopper
a) √(1-α/α)
b) α2/2
c) √α/2
d) √(1-α)

Explanation: RF = Vr/Vo
Vr = √α x Vs = √(αVs2 – α2Vs2) = Vs √(α- α2)
RF = Vs.√(α- α2)/Vs.α = √(1-α/α).

8. The AC ripple voltage is given by
a) Vrms/Vo
b) Vrms2
c) √(Vrms2 – Vo2)
d) √(Vrms + Vo)

Explanation: The term AC ripple voltage is used for knowing the harmonic content of a waveform, without calculating its harmonic components. It is the rms difference of ac and dc voltage.

9. The expression for the thyristor (chopper switch) current is given by
I = (α2.Vs – αE)/R
Find the value of firing angle α, for which the thyristor current is maximum.
a) 2Vs/π
b) 180°
c) E/2Vs
d) Vs/R

Explanation: In the given expression, to find the maximum value of α
dI/dα = 0
Therefore, dI/dα = (2αVs – E)/R = 0
α = E/2Vs.

10. In a type E chopper, if all the four chopper switches are closed simultaneously then
b) supply is short circuited
c) both load and supply are shorted
d) none of the mentioned.

Explanation: Before anything, the supply will be shorted by CH1 and CH2 first.

11. A type A step down chopper has Vs = 220 V and is connected to RLE load. With R = 1 Ω, E = 24 V and L large enough to maintain continuous conduction. Find the average value of load current for a duty cycle of 30 %.
a) 100 A
b) 22 A
c) 42 A
d) 16.5 A

Explanation: I = (αVs – E)/R = 42 A.

12. A step down chopper has Vs = 220 V and is connected to RLE load. With R = 1 Ω, E = 24 V and L = 5 mH. The chopping period is 2000 μs and the on-period is 600 μs. Find the value of minimum steady state output current.
a) 33 A
b) 42 A
c) 0 A
d) 51 A

Explanation:
I(min) = (Vs/R) x [ (1 – e-m)/(1 – e-n) ] – (E/R)
m = Ton / (L/R) = R.Ton/L = 0.12
n = T/(L/R) = R.T/L
T = 2000μs
n = 0.4
I (min) = (220/1) x [ (1 – e-0.12)/(1 – e-0.4) ] – (24/1) = 51.46 A.

Sanfoundry Global Education & Learning Series – Power Electronics.

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