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# Multiple choice question for engineering

## Set 1

1. Identify the below given circuit. a) Three-phase bridge regulator
b) Three-phase bridge type semi-converter circuit
c) Three-phase bridge thyristor inverter
d) Three-phase bridge IGBT inverter

Answer: c [Reason:] It is a three-phase bridge type inverter. As SCRs are used as the switching device, it is called as a thyristor inverter.

2. A three-phase bridge inverter requires minimum of _____________ switching devices.
a) 3
b) 4
c) 6
d) 8

Answer: c [Reason:] It requires a min. 6 devices, 2 in each leg. Switching devices could be anything BJT, MOSFET or an IGBT. SCRs are used when very high power ratings are required.

3. The below given inverter circuit is a __ step inverter. a) 3
b) 2
c) 6
d) none of the mentioned

Answer: c [Reason:] The three-phase bridge type is a 6-step inverter. That means, the firing changes from one SCR to another 6 times per cycle.

4. In the three-phase bridge inverter, each step consists of
a) 30°
b) 60°
c) 90°
d) will depend on the value of the firing angle

Answer: b [Reason:] It is a 6 step inverter. Hence, 360/6 = 60°. This means that the SCRs are gated every 60° in proper sequence.

5. In inverters, to make the supply voltage constant
a) an inductor is placed in series with the load
b) capacitor is connected in parallel to the load side
c) capacitor is connected in parallel to the supply side
d) none of the mentioned

Answer: c [Reason:] A large C connected across the input terminal keep the supply voltage from altering.

6. In the 180° mode VSI, ___________ devices conduct at a time.
a) 5
b) 2
c) 3
d) 4

Answer: c [Reason:] Three devices conduct at a time. One from the upper pair and two from the lower pair or vice-versa.

7. In the figure given below, for 180° mode of operation if T1 is fired at 0°. Then SCRs T3 and T5 should be fired at _________ and _________ respectively. a) 180°, 360°
b) 90°, 180°
c) 120°, 240°
d) none of the mentioned

Answer: c [Reason:] T1-T4 form the first pair. T3-T6 form the second pair, and like-wise. For the 180° mode, each SCR conducts for 180°, but the groups of SCRs lag the prior group by an angle of 120°. e.g. If T1 is fired at 0 then T3 must be fired at an angle of 0 + 120° and T5 at 120 + 120 = 240°.

8. For a three phase bridge inverter in the 180° mode, ___________ devices are conducting from 120° to 180°. a) T1, T6, T5
b) T2, T6, T5
c) T1, T6, T5
d) T1, T2, T3

 Group I T1 T1 T1 T4 T4 T4 T1 T4 Group II T6 T6 T3 T3 T3 T6 T6 T6 Group III T5 T2 T2 T2 T5 T5 T5 T2 Step No I II III IV V VI I II

Each step consists of 60°. 120° to 180° will be step III.

9. _________ SCRs conduct from 300° to 360°. a) T1, T2, T3
b) T4, T5, T6
c) T4, T3, T2
d) T1, T6, T5

 Group I T1 T1 T1 T4 T4 T4 T1 T4 Group II T6 T6 T3 T3 T3 T6 T6 T6 Group III T5 T2 T2 T2 T5 T5 T5 T2 Step No I II III IV V VI I II

Each step consists of 60°. 300° to 360° will be step VI.

10. The diodes D4 and D1 will conduct from a) they will never conduct
b) 300° to 360°
c) 120° to 180°
d) insufficient information

Answer: d [Reason:] This will depend on the nature on the load, which is not mentioned in the above problem.

## Set 2

1. The conducting SCRs from 180 to 240 degrees would be a) 6, 1, 2
b) 2, 3, 4
c) 3, 4, 5
d) 5, 6, 1

Answer: b [Reason:] 180 to 240 degrees i.e. step IV.

2. What is the peak value of phase voltage in case of 3-phase VSI with 180° mode. The supply side consists of a constant dc voltage source of Vs.
a) Vs
b) 3Vs/2
c) 2Vs/3
d) 3Vs

Answer: c [Reason:] Apply KVL to the equivalent circuit of the inverter by closing and opening the proper switches. The phase voltage is a stepped sine-wave with peak value 2Vs/3.

3. What is the R phase voltage when T1, T6 and T5 are conducting? Consider a star connected R load. a) Vs
b) 2Vs/3
c) Vs/3
d) –Vs/3

Answer: c [Reason:] They conducting SCRs can be represented as closed switches. The load terminals R and B are connected to the positive bus but the terminal Y is connected to the negative bus. VRN = VBN = Vs/3. VYN = -2Vs/3.

4. What are the phase voltages when T6, T1 and T2 are conducting? Consider a star connected R load. a) VYN = VBN = Vs/3, VRN = 0
b) VYN = VRN = – Vs/3, VBN = 2Vs/3
c) VRN = VBN = – Vs/3, VYN = 2Vs/3
d) VYN = VBN = – Vs/3, VRN = 2Vs/3

Answer: c [Reason:] When 6, 1 and 2 are conducting and the others are off. The conducting SCRs can be represented as closed switches. The load terminals Y and B are connected to the negative bus but the terminal R is connected to the positive bus. Supply voltage is Vs. VYN = VBN = -Vs/3. VRN = 2Vs/3. It should be noted that at any time, the summation of these phase voltages must be zero.

5. What are the phase voltages when T1, T2 and T3 are conducting? Consider a star connected R load. a) VYN = VBN = VRN = 0
b) VYN = VRN = Vs/3, VBN = – 2Vs/3
c) VRN = VBN = – Vs/3, VYN = 2Vs/3
d) VYN = VBN = Vs/3, VRN = – 2Vs/3

Answer: b [Reason:] When 6, 1 and 2 are conducting and the others are off. They conducting SCRs can be represented as closed switches. The load terminals R and Y are connected to the positive bus and the terminal B is connected to the nrgative bus. Supply voltage is Vs. VRN = VYN = Vs/3. VBN = -2Vs/3. It should be noted that at any time, the summation of these phase voltages must be zero.

6. What is the maximum line voltage value in case of a three-phase VSI in 180° mode?
a) 2Vs
b) Vs
c) 3Vs
d) 2Vs/3

Answer: b [Reason:] The line voltage waveform has a peak value of Vs. Any line voltage value can be found by just adding the two phase voltage value. e.g. Vab = Vrn – Vbn Vab = (2Vs/3) – (-Vs/3) = Vs.

7. Find VYB, when T6, T1 and T2 are conducting. Consider a star connected R load. a) Vs
b) Vs/3
c) 0
d) 2Vs/3

Answer: c [Reason:] When 6, 1 and 2 are conducting and the others are off. They conducting SCRs can be represented as closed switches. The load terminals Y and B are connected to the negative bus but the terminal R is connected to the positive bus. Supply voltage is Vs. VYN = VBN = -Vs/3. VYB = VYN – VBN = (-Vs/3) – (-Vs/3) = -Vs/3 + Vs/3 = 0.

8. Find VRY, when T6, T1 and T2 are conducting. Consider a star connected R load. a) Vs
b) Vs/3
c) 0
d) 2Vs/3

Answer: a [Reason:] When 6, 1 and 2 are conducting and the others are off. They conducting SCRs can be represented as closed switches. The load terminals Y and B are connected to the negative bus but the terminal R is connected to the positive bus. Supply voltage is Vs. VYN = VBN = -Vs/3. VRN = 2Vs/3 VRY = VRN – VYN = (2Vs/3) – (-Vs/3) = 3Vs/3 = Vs.

9. What is the Y phase voltage, when T2, T3 and T4 are conducting? a) 0
b) Vs/3
c) 2Vs/3
d) -Vs/3

Answer: c [Reason:] Construct the equivalent circuit, considering the conducting devices as S.C and the non-conducting devices as open circuit. When 2, 3 and 4 are conducting and the others are off. They conducting SCRs can be represented as closed switches. The load terminals R and B are connected to the negative bus but the terminal Y is connected to the positive bus. Supply voltage is Vs. VRN = VBN = -Vs/3. VYN = 2Vs/3.

10. What is the voltage between the B phase and the neutral, when T2, T3 and T4 are conducting? a) -2Vs/3
b) Vs/3
c) 2Vs/3
d) -Vs/3

Answer: d [Reason:] Construct the equivalent circuit, considering the conducting devices as S.C and the non-conducting devices as open circuit. When 2, 3 and 4 are conducting and the others are off. They conducting SCRs can be represented as closed switches. The load terminals R and B are connected to the negative bus but the terminal Y is connected to the positive bus. Supply voltage is Vs. VRN = VBN = -Vs/3. VYN = 2Vs/3.

## Set 3

1. The 120° mode of operation of a three phase bridge inverter requires ___________ number of steps.
a) 2
b) 4
c) 6
d) 8

Answer: c [Reason:] Like the 180 mode, the 120° mode also requires six steps, each of 60° duration.

2. In case of the 120° mode of operation, __________ devices conduct at a time.
a) 2
b) 3
c) 4
d) none of the mentioned

Answer: a [Reason:] Unlike the 180° mode, in the 120° mode one 2 devices are conducting at a time as each conduct for 120°.

3. Safe commutation can be achieved in case of the ____________ operating mode.
a) 180°
b) 120°
c) 360°
d) none of the mentioned

Answer: b [Reason:] In the 120 mode of operation, there is a 60° interval between turning off of T1 and turning on of T4 (or any two SCRs belonging to the same leg), hence SCRs are commutated safely.

4. What is the R phase voltage when only T1 and T2 are conducting from 60° to 120°. Consider star connected R load. a) Vs
b) Vs/2
c) –Vs/2
d) 0

Answer: b [Reason:] As only 2 SCRs are conducting at a time, it is 120 mode. When T1 and T2 are conducting, R phase is connected to the positive bus through the T1 SCR and B phase is connected to the negative bus through the T2 SCR. Considering that the load is a balanced R load. VRN = Vs/2 VYN = -Vs/2.

5. If T1 is gated at 0 °, T3 and T5 will start conducting at _______ and _________ respectively.
a) 180°, 270°
b) 120°, 240°
c) 180°, 300°
d) 240°, 360°

Answer: b [Reason:] As T1 is gated at 0, T3 will be gated after 120° and T5 after 120+120 = 240°.

 Devices T1 T1 0 T4 T4 0 T1 T1 0 T6 0 T3 T3 0 T6 T6 0 T3 0 T2 T2 0 T5 T5 0 T2 T2

6. What is the B phase voltage when only T1 and T2 are conducting from 60° to 120°. Consider star connected R load. a) Vs
b) Vs/2
c) –Vs/2
d) 0

Answer: c [Reason:] When T1 and T2 are conducting, R phase is connected to the positive bus through the T1 SCR and B phase is connected to the negative bus through the T2 SCR. Considering that the load is a balanced R load.

 Devices T1 T1 0 T4 T4 0 T1 T1 0 T6 0 T3 T3 0 T6 T6 0 T3 0 T2 T2 0 T5 T5 0 T2 T2

VRN = Vs/2 VBN = -Vs/2.

7. What is the R phase voltage and Y phase voltage when only T3 and T4 are conducting? Consider a star connected balanced R load. a) Vs, -Vs
b) Vs/2, -Vs/2
c) –Vs/2, Vs/2
d) –Vs, Vs

Answer: c [Reason:] When T3 and T4 are conducting, R phase is connected to the negative bus through the T4 SCR and the Y phase is connected to the positive bus through the T3 SCR. Considering that the load is a balanced R load.

 Devices T1 T1 0 T4 T4 0 T1 T1 0 T6 0 T3 T3 0 T6 T6 0 T3 0 T2 T2 0 T5 T5 0 T2 T2

VYN = Vs/2 VRN = -Vs/2.

8. If the SCR T1 is gated at 0°, then for 120° mode of operation, from ωt = 240° to 300° __________ devices would conduct. a) T3, T4
b) T4, T5
c) T1, T6
d) T5, T6

 Devices T1 T1 0 T4 T4 0 T1 T1 0 T6 0 T3 T3 0 T6 T6 0 T3 0 T2 T2 0 T5 T5 0 T2 T2

ωt = 240° to 300°, would form the 5th step.

9. Find the line voltage VYR when only T3 and T4 are conducting? Consider a star connected balanced R load. a) 2Vs/3
b) Vs/2
c) Vs
d) –Vs

Answer: c [Reason:] When T3 and T4 are conducting, R phase is connected to the negative bus through the T4 SCR and the Y phase is connected to the positive bus through the T3 SCR. Considering that the load is a balanced R load. VYN = Vs/2 VRN = -Vs/2. VYR = VYN – VRN = Vs.

10. The peak value of the line voltage in case of 120° mode of operation of a three-phase bridge inverter is
a) Vs/2
b) 3Vs/2
c) Vs/√2
d) Vs

Answer: d [Reason:] The peak value for 120° mode is Vs. The line voltage waveform is a sine wave with a peak value of Vs ( = supply voltage).

## Set 4

1. AC voltage controllers convert
a) fixed ac to fixed dc
b) variable ac to variable dc
c) fixed ac to variable ac
d) variable ac to fixed ac

Answer: c [Reason:] Voltage controllers convert the fixed ac voltage to variable ac by changing the values of the firing angle.

2. In AC voltage controllers the
a) variable ac with fixed frequency is obtained
b) variable ac with variable frequency is obtained
c) variable dc with fixed frequency is obtained
d) variable dc with variable frequency is obtained

Answer: a [Reason:] Voltage controllers convert the fixed ac voltage to variable ac by changing the values of the firing angle. The available ac obtained has the same fixed frequency as the input ac.

3. Earlier then the semiconductor technology, ___________ devices were used for voltage control applications.
a) cycloconverters
b) vacuum tubes
c) tap changing transformer
d) induction machine

Answer: c [Reason:] A tap changing transformer can give variable ac from fixed ac without a change in frequency.

4. The AC voltage controllers are used in __________ applications.
a) power generation
b) electric heating
c) conveyor belt motion
d) power transmission

Answer: b [Reason:] In electric heating, variable ac supply is needed. The devices are fired appropriately to apply enough temperature.

5. In the principle of phase control
a) the load is on for some cycles and off for some cycles
b) control is achieved by adjusting the firing angle of the devices
c) control is achieved by adjusting the number of on off cycles
d) control cannot be achieved

Answer: b [Reason:] Switching devices is so operated that the load gets connected to ac source for a part of each half cycle.

6. A single-phase half wave voltage controller consists of
a) one SCR is parallel with one diode
b) one SCR is anti parallel with one diode
c) two SCRs in parallel
d) two SCRs in anti parallel

Answer: b [Reason:] As it is half wave, it consists of one SCR ( the control element) in anti parallel with one diode.

7. In the below given voltage controller circuit a) the positive half cycle at the load is same as the supply Vs
b) the negative half cycle at the load is same as the supply Vs
c) the positive and negative half cycles at the load are identical to the supply
d) none of the mentioned

Answer: b [Reason:] As T1 is triggered at an angle α, the conduction in the positive have cycle will start at α. In the negative half cycle, the diode is forward biased and load is connected as it is to the supply.

8. The below shown controller circuit is a a) half wave controller
b) full wave controller
c) none of the mentioned
d) will depend upon the firing angle

Answer: a [Reason:] As it consists one diode and one SCR only, the control is only in one cycle (the positive half in this case), hence it is a half wave controller.

9. The below given controller circuit is a a) half wave controller
b) full wave controller
c) none of the mentioned
d) will depend upon the firing angle

Answer: b [Reason:] As it consists of two SCRs, it is a full wave controller.

10. In the below given voltage controller circuit a) only the negative cycle can be controlled
b) only the positive cycle can be controlled
c) both the cycles can be controlled
d) none of the mentioned

Answer: c [Reason:] As it consists of two SCRs, it is a full wave controller and both the half cycles can be controlled by varying their respective firing angles.

## Set 5

1. In the below shown circuit, the diode conducts for a) 90°
b) > 90°
c) < 90°
d) 0°

Answer: a [Reason:] The diode conducts from π to 2π in the negative half cycle. If it were a non resistive load, the diode would conduct for less than 90°, as the inductor would force conduct the SCR for some time.

2. The SCR T1 is fired at an angle of α, and the supply Vs = Vm sinωt. Find the average value of the output voltage. a) (Vm/2π) (cosα + )
b) (Vm/2π) (cosα)
c) (Vm/2π) (cosα – 1)
d) Vm

Answer: c [Reason:] Vo = (1/2π) ∫Vm sinωt d(ωt). Where the integration would be run from α to 2π as the conduction takes place from α to 2π. Vo = (Vm/2π) (cosα – 1).

3. The below given output voltage waveform can be obtained by a a) half wave ac voltage controller
b) full wave ac voltage controller
c) half wave controller with firing angle = 0° for T1
d) full wave controller with firing angle = 0° for both T1 and T2

Answer: a [Reason:] As the positive half is chopped off due to some value of firing angle, the firing angle is not equal to zero. As the negative half is a half sine wave, either it is a full wave controller with firing angle for T2 set to zero or it is a half wave controller with a Thyristor (T1) and a diode.

4. In the below shown AC converter circuit with firing angle = α for both the devices, T2 will conduct from a) α to π
b) π + α to 2π
c) π to 2π
d) α to 2π

Answer: b [Reason:] T2 is triggered at π + α, it conducts from π + α to 2π after which it is line commutated.

5. The below given output voltage waveform can be obtained by a a) half wave ac voltage controller
b) full wave ac voltage controller
c) full wave inverter circuit
d) none of the mentioned

Answer: b [Reason:] As the control is in both the directions or cycles, it is a full wave ac voltage controller circuit.

6. In the integral cycle control method of ac voltage controller
a) the average power delivered to the load is controlled
b) the instantaneous power delivered to the load is controlled
c) the frequency of output voltage is controlled
d) none of the mentioned

Answer: a [Reason:] In integral cycle control, power is delivered for m cycles and not delivered for n cycles. Hence, the average value of the power delivered is controlled by manipulating m and n.

7. In the integral cycle control of ac voltage controller, is the load is on for n cycles and off for m cycles, then the periodicity is given by? Consider the output is sinusoidal.
a) m/2π(m+n)
b) n/2π(m+n)
c) m/π(m+n)
d) n/π(m+n)

Answer: b [Reason:] Over a complete cycle of 2π x (on cycles + off cycles) the power is delivered for n cycles. Hence, P = n/2π(m+n).

8. If k is the duty cycles of the controller, then the rms value of the output voltage in case of a integral cycle control circuit will be?
Consider the input to be sinusoidal with peak value Vm and rms value Vs.
a) Vs x k
b) Vs/k
c) Vs x √k
d) Vs

Answer: c [Reason:] Vrms = [ (k/2π) x ∫Vm2 sin⁡2 ωt d(ωt) ] 1/2 Where the integral runs from 0 to 2π. Vrms = (Vm/√2) k = Vs x √k.

9. Find the power delivered to the load in the integral cycle control method of ac voltage control, having a sine input of Vs, R load and duty cycle = k. a) Vs2/R
b) k.Vs2/R
c) √k .Vs2/R
d) 0

Answer: b [Reason:] Output voltage = Vs x √k P = (Vs x √k)2/R.

10. In the integral cycle control method with duty cycle = k and maximum load current = Im. Find the value of average SCR current. a) Im/k.π
b) Im
c) k.Im
d) k.Im/π