# Multiple choice question for engineering

## Set 1

1. A step-down delta-star transformer, with per-phase turns ratio of 5 is fed from a 3-phase 1100 V, 50 Hz source. The secondary of this transformer is connected through a 3-pulse type rectifier, which is feeding feeding an R load. Find the average value of output voltage.

a) 220 V

b) 257 V

c) 1100/√3 V

d) 206 V

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2. The circuit shown below is that of a

a) 3-phase, 6-pulse, diode rectifier

b) 3-phase, 6-pulse, diode inverter

c) 3-phase, 3-pulse, diode rectifier

d) 3-phase, 3-pulse, diode inverter

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3. A step-down delta-star transformer, with per-phase turns ratio of 5 is fed from a 3-phase 1100 V source. The secondary of this transformer is connected through a 3-pulse type rectifier, which is feeding an R load.

The power delivered to the load is 6839.3 Watts.

The maximum value of the load current is √2 x 22 A.

Fin, the rms value of output voltage Vo (rms)

a) 257.3 V

b) 220 V

c) 261.52 V

d) 248.32 V

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^{2}/R (i) Imp = Vmp/R Therefore, R = Vmp/Imp = (1100 x √2)/(5 x √2 x 22) = 10 Ω Put R in equation (i) & find the required R.M.S voltage.

4. From the diode rectifier circuit shown below, with phase sequence R-Y-B, diodes D3 & D5 conduct when

a) R is the most positive & B is the most negative

b) R is the most positive & Y is the most negative

c) R is the most negative & B is the most positive

d) R is the most negative & Y is the most positive

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5. From the diode rectifier circuit shown below, with phase sequence R-Y-B, from ωt = 150° to 270°

a) D1

b) D2

c) D3

d) None of the diodes conduct

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6. A 3-phase 6-pulse diode rectifier is shown below with phase sequence R-Y-B. The negative group of diodes (D4, D5, D6) conduct in sequence (from ωt = 0°)

a) D4-D5-D6

b) D5-D6-D4

c) D6-D5-D4

d) D6-D4-D5

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7. For a 3-phase 6-pulse diode rectifier, has Vml as the maximum line voltage value on R load. The peak current through each diode is

a) Vml/2R

b) 2Vml/R

c) Vml/R

d) Insufficient Data

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8. A 3-phase bridge rectifier, has the average output voltage as 286.48 V. Find the maximum value of line voltage

a) 100 V

b) 200 V

c) 300 V

d) 400 V

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9. A 3-phase bridge rectifier charges a 240 V battery. The rectifier is given a 3-phase, 230 V supply. The current limiting resistance in series with the battery is of 8 Ω.

Find the average value of battery charging current.

a) 12.56 A

b) 8.82 A

c) 9.69 A

d) 6.54 A

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10. A 3-phase bridge rectifier charges a 240-V battery. The rectifier is given a 3-phase 230 V supply. The current limiting resistance in series with the battery is 8 Ω.

Find the power delivered to the battery (Pdc).

a) Pdc = 2000 W

b) Pdc = 1226 W

c) Pdc = 2356 W

d) Pdc = 2116 W

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## Set 2

1. A single-phase full wave mid-point type diode rectifier requires __________ number of diodes whereas bridge type requires _________

a) 1,2

b) 2,4

c) 4,8

d) 3,2

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2. A single-phase full wave rectifier is a

a) single pulse rectifier

b) multiple pulse rectifier

c) two pulse rectifier

d) three pulse rectifier

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3. The below shown circuit is that of a

a) full wave B-2 type connection

b) full wave M-2 type connection

c) half wave B-2 type connection

d) half wave M-2 type connection

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4. In a 1-phase full wave bridge rectifier with M-2 type of connection has secondary side voltage Vs = Vm sin ωt, with R load & ideal diodes.

The expression for the average value of the output voltage can be given by

a) 2Vm/π

b) Vm/π

c) Vm/√2

d) 2Vm/√2

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^{π}Vm sinωt d(ωt)

5. The below shown circuit is that of a

a) full wave B-2 type connection

b) full wave M-2 type connection

c) half wave B-2 type connection

d) half wave B-2 type connection

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6. In a 1-phase full wave bridge rectifier with M-2 type of connection has secondary side voltage Vs = Vm sin ωt,

with R load & ideal diodes.

The expression for the rms value of the output voltage can be given by

a) Vm/π

b) Vm/√2

c) Vm

d) Vm^{2}

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^{π}Vm

^{2}sin

^{2}ωt d(ωt) = Vm/√2.

7. For the circuit shown below, find the power delivered to the R load

Where,

Vs = 230V

Vs is the secondary side single winding rms voltage.

R = 1KΩ

a) 46 W

b) 52.9 W

c) 67.2 W

d) 69 W

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^{2}/R Where, for the given circuit, V(rms) = Vs.

8. The PIV experienced by the diodes in the mid-point type configuration is

a) Vm

b) 2Vm

c) 4Vm

d) Vm/2

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9. For the circuit shown below, find the value of the average output current.

Where,

Vs = 230V

R = 1KΩ

Vs is the secondary side single winding rms voltage.

a) 100mA

b) 107mA

c) 200mA

d) 207mA

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10. In the circuit, let Im be the peak value of the sinusoidal source current. The average value of the diode current for the below given configuration is

a) Im

b) Im/2

c) Im/π

d) Im/√2

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## Set 3

1. The PIV experienced by each of the diodes for the below shown rectifier configuration is

a) Vm

b) 2Vm

c) 3Vm

d) Vm/2

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2. For the circuit shown in the figure below,

Vs = 230 V

R = 10Ω

Find the average value of output current.

a) 207.04 A

b) 20.704 A

c) 2.0704 A

d) 207.04 mA

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3. Choose the correct statement regarding the below given circuit.

a) The load current is never negative

b) The load current is never zero

c) The load current is never positive

d) The load voltage is never negative

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4. For a single phase, full bridge, diode rectifier excited from a 230 V, 50 Hz source. With R = 10 Ω & the inductance(L) large enough to maintain continues conduction, the average and rms values of diode currents will be

a) 7.85 A, 8 A

b) 10.35 A, 7.85 A

c) 10.35 A, 14.6 A

d) 8 A, 8 A

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5. The circuit shown below, will have the output voltage waveform similar to that of a

a) half wave rectifier with RL load

b) full wave bridge rectifier with RL load

c) full wave bridge rectifier with R load

d) full wave bridge rectifier with RC load

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6. For the circuit shown below, the load current attains the maximum value at ωt =

a) 0

b) π

c) 2π

d) none of the mentioned

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7. For a single phase, full bridge, diode rectifier excited from a 230 V, 50 Hz source. With R = 10 Ω & the inductance(L) large enough to maintain continuous conduction, the value of the supply power factor will be

a) 0.707 lag

b) 0.9 lag

c) 0.86 lag

d) Unity

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8. The rectification efficiency for B-2 type & M-2 type full wave diode rectifiers are ___ & ___ respectively.

a) 8/π & 4/π

b) 4/π & 8/π

c) 8/π & 8/π

d) 4/π & 4/π

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9. A load of R = 60 Ω is fed from 1phase, 230 V, 50 Hz supply through a step-up transformer & than a diode. The transformer turns ratio = 2. The power delivered to the load is

a) 614 Watts

b) 714 Watts

c) 814 Watts

d) 914 Watts

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^{2}/R Vo = Vm/π AC supplied to the rectifier is 2 x 230 = 460 V (rms) Therefore, Vo = √2 x 460 / π = 207.04 P = 714.43 W.

10. For the circuit shown below, D11 & D14 conduct from?

Assume that anode of D12 is positive at ωt = 0 and likewise.

a) 0 to π

b) π to 2π

c) 2π to 3π

d) 0 to π/2

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## Set 4

1. In the process of diode based rectification, the alternating input voltage is converted into

a) an uncontrolled alternating output voltage

b) an uncontrolled direct output voltage

c) a controlled alternating output voltage

d) a controlled direct output voltage

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2. In a half-wave rectifier, the

a) current & voltage both are bi-directional

b) current & voltage both are uni-directional

c) current is always uni-directional but the voltage can be bi-directional or uni-directional

d) current can be bi-directional or uni-directional but the voltage is always uni-directional

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3. For a certain diode based rectifier, the output voltage (average value) is given by the equation

1/2π [ ∫Vm sin ωt d(ωt) ]

Where the integral runs from 0 to π

The rectifier configuration must be that of a

a) single phase full wave with R load

b) single phase full wave with RL load

c) single phase half wave with R load

d) single phase half wave with RL load

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4. For a single phase half wave rectifier, with R load, the diode is reversed biased from ωt =

a) 0 to π, 2π to 2π/3

b) π to 2π, 2π/3 to 3π

c) π to 2π, 2π to 2π/3

d) 0 to π, π to 2π

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5. For the circuit shown below,

The secondary transformer voltage Vs is given by the expression

Vs = Vm sin ωt

Find the PIV of the diode.

a) √2

b) Vs

c) Vm

d) √2 Vm

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6. For the circuit shown below,

The peak value of the load current occurs at ωt = ?

a) 0

b) π

c) 2π

d) Data is insufficient

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7. Find the rms value of the output voltage for the circuit shown below.

Voltage across the secondary is given by Vm sinωt.

a) Vm

b) 2Vm

c) Vm/2

d) Vm^{2}/2

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^{π}Vm

^{2}sin

^{2}ωt. d(ωt) ] Solving above equation we get, Vor = Vm/2.

8. In a 1-Phase HW diode rectifier with R load, the average value of load current is given by

Take Input (Vs) = Vm sinωt

a) Vm/R

b) Vm/2R

c) Vm/πR

d) Zero

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^{π}Vm sinωt. d(ωt)] Vo = Vm/π I = Vo/R = Vm/πR.

9. In the circuit shown below,

The switch (shown in green) is closed at ωt = 0. The load current or capacitor current has the maximum value at ωt =

a) 0

b) π

c) 2π

d) none of the mentioned

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10. Find the average value of output current for a 1-phase HW diode rectifier with R load, having RMS output current = 100A.

a) 200R A

b) 100/R√2 A

c) 200/R√2 A

d) 200/Rπ A

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## Set 5

1. A 1-phase 230V, 1KW heater is connected across a 1-phase HW rectifier (diode based). The power delivered to the heater is

a) 300 W

b) 400 W

c) 500 W

d) 600 W

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^{2}/R = 500W.

2. A 1-phase half wave diode rectifier with R load, has input voltage of 240 V. The input power factor is

a) Unity

b) 0.707 lag

c) 0.56 lag

d) 0.865 lag

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3. A 1-phase half wave diode rectifier with R = 1 KΩ, has input voltage of 240 V. The diode peak current is

a) Zero

b) 240mA

c) 24mA

d) 0.24mA

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4. For the below given circuit, after the switch is closed the voltage across the load (shown open) remains constant.

Assuming that all initial conditions are zero. The element across the load would be a/an

a) resistor

b) capacitor

c) inductor

d) data not sufficient

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5. For the below given circuit,

After the supply voltage (Vs) is given the

a) diode starts conducting

b) diode starts conducting only when Vs exceeds Vdc

c) diode never conducts

d) diode stops conducting only when Vs exceeds Vdc

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6. For the below given circuit,

With Vs = Vm sin ωt (secondary side). The expression for the average voltage is

a) Vm

b) Vm/2π

c) Vm/π

d) Vm/2

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7. For the below given circuit, the

a) output voltage is never positive

b) output current is never positive

c) output current is never zero

d) output voltage is never zero

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8. For the below given circuit,

Vs = 325 sin ωt (secondary side)

The ripple voltage is

a) 125.32 V

b) 255.65 V

c) 325 V

d) 459.12 V

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^{2}+ Vavg

^{2}) Vrms = Vm/2 Vavg = Vm/π

9. For a single phase half wave rectifier, the rectifier efficiency is always constant & it is

a) 4/π^{2}

b) 8/π^{2}

c) 100

d) 2/π^{2}

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^{2}Pac = 4/(Vm x Im).

10. For the below given circuit,

The power delivered to the load in Watts is

a) I(avg).Vdc

b) I(avg).Vdc + I(rms)^{2}.R

c) I(avg).Vdc – I(rms)^{2}.R

d) I(avg).Vdc + I(avg)^{2}.R