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

Answer

Answer: b [Reason:]
Vph = 1100/5 = 220 V (Transformer ratio = 5)
Vmp = √2 x 220 V
Vo = 3√3/2π x Vmp = (√2 x √3 x 3 x 220)/(2 x π).

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

Answer

Answer: a [Reason:] A 3-phase, 6-pulse rectifier consists of 6 diodes connected in 3 legs. Two diodes conduct at a time.

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

Answer

Answer: c [Reason:] Power delivered to the load (Pdc) = Vo(rms)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

Answer

Answer: b [Reason:] Which diode will conduct depends on where is it in connected? as in in which phase?. D3’s anode is connected to the R phase, hence it will turn on when R is the most positive.

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

Answer

Answer: b [Reason:] Construct the phase voltage waveforms on a graph. At 150 degree, D2 is forward biased while the other positive group diodes i.e. D2 and D3 remain reserved biased.

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

Answer

Answer: b [Reason:] The conduction sequence always depends on the phase sequence, which diode is conducting will depend upon which phase voltage is active at that moment.

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

Answer

Answer: c [Reason:] Two diodes conduct at a time, constructing the equivalent circuit with supply, R & replacing the conducting diodes by S.C & non-conducting as O.C, the required value can be found out.

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

Answer

Answer: c [Reason:] Vo = 3Vml/π
Vml = (π x Vo)/3 = 300 V.

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

Answer

Answer: b [Reason:] Vo = (3√2 x 230)/π = 310.56 V
Draw the battery charging circuit,
Vo = E + (Io x R)
Io = (Vo – E)/R = (310.56 – 240)/8.

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

Answer

Answer: d [Reason:] Vo = (3√2 x 230)/π = 310.56 V
Draw the battery charging circuit,
Vo = E + (Io x R)
Io = (Vo – E)/R = (310.56 – 240)/8 = 8.82A
Pdc = 240 x 8.82 = 2116 W.

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

Answer

Answer: b [Reason:] A bridge type requires 4 diodes which are connected in a bridge, and the mid-point has 2 diodes.

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

Answer

Answer: c [Reason:] It is a two-pulse rectifier as it generates 2 pulses per cycle.

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

Answer

Answer: b [Reason:] Full wave M-2 type employs a transformer and two diodes.

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

Answer

Answer: a [Reason:] The voltage waveform is a pulsating voltage with peak value Vm & symmetrical about π.
Vo = (1/π) ∫^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

Answer

Answer: a [Reason:] Full wave B-2 type uses 4 diodes in a bridge connection.

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) Vm²

Answer

Answer: b [Reason:] The voltage waveform is a pulsating voltage with peak value Vm & symmetrical about π.
Vo = (1/π) ∫^π Vm² sin²ωt 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

Answer

Answer: b [Reason:] Power delivered to the load is V(rms).I(rms) = V(rms)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

Answer

Answer: b [Reason:] In the m-2 type connection, each diode experiences a reverse voltage of 2Vm.

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

Answer

Answer: d [Reason:] I = Vo/R
Vo = 2Vm/π.

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

Answer

Answer: b [Reason:] The peak current through the diodes is same as that passing from the load. Each diode pair conducts for 180 degrees. Hence, 1/2 of a cycle. Average current = Im/2.

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

Answer

Answer: a [Reason:] When any diode is reversed biased due to the negative half cycle, the maximum peak value through it will be Vm.

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

Answer

Answer: b [Reason:] I = Vo/R
Vo = 2Vm/π.

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

Answer

Answer: a [Reason:] Due to the inductive nature of the load, the Diodes are force conducted & voltage goes negative but the current can never fall below zero.

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

Answer

Answer: c [Reason:]
Id(avg) = Io/2 = Vo/2R
Id(rms) = Io/√2 = Vo/R√2.

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

Answer

Answer: c [Reason:] The FD short circuits the load & voltage waveform is similar to that of a Full wave bridge rectifier with R load.

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

Answer

Answer: b [Reason:] As the load is RL, the load current will be maximum when the output voltage waveform falls to zero i.e. at π. At π the inductor is charged to its maximum value and starts delivering power to the source.

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

Answer

Answer: b [Reason:]
Pf = Vs.Is.cosθ/Vo.Io
Io = Vo/R A
Vo = 2Vm/π Volts.

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

Answer

Answer: c [Reason:] B-2 type has efficiency 8/π. M-2 type has efficiency half of that of a B-2 type.

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

Answer

Answer: b [Reason:] P = Vo²/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

Answer

Answer: b [Reason:] In the first cycle i.e. 0 to π, D12 and D13 conduct. In the next cycle i.e. π to 2π, D11 and D14 conduct.

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

Answer

Answer: b [Reason:] Rectification is AC to DC. In DIODE biased rectification, control is not possible.

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

Answer

Answer: c [Reason:] Current is always in one direction only, but voltage can be bi-directional in case of an L load.

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

Answer

Answer: c [Reason:] Integration is 0 to π from base period of 1/2π so it is a half wave R load.

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

Answer

Answer: b [Reason:] Diode will be reversed biased in the negative half cycles.

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

Answer

Answer: c [Reason:] PIV = √2 Vs = Vm.

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

Answer

Answer: b [Reason:] Due to the L nature, load current is maximum when the diode will be com-mutated i.e at π.

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) Vm²/2

Answer

Answer: c [Reason:] The above is a HW diode rectifier, the RMS o/p voltage equation is given by
Vor = √ [ (1/2π) ∫^π Vm²sin²ωt. 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

Answer

Answer: c [Reason:] Vo = √ [(1/2π) ∫^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

Answer

Answer: a [Reason:] The instant switch is closed the load current will be zero due to the nature of the capacitor.

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

Answer

Answer: d [Reason:] I(rms) = Vm/2R
Therfore, Vm = 200R
I(avg) = Vm/πR = 200R/πR.

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

Answer

Answer: c [Reason:] R = (230 x 230)/1000
V(rms) = (√2 x 230)/2
P = V(rms)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

Answer

Answer: c [Reason:] Input p.f = V(rms)/Vs
Vrms is the RMS value of output voltage. Vrms = (√2 x 230)/2
Vs = 230
pf = 0.707.

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

Answer

Answer: b [Reason:] Diode peak current = peak current through the load = Vo/R = Vm/2R.

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

Answer

Answer: b [Reason:] As the voltage remains constant as soon as the switch is closed, the element is most likely to be a capacitor.

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

Answer

Answer: b [Reason:] The diode will be forward biased only when Vs will be greater than Vdc.

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

Answer

Answer: c [Reason:] Due to the FD, we get 1st quadrant operation.
Where, output voltage (Avg) = 1/2π [ ∫Vm sin ωt d(ωt) ], integration runs from 0 to 180 degrees.

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

Answer

Answer: d [Reason:] The output voltage is the voltage across the resister and Vdc. Even if the current falls to zero, the output voltage will be equal to Vdc.

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

Answer

Answer: a [Reason:] Ripple voltage = √(Vrms² + Vavg²)
Vrms = Vm/2
Vavg = Vm/π

9. For a single phase half wave rectifier, the rectifier efficiency is always constant & it is
a) 4/π²
b) 8/π²
c) 100
d) 2/π²

Answer

Answer: a [Reason:]
Rectifier efficiency = Pdc/Pac
Pdc = (Vm x Im)/π²
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

Answer

Answer: b [Reason:] P = power delivered to the resister + power delivered to the emf source.