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QUESTION: 1

In a three-phase half wave rectifier usually, the primary side of the transformer is delta connected because

Solution:

The delta connected winding help circulating and eliminating the triplen (3rd order) harmonics.

QUESTION: 2

The diode rectifier circuit given below is that of a

Solution:

Common anode arrangement because all the anodes are connected to the load side.

QUESTION: 3

In a three-phase half wave diode rectifier using 3 diodes, each diode conducts for

Solution:

Each diode conducts for 120 degrees, starting from ωt = 30 degrees.

QUESTION: 4

In the below shown diode rectifier circuit,

The diodes D1, D2 & D3 are connected to phases R,Y and B respectively as shown

The phase sequence is R-Y-B.

The diode D1 would conduct from

Solution:

It conducts from 30 to 150, for 90 degrees. D1 starts conducting first as it will be the most positive as it is connected to the R phase.

QUESTION: 5

In the below shown diode rectifier circuit,

The diodes D1, D2 & D3 are connected to phases R,Y and B receptively as shown

The phase sequence is R-Y-B.

The diode D3 conducts from

Solution:

It conducts from 270 to 450, for 120 degrees. D1 starts conducting first (from 30 degrees) as it will be the most positive as it is connected to the R phase & likewise.

QUESTION: 6

In a three-phase half wave diode rectifier using 3 diodes,

Solution:

3 diodes, each conduct for 120 degree at a time.

QUESTION: 7

In a three-phase half wave diode rectifier, if Vmp is the maximum phase voltage, then the output voltage on a R load varies from

Solution:

The voltage value is positive and varies from (1/2)Vmp to Vmp.

QUESTION: 8

The average value of the output voltage, in a 3-phase half wave diode rectifier with Vml as the maximum line voltage value, is given by the expression

Solution:

The average value can be obtained by

3 x [ 1/2π x Vml sin ωt d(ωt) ] The integration runs from π/6 to 5π/6 as the diode is conducting for 120 degrees each.

QUESTION: 9

In a three-phase half wave 6-pulse mid-point type diode rectifier, each diode conducts for

Solution:

In a six-pulse rectifier, each diode conducts once every one cycle, 60° x

6 diodes = 360°.

QUESTION: 10

A step-down delta-star transformer, with per-phase turns ration of 5, is fed from a 3-phase, 1100 V, 50 Hz source. The secondary of this transformer through a 3-pulse type rectifier feeds a R load of 10 Ω. Find the maximum value of the load current (phase).

Solution:

Vph = 1100/5 = 220 V (Transformer ratio = 5)

Vmp = √2 x 220 V

Imp = Vmp/R.

QUESTION: 11

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.

Solution:

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).

QUESTION: 12

The circuit shown below is that of a

Solution:

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

QUESTION: 13

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)

Solution:

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.

QUESTION: 14

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

Solution:

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.

QUESTION: 15

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

Solution:

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.

QUESTION: 16

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°)

Solution:

The conduction sequence always depends on the phase sequence, which diode is conducting will depend upon which phase voltage is active at that moment.

QUESTION: 17

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

Solution:

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.

QUESTION: 18

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

Solution:

Vo = 3Vml/π

Vml = (π x Vo)/3 = 300 V.

QUESTION: 19

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.

Solution:

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.

QUESTION: 20

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).

Solution:

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.

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