Test: Three Phase Rectifiers
20 Questions MCQ Test Power Electronics | Test: Three Phase Rectifiers
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In a three-phase half wave rectifier usually, the primary side of the transformer is delta connected because
The delta connected winding help circulating and eliminating the triplen (3rd order) harmonics.
Common anode arrangement because all the anodes are connected to the load side.
In a three-phase half wave diode rectifier using 3 diodes, each diode conducts for
Each diode conducts for 120 degrees, starting from ωt = 30 degrees.
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
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.
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
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.
In a three-phase half wave diode rectifier using 3 diodes,
3 diodes, each conduct for 120 degree at a time.
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
The voltage value is positive and varies from (1/2)Vmp to Vmp.
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
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.
In a three-phase half wave 6-pulse mid-point type diode rectifier, each diode conducts for
In a six-pulse rectifier, each diode conducts once every one cycle, 60° x
6 diodes = 360°.
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).
Vph = 1100/5 = 220 V (Transformer ratio = 5)
Vmp = √2 x 220 V
Imp = Vmp/R.
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.
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).
A 3-phase, 6-pulse rectifier consists of 6 diodes connected in 3 legs. Two diodes conduct at a time.
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)
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.
From the diode rectifier circuit shown below, with phase sequence R-Y-B, diodes D3 & D5 conduct when
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.
From the diode rectifier circuit shown below, with phase sequence R-Y-B, from ωt = 150° to 270°
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.
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°)
The conduction sequence always depends on the phase sequence, which diode is conducting will depend upon which phase voltage is active at that moment.
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
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.
A 3-phase bridge rectifier, has the average output voltage as 286.48 V. Find the maximum value of line voltage
Vo = 3Vml/π
Vml = (π x Vo)/3 = 300 V.
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.
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.
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).
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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