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# Test: 1P Diode Rectifiers FW

## 20 Questions MCQ Test Power Electronics | Test: 1P Diode Rectifiers FW

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Attempt Test: 1P Diode Rectifiers FW | 20 questions in 20 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Power Electronics for Electrical Engineering (EE) Exam | Download free PDF with solutions
QUESTION: 1

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

Solution:

A bridge type requires 4 diodes which are connected in a bridge, and the mid-point has 2 diodes.

QUESTION: 2

### A single-phase full wave rectifier is a

Solution:

It is a two-pulse rectifier as it generates 2 pulses per cycle.

QUESTION: 3

### The below shown circuit is that of a

Solution:

Full wave M-2 type employs a transformer and two diodes.

QUESTION: 4

A single-phase full wave rectifier is a

Solution:

It is a two-pulse rectifier as it generates 2 pulses per cycle.

QUESTION: 5

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

Solution:

The voltage waveform is a pulsating voltage with peak value Vm & symmetrical about π.
Vo = (1/π) ∫π Vm sinωt d(ωt)

QUESTION: 6

The below shown circuit is that of a​

Solution:

Full wave B-2 type uses 4 diodes in a bridge connection.

QUESTION: 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

Solution:

Power delivered to the load is V(rms).I(rms) = V(rms)2/R
Where, for the given circuit, V(rms) = Vs.

QUESTION: 8

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

Solution:

In the m-2 type connection, each diode experiences a reverse voltage of 2Vm.

QUESTION: 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.

Solution:

I = Vo/R
Vo = 2Vm/π.

QUESTION: 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

Solution:

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.

QUESTION: 11

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

Solution:

When any diode is reversed biased due to the negative half cycle, the maximum peak value through it will be Vm.

QUESTION: 12

For the circuit shown in the figure below,

Vs = 230 V
R = 10Ω
Find the average value of output current.

Solution:

I = Vo/R
Vo = 2Vm/π.

QUESTION: 13

Choose the correct statement regarding the below given circuit.

Solution:

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

QUESTION: 14

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

Solution:

Id(avg) = Io/2 = Vo/2R
Id(rms) = Io/√2 = Vo/R√2.

QUESTION: 15

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

Solution:

The FD short circuits the load & voltage waveform is similar to that of a Full wave bridge rectifier with R load.

QUESTION: 16

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

Solution:

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.

QUESTION: 17

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

Solution:

Pf = Vs.Is.cosθ/Vo.Io
Io = Vo/R A
Vo = 2Vm/π Volts.

QUESTION: 18

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

Solution:

B-2 type has efficiency 8/π. M-2 type has efficiency half of that of a B-2 type.

QUESTION: 19

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

Solution:

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.

QUESTION: 20

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

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

Solution:

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.

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