Test: Single Phase Full wave mid point converters - Electrical Engineering (EE) MCQ

# Test: Single Phase Full wave mid point converters - Electrical Engineering (EE) MCQ

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## 10 Questions MCQ Test Power Electronics - Test: Single Phase Full wave mid point converters

Test: Single Phase Full wave mid point converters for Electrical Engineering (EE) 2024 is part of Power Electronics preparation. The Test: Single Phase Full wave mid point converters questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Single Phase Full wave mid point converters MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Single Phase Full wave mid point converters below.
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*Answer can only contain numeric values
Test: Single Phase Full wave mid point converters - Question 1

### SCRs with a peak forward rating of 2 kV and an average on-state current rating of 50 A are used in a single-phase mid-point converter. If the factor of safety is 2.5, the power that can be handled by this converter is ______kW

Detailed Solution for Test: Single Phase Full wave mid point converters - Question 1

The maximum voltage across SCRs in mod point connection is 2 Vm. So the maximum voltage of SCR = = 400 V

The maximum average power that the midpoint converter can handle is

= 12.732 kW

Test: Single Phase Full wave mid point converters - Question 2

### In a center tap full wave rectifier, if the peak voltage applied between the center tap and one of the secondary 200 V, then the reverse biased diode is applied with a maximum voltage of:

Detailed Solution for Test: Single Phase Full wave mid point converters - Question 2

Concept:

Peak inverse voltage: It is a voltage across the diode when it is reversed bias.

The following circuit is the center tapped full-wave rectifier.

Consider for positive half cycle, D1 is FB and D2 is reversed biased.

In loop 1, Apply KVL

⇒ V0 = Vin

Now, apply KVL in loop 2

Vin + V0 = VD2

⇒ VD2 = Vin + Vin = 2Vin

PIV of the centre tapped full wave rectifier = 2Vin

Calculation:

Given that, the center-tapped full-wave rectifier and Vin = 200 V (Peak) between the centre tap and one end of secondary.

Peak inverse voltage = 2Vin = 400 V

Test: Single Phase Full wave mid point converters - Question 3

### A full-wave rectifier uses 2 diodes. The internal resistance of each diode is 20 Ω. The transformer RMS secondary voltage from centre tap to each end of the secondary is 50 V and the load resistance is 980 Ω. Mean load current will be

Detailed Solution for Test: Single Phase Full wave mid point converters - Question 3

Concept:

Center tapped full wave rectifier:

• The Center tapped full-wave rectifier is a device used to convert the AC input voltage into DC voltage at the output terminals.
• It employs a transformer with the secondary winding tapped at the center point. And it uses only two diodes, which are connected to the opposite ends of a center-tapped transformer as shown in the figure below.
• The center tap is usually considered as the ground point or the zero voltage reference point.

Analysis: The DC output voltage or average output voltage can be calculated as follows,

V0 = 2Vm / π

Now we can calculate the average or mean current of load by dividing the average load voltage by load resistance RL. Therefore mean load current is given by

I0 = V0 / RL

If the internal resistance of the diode is given in that case mean load current I0 = V0 / (RL + r)

Where r = internal resistance of the diode.

Calculation:

Given that

Rms value of supply voltage V = 50 V

The internal resistance of diode r = 20 Ω

The load resistance RL = 980 Ω

Maximum voltage on the secondary side Vm = √2 V = √2 × 50 = 70.7 V

Average or DC output voltage V0 = (2 × 70.7) / π = 45 V

Average or mean load current is

I0 = V0 / (RL + r) = 45 /(980 + 20) = 45 mA

Test: Single Phase Full wave mid point converters - Question 4

In a two-diode full wave rectifier, with a load current requirement of 4.2 A, what should be the current ratings of the diodes used?

Detailed Solution for Test: Single Phase Full wave mid point converters - Question 4

Concept:

Center tapped full wave rectifier:

• The Center tapped full-wave rectifier is a device used to convert the AC input voltage into DC voltage at the output terminals.
• It employs a transformer with the secondary winding tapped at the center point. And it uses only two diodes, which are connected to the opposite ends of a center-tapped transformer as shown in the figure below.
• The center tap is usually considered as the ground point or the zero voltage reference point.

Analysis:

The DC output voltage or average output voltage can be calculated as follows,

V0 = 2Vm / π

Now we can calculate the average or mean current of load by dividing the average load voltage by load resistance RL. Therefore mean load current is given by

I0 = V0 / RL

Current through each diode will be half of the load current, as each diode conducts only for the half cycle.

Calculation:

Load current = I0 = 4.2 A

Current through each diode = I0 / 2 = 4.2 / 2 = 2.1 A

∴The current ratings of the each diode is 2.1 A.

Test: Single Phase Full wave mid point converters - Question 5

A single-phase mid-point full-wave SCR converter with maximum mid-point voltage of Vm volts develops an average output voltage across a resistive load at firing delay angles of 0 and π/2 rad., respectively, as

Detailed Solution for Test: Single Phase Full wave mid point converters - Question 5

A single-phase mid-point full-wave SCR converter:

In this circuit, two thyristors are used and are connected as shown below

Where,

Vs is the source voltage

Vm is the maximum voltage of the midpoint

Let α be the firing angle

V0 is the output voltage

When the thyristors are triggered with a firing angle 0 and π/2

Then the output waveform will be as follows

Let the average output voltage of the circuit be Va

The value of source voltage is Vm sin ωt

Calculation:

Given,

α1 = 0, α2 = π/2

For the first thyristor, the average output voltage is

Va = 2Vm / π

For the second thyristor, the average output voltage is

Va = Vm / π

*Answer can only contain numeric values
Test: Single Phase Full wave mid point converters - Question 6

In a 12 phase full wave rectifier if source frequency is 30 Hz, then the ripple frequency will be_____Hz

Detailed Solution for Test: Single Phase Full wave mid point converters - Question 6

We know that

Ripple frequency = 2nf = 2 × 12 × 30 = 720 Hz

*Answer can only contain numeric values
Test: Single Phase Full wave mid point converters - Question 7

A single-phase AC supply is connected to a full wave rectifier through a transformer. The rectifier is required to supply an average DC output voltage of Vdc = 400 V to a resistive load of R = 10 Ω. The kVA rating of the transformer is __________ (in kVA)

Detailed Solution for Test: Single Phase Full wave mid point converters - Question 7

Concept:

In a full wave rectifier though a transformer,

Average output voltage, V0 = 2Vm

Average output current, I0 = V0/R

RMS value of output voltage Vor = Vs

RMS value of load current, Ior = Vs/R

Average value of diode current, Id = Im/2

RMS value of diode current, Idr = Ior

Peak value of diode current, Idm = √2 Ior

Power delivered to the load = Vor Ior

Input voltamperes = Vs Ior

Calculation:

Given that, DC output voltage (VDC) = 400 V

Average output voltage of rectifier (V0) = 400 V

⇒ Vs = 444.28 V

Load resistance (R) = 10 Ω

RMS value of load current,  Ior = 444.2810 = 44.428A

kVA rating of transformer = 444.28 × 44.428 = 19.738 kVA

Test: Single Phase Full wave mid point converters - Question 8

A single-phase full-wave controlled rectifier, operating at 120 V rms and 60 Hz ac supply, has a firing angle of 60°. The average value of its output voltage is

Detailed Solution for Test: Single Phase Full wave mid point converters - Question 8

Concept:

The average output voltage of the single-phase full-wave rectifier with a resistive load is given by
V0 = 2Vm/π cos α

Where,

Vm = peak value of source voltage

α = firing angle

Calculation:

Given that:

Source voltage Vs = 120 V rms

firing angle α = 600

From the source voltage peak value Vm = √2 × 120 V

The average output value of a single-phase full-wave rectifier is given by

Test: Single Phase Full wave mid point converters - Question 9

A 1ϕ full wave converter with RLE load wit R = 10Ω, L = 8mH and E = 150V. The AC voltage source is 230V, 50Hz for continuous conduction. Find the average value of load current for firing angle delay of 120°.

Detailed Solution for Test: Single Phase Full wave mid point converters - Question 9

Concept:

The average output voltage of a 1ϕ full wave converter is given by:

Case 1: α ≤ 90°

Vo(avg) = IoR + E

This is a motoring mode of operation of a rectifier.

Case 2: α ≥ 90°

Vo(avg) = IoR - E

This is a generating mode of operation of a rectifier.

In generating mode, the polarity of battery voltage gets reversed.

Calculation:

Given, α = 120°

Io = 4.64 A

Test: Single Phase Full wave mid point converters - Question 10

An AC voltage of maximum value equal to 100V is applied to a single-phase fully controlled bridge circuit. The peak inverse voltage rating of each SCR used will be ______.

Detailed Solution for Test: Single Phase Full wave mid point converters - Question 10

Concept:

Single-phase fully controlled bridge circuit:

• The bridge rectifier circuit is made of four SCRs T1, T2, T3, T4, and a load resistor RL.
• The four SCRs are connected in a closed-loop configuration to effectively convert the alternating current (AC) into Direct current (DC).
• The input signed is applied across terminals A and B and output DC signal is obtained across the load resistor RL connected across the terminals C and D.
• The four SCRs are arranged in such a way that only two SCRs conduct electricity during each half cycle.
• SCRs T1 and T3 are pairs that conduct electric current during the positive half cycle.
• Similarly, SCRs T2 and T4 conduct electric current during a negative half cycle.
• During the positive half cycle, SCRs T1 and T3 become forward biased while SCRs T2 and T4 become reverse biased.
• During the negative half-cycle, SCRs T2 and T4 become forward biased while SCRs T1 and T3 become reverse biased.
• The current flow across load resistor RL is the same during positing half-cycles and the negative half cycles.
• The output DC signal polarity may be either completely positive or negative.
• The bridge rectifier allows electric current during both positive and negative half cycles of the input AC signal.
• Peak inverse voltage is the maximum voltage that an SCR can withstand in a reverse bias condition.
• During the positive half cycle, SCRs T1 and T3 are conducting.
• Similarly, during the negative half cycle, SCR T2 and T3 are in the conducting stage while SCRs T1 and T3 are in a non-conducting state.

Calculation:

Given that

Vm = 100 V (AC signal)

PIV of bridge Rectifier =  PIV of SCR T1 and T3 = Vm = 100 V

Similarly, PIV of SCR T2 and T2 is Vm = 100 V.

i.e. PIV of each SCR T1, T2, T3, T4 is Vm = 100 V

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