Three Phase Full Converters MCQ Quiz - Objective Question with Answer for Three Phase Full Converters - Download Free PDF
Last updated on Jun 11, 2025
Latest Three Phase Full Converters MCQ Objective Questions
Three Phase Full Converters Question 1:
A 750 Hp, 250 V, 1200 r/min DC motor is connected to a 208 V, 3 phase, 60 Hz, line using a 3 phase bridge converter. The full load armature current is 2500 A and the armature resistance is 4 m ohm. Calculate the required firing angle under rated full load conditions.
Answer (Detailed Solution Below)
Three Phase Full Converters Question 1 Detailed Solution
Calculation of the Required Firing Angle:
The given problem involves a 750 Hp, 250 V, 1200 r/min DC motor connected to a 208 V, 3-phase, 60 Hz line through a 3-phase bridge converter. We aim to calculate the required firing angle under rated full load conditions.
Given Data:
- Motor Power (P) = 750 Hp (1 Hp = 746 W, so P = 750 × 746 W = 559500 W)
- Armature Voltage (Va) = 250 V
- Armature Current (Ia) = 2500 A
- Armature Resistance (Ra) = 4 mΩ = 0.004 Ω
- Line Voltage (VL) = 208 V
- Frequency (f) = 60 Hz
- Motor Speed = 1200 r/min
Step-by-Step Solution:
1. Calculate the Back EMF (E):
The back EMF (E) can be calculated using the voltage equation of the motor:
Va = E + Ia × Ra
Rearranging for E:
E = Va - Ia × Ra
Substitute the values:
E = 250 - (2500 × 0.004) = 250 - 10 = 240 V
2. RMS Line Voltage of the Converter:
The RMS line voltage of the converter is given as VL = 208 V. The peak line voltage (Vm) can be calculated as:
Vm = √2 × VL
Substitute the value of VL:
Vm = √2 × 208 = 294.96 V
3. Average Output Voltage of the Converter:
The average output voltage (Vdc) of a 3-phase bridge converter is given by:
Vdc = (3√2 / π) × Vm × cos(α)
Here, α is the firing angle. Rearranging to solve for cos(α):
cos(α) = (Vdc × π) / (3√2 × Vm)
Since Vdc is the back EMF (E), we use E = 240 V:
cos(α) = (240 × π) / (3√2 × 294.96)
Substitute the values:
cos(α) = (240 × 3.1416) / (3 × 1.414 × 294.96)
cos(α) = 753.984 / 1250.89 = 0.6026
4. Calculate the Firing Angle (α):
Take the inverse cosine of 0.6026 to find α:
α = cos-1(0.6026)
α ≈ 52.75°
5. Adjust the Angle to the Closest Option:
The closest option to the calculated firing angle is 45°. Therefore, the correct option is:
Option 1: 45°
Additional Information
Analysis of Other Options:
Option 2 (208°): This option is far beyond the practical firing angle range for a 3-phase bridge converter. Firing angles typically range between 0° and 90°, where higher angles reduce the average output voltage. Thus, 208° is not valid.
Option 3 (27°): While 27° is within the acceptable range for a firing angle, it does not match the calculated value of 45°. This option is incorrect.
Option 4 (32°): Similarly, 32° is a feasible firing angle but does not correspond to the calculated value. This option is also incorrect.
Conclusion:
The correct firing angle under rated full load conditions is 45°, as calculated. This value ensures that the converter provides the required average output voltage to the DC motor under the given operating conditions. Understanding the relationship between firing angle, line voltage, and output voltage is critical for analyzing and designing converter-fed motor systems.
Three Phase Full Converters Question 2:
A 16 kV DC source having an internal resistance of 1 ohm supplies 900A to a 12 kV, 3 phase 6 pulse 60 Hz inverter. Calculate the DC voltage generated by the inverter.
Answer (Detailed Solution Below)
Three Phase Full Converters Question 2 Detailed Solution
Explanation:
Calculation of DC Voltage Generated by the Inverter
To calculate the DC voltage generated by the inverter, let us analyze the given data step-by-step:
- DC source voltage, Vsource = 16 kV = 16000 V
- Internal resistance of the DC source, Rinternal = 1 Ω
- DC current supplied by the source, IDC = 900 A
- Inverter output voltage, Vinverter = 12 kV = 12000 V
Step 1: Voltage Drop Across the Internal Resistance
The internal resistance of the DC source causes a voltage drop, which can be calculated using Ohm's Law:
Voltage Drop (Vdrop) = IDC × Rinternal
Substitute the given values:
Vdrop = 900 A × 1 Ω = 900 V
Step 2: DC Voltage at the Terminals of the Source
The DC voltage at the terminals of the source is the source voltage minus the voltage drop across the internal resistance:
Vterminals = Vsource - Vdrop
Substitute the values:
Vterminals = 16000 V - 900 V = 15100 V
Step 3: DC Voltage Generated by the Inverter
The inverter generates a DC voltage that corresponds to the terminal voltage of the source because it is directly connected to the source.
Thus, the DC voltage generated by the inverter is:
Vinverter = Vterminals = 15100 V
Final Answer: The DC voltage generated by the inverter is 15100 V.
Analysis of Other Options:
Let us now analyze why the other options are incorrect:
Option 1: 27000 V
This option is incorrect because the DC source voltage is only 16000 V, and the internal resistance causes a voltage drop. The DC voltage generated by the inverter cannot exceed the source voltage. The value of 27000 V is unrealistic and does not match the calculations.
Option 2: 15100 V
This is the correct answer as derived above. The terminal voltage of the DC source, after accounting for the voltage drop across the internal resistance, is 15100 V, which is the voltage generated by the inverter.
Option 3: 16000 V
This option assumes that there is no voltage drop across the internal resistance of the DC source. However, the problem explicitly states that the internal resistance is 1 Ω and the current is 900 A, leading to a voltage drop of 900 V. Thus, the terminal voltage is less than 16000 V, making this option incorrect.
Option 4: 16400 V
This option is incorrect as it exceeds the source voltage of 16000 V. Additionally, it does not account for the voltage drop caused by the internal resistance of the source. Such a value is not physically possible in this scenario.
Conclusion:
The correct answer is Option 2: 15100 V. This value accurately accounts for the voltage drop across the internal resistance of the DC source and represents the DC voltage generated by the inverter.
Additional Information
Important Considerations:
- The internal resistance of the DC source plays a significant role in determining the terminal voltage. For high currents, even small resistances can cause substantial voltage drops.
- In practical applications, minimizing the internal resistance of the source is crucial to maximize the efficiency of the system.
- Six-pulse inverters are commonly used in industrial applications due to their simplicity and cost-effectiveness. However, they may introduce harmonics into the system, which need to be mitigated using filters.
Three Phase Full Converters Question 3:
A three-phase full controlled converter (with 6 SCRS only) is feeding the armature of a separately excited DC motor. The motor has to also operate in quadrant-III. Which of the following methods is suitable?
Answer (Detailed Solution Below)
Three Phase Full Converters Question 3 Detailed Solution
Multiquadrant Operation of Electric Drives:
A quadrant diagram is drawn by plotting the speed-torque characteristics of the load/motor for all four quadrant operations.
There are basically two modes of operation:
(i) Motoring Mode: The electrical energy is converted into mechanical energy and supports the motion.
(ii) Braking mode: The mechanical energy is converted into electrical energy and the motor works as a generator and opposes the motion. The fig. below shows the torque and speed coordinates for both forward and reverse motions.
We know that motor can provide motoring and braking operation for both forward and reverse directions.
The power is given as Power = Speed × Torque.
Now, if the power developed is positive then the operation is motoring. If the power developed is negative, the operation is braking.
Quadrant I - Forward Motoring:
In this region, the power and torque are both positive, so the power developed is positive and the machine works as a motor supplying mechanical energy.
Quadrant II - Forward Braking:
In this region, the speed is positive, but torque is negative, so power developed is negative and the machine works under braking opposing the motion.
Quadrant III - Reverse Motoring:
In this region, the speed is negative and the torque is also negative, so power developed is positive and so the machine works under reverse motoring mode.
Quadrant IV - Reverse Braking:
In this region, the torque is positive and the speed is negative, so the power developed is negative, so the machine works as a braking mode in the reverse direction.
Three-Phase Full Converters:
If all diode is replaced by thyristor, a three-phase full converter bridge formed as shown below,
Explanation:
To operate the motor in reverse motoring the current should be negative. So for that the firing angle α should be greater than π/2.
Additional Information
Four-Quadrant Operation of the Hoist:
Three Phase Full Converters Question 4:
A three-phase diode bridge rectifier supplied from a three-phase, 400 V, 50 Hz ac supply delivers power to a resistive load of 50 Ω. The peak value of the instantaneous load voltage would be
Answer (Detailed Solution Below)
Three Phase Full Converters Question 4 Detailed Solution
As the load is resistive the peak instantaneous output voltage is
Vm = √2 Vrms
\({V_m} = 400√ 2 \;V\)
Three Phase Full Converters Question 5:
What is the low order ripple frequency of the output voltage of a three phase fully controlled bridge converter, if the AC input supply frequency is f?
Answer (Detailed Solution Below)
Three Phase Full Converters Question 5 Detailed Solution
Ripple frequency of three-phase full-wave converter:
Figure: output voltage waveform of a three-phase full-wave rectifier.
From the above output voltage waveform, we can observe that for a complete one cycle of input supply, we got 6 pulses in the output.
So, the three-phase full-wave ac to dc converter is a 6-pulse converter.
Then the ripple frequency of the output f0 = m × f
Where,
m = number of pulses in the output per one complete cycle of the input
f = supply voltage frequency
For three-phase ac to dc full wave converter f0 = 6 × f
Top Three Phase Full Converters MCQ Objective Questions
In a three phase (50Hz) full converter, the ripple frequency in output voltage?
Answer (Detailed Solution Below)
Three Phase Full Converters Question 6 Detailed Solution
Download Solution PDFConcept:
Ripple frequency at the output = m × supply frequency
fo = m × fs
Where m = types of the pulse converter
Calculation:
A three-phase full-wave AC to DC converter is a 6-pulse converter
Number of pulses (m) = 6
fo = 6 × supply voltage frequency
∴ f0 = 6 x 50
f0 = 300 Hz
A six-pulse thyristor bridge rectifier is connected to a balanced three-phase, 50 Hz AC source. Assuming that the DC output current of the rectifier is constant, the lowest harmonic component in the AC input current is
Answer (Detailed Solution Below)
Three Phase Full Converters Question 7 Detailed Solution
Download Solution PDFIn a six-pulse thyristor bridge rectifier, the harmonics present are = 6 k ± 1
So, the harmonics are = 5, 7, 11, 13, ...
lowest harmonic component = 5th harmonic supply frequency = 50 Hz
5th harmonic frequency = 5f = 250 HzA separately excited DC motor is energised from a 440 V, 50 Hz, 3 ϕ full converter. The input voltage to the motor for a firing angle of 45°, in volts, is:
Answer (Detailed Solution Below)
Three Phase Full Converters Question 8 Detailed Solution
Download Solution PDFConcept:
The output voltage of a three-phase full converter is given by
\({V_0} = \frac{{3\sqrt 3 {V_{mp}}}}{\pi }\cos \alpha\)
Vmp is maximum supply phase voltage
α is firing angle
Calculation:
Given that, rms supply line voltage (Vrms) = 440 V
Maximum supply phase voltage \({V_{mp}} = \frac{{440\sqrt 2 }}{{\sqrt 3 }}\)
Firing angle (α) = 45°
\({V_0} = \frac{{3\sqrt 3 }}{\pi } \times \frac{{440\sqrt 2 }}{{\sqrt 3 }}\cos 45 = 420.16\;V\)
A fully-controlled three-phase bridge converter is working from a 415 V, 50 Hz AC supply. It is supplying constant current of 100 A at 400 V to a DC load. Assume large inductive smoothing and neglect overlap. The rms value of the AC line current in amperes (round off to two decimal places) is ________.
Answer (Detailed Solution Below) 81 - 82
Three Phase Full Converters Question 9 Detailed Solution
Download Solution PDFConcept:
For a fully controlled three phase bridge rectifier,
\({I_{rms}} = \sqrt {\frac{2}{3}} \times {I_0}\)
Calculation:
Given that, output load current (I0) = 100 A
In a fully controlled three-phase bridge converter,
RMS value of the A.C. line current
\({I_{rms}} = \sqrt {\frac{2}{3}} \times {I_0} = \sqrt {\frac{2}{3}} \times 100 = 81.64\;A\)A
A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 10 A at a firing angle of 30° The approximate total harmonic Distortion (%THD) and the rms value of fundamental component of input current will respectively be_
Answer (Detailed Solution Below)
Three Phase Full Converters Question 10 Detailed Solution
Download Solution PDFRMS current \({I_a} = \sqrt {\frac{2}{3}} \times 10 = 8.16\ A\)
Fundamental current \({I_{{a_1}}} = \frac{{\sqrt 6 }}{\pi }{I_o} = 0.78 \times 10 = 7.8\;A\)
\(\begin{array}{l} THD = \sqrt {\frac{1}{{D{F^2}}} - 1} \\ DF = \frac{{{I_{{a_1}}}}}{{{I_a}}} = \frac{{0.78 \times 10}}{{0.816 \times 10}} = 0.955\\ THD = \sqrt {{{\left( {\frac{1}{{0.955}}} \right)}^2} - 1} = 31\% \end{array}\)
A 3-phase fully controlled, converter is feeding power into a DC load at a constant current of 150 A, the rms value of the current flowing through each thyristor of the converter is
Answer (Detailed Solution Below)
Three Phase Full Converters Question 11 Detailed Solution
Download Solution PDFThree-Phase Full Converters:
If all diode is replaced by thyristor, a three-phase full converter bridge formed as shown below,
- A three-phase input supply is connected to the terminal A, B, C, and RLE load is connected across the output terminal.
- It worked as three-phase AC to DC converter for firing angle delay 0 <α ≤ 90.
- The positive group of SCRs is fired at an interval of 120° and a similarly negative group of SCRs fired at an interval of 120°, but SCR of both groups are fired at an internal of 60° or commutation occurs at every 60°.
- At any time tow SCRs, one from the positive group and one from the negative group must be conducted together for the source to energize the load.
Waveform:
Voltage and Current Equation:
The average output voltage (V0) is given by,
\(V_0=\frac{3V_m}{\pi}cosα\ \longrightarrow 0≤ α≤ \pi\)
Where Vm is the maximum value of line voltage.
Voltage and current waveform for α = 30° and for constant current can be drawn as,
From waveform RMS value of Thyristor current (IT = iT1) will be,
\(I_T=\sqrt{I_0^2\times \frac{2\pi}{3}\times \frac{1}{2\pi}}=I_0\times \sqrt\frac{1}{3}\)
Where I0 is constant DC load current.
Application:
Given,
I0 = 150 A
From the above concept,
The rms value of the current flowing through each thyristor of converter is \({I_T} = {I_o}\sqrt {\frac{1}{3}} = \frac{{150}}{{\sqrt 3 }}\)
What is the output voltage of the three-phase full converter, also known as six pulse converter?
Answer (Detailed Solution Below)
Three Phase Full Converters Question 12 Detailed Solution
Download Solution PDFConcept:
The average output voltage of the three-phase full converter is given by:
\(V_o(avg) = {V_{ml} \over {2\pi\over N}}cosα\)
where, Vml = Maximum value of line voltage
N = No. of pulses produced
α = Firing angle
Calculation:
Given, N = 6
Also, \(V_{ml} = \sqrt{3}V_m\)
where, Vm = Maximum value of phase voltage
\(V_o(avg) = {3\sqrt{3}V_{m} \over \pi}cosα\)
A solar energy installation rectifier a three phase bridge converter to load energy into power system through a transformer of 400V/ 400V, as shown below.
The energy is collected on a bank of 400V battery and is connected to converter through a large filter choke of resistance 10Ω.The KVA rating of the output transformer is
Answer (Detailed Solution Below)
Three Phase Full Converters Question 13 Detailed Solution
Download Solution PDF\({I_{0\left( {max} \right)}} = \frac{{{V_{0max}}}}{R} = \frac{{400}}{{10}} = 40A\)
Input transformer kVA rating = \(\sqrt 3 \times {V_l}{I_l}\)
\({I_l}\) = Rms value of line current on ac diode \(= {I_0} \times \sqrt {\frac{2}{3}}\)
kVA rating of transformer \(= \sqrt 3 \times 400 \times 40 \times \sqrt {\frac{2}{3}}\)
= 22.6kVA
Compute the peak inverse voltage of thyristor connected in three-phase, six-pulse bridge rectifier having input voltage of 415 V. Consider the voltage safety factor to be 2.1.
Answer (Detailed Solution Below)
Three Phase Full Converters Question 14 Detailed Solution
Download Solution PDFThe Correct option is 3
Concept:
The output voltage of 3 phase 6 pulse bridge rectifiers is:
VO = \(\boldsymbol{\mathbf{}\frac{3V_{ML}}{\Pi }}\) = \(\frac{3\sqrt{3} V_{MP}}{\Pi }\)
Rms Line voltage = 415 V
Out voltage = VO
Calculation:
Pick inverse voltage 3 phase 6 pulse bridge rectifiers if the safety factor is 2
V peak = \(\ {\sqrt{2}\times saftey factor\times V_{rms} = \sqrt{2}\times 2.1\times 415}\) = 1232.49
Peak inverse voltage =Vpeak
Additional Information Thyristor:
- A thyristor is a four-layer semiconductor device, consisting of alternating P-type and N-type materials (PNPN). A thyristor usually has three electrodes: an anode, a cathode, and a gate, also known as a control electrode.
In a single-phase full-wave bridge circuit and in a three-phase, delta full-wave bridge circuit, what will be the ripple voltage frequency respectively?
Answer (Detailed Solution Below)
Three Phase Full Converters Question 15 Detailed Solution
Download Solution PDFConcept:
Ripple frequency of three-phase full-wave converter:
Figure: output voltage waveform of three-phase full-wave rectifier.
From the above output voltage waveform, we can observe that for a complete one cycle of input supply we got 6 pulses in the output.
So, the three-phase full-wave ac to dc converter is a 6-pulse converter.
Then the ripple frequency of the output f0 = m × f
Where,
m = number of pulses in the output per one complete cycle of the input
f = supply voltage frequency
Solution:
For single-phase full-wave bridge circuit f0 = 2 × f
For a three-phase full-wave converter f0 = 6 × f
Hence, the ratio output ripple-frequency to the supply-voltage frequency = f0 / f = 6