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All questions of Power Electronics for Electrical Engineering (EE) Exam
If T is the chopping period, Ton the on-time period of chopper and Toff the off-time period of the chopper, then consider the following statements associated with the methods of controlling the output voltage of a chopper:
1. In pulse-width modulation (PWM) scheme, T is kept constant while Ton is varied.
2. In variable-frequency modulation (VFM) scheme, T is varied while either of Ton or Toff is kept constant.
3. In PWM scheme, the output voltage can be varied between zero to source voltage.
4. The large Toff in VFM scheme may make the load current discontinuous
5. PWM scheme is better than VFM scheme.Which of the statements given above are correct?- a)1,2,3, 4, and 5
- b)2, 3 and 5
- c)1,3 and 4
- d)2, 3, 4 and 5
Correct answer is option 'A'. Can you explain this answer?
If T is the chopping period, Ton the on-time period of chopper and Toff the off-time period of the chopper, then consider the following statements associated with the methods of controlling the output voltage of a chopper:
1. In pulse-width modulation (PWM) scheme, T is kept constant while Ton is varied.
2. In variable-frequency modulation (VFM) scheme, T is varied while either of Ton or Toff is kept constant.
3. In PWM scheme, the output voltage can be varied between zero to source voltage.
4. The large Toff in VFM scheme may make the load current discontinuous
5. PWM scheme is better than VFM scheme.
1. In pulse-width modulation (PWM) scheme, T is kept constant while Ton is varied.
2. In variable-frequency modulation (VFM) scheme, T is varied while either of Ton or Toff is kept constant.
3. In PWM scheme, the output voltage can be varied between zero to source voltage.
4. The large Toff in VFM scheme may make the load current discontinuous
5. PWM scheme is better than VFM scheme.
Which of the statements given above are correct?
a)
1,2,3, 4, and 5
b)
2, 3 and 5
c)
1,3 and 4
d)
2, 3, 4 and 5
 | Luminary Institute answered |
For dc – dc converter the critical inductance value.
V0 = a Vs = 0.8 × 100 = 80 V
I0 = 2A Power P0 = V0 I0 = 160 Watt
f = 500
= 8 × 10–3 H = 8 mH.
V0 = a Vs = 0.8 × 100 = 80 V
I0 = 2A Power P0 = V0 I0 = 160 Watt
f = 500
= 8 × 10–3 H = 8 mH.
In current source inverters load voltage waveform V0 and load current waveform i0 respectively- a)depends on load impedance Z, does not depends on Z.
- b)depends on Z, depends on Z.
- c)does not depend on Z, does not depend on Z.
- d)does not depend on Z, depends on Z.
Correct answer is option 'A'. Can you explain this answer?
In current source inverters load voltage waveform V0 and load current waveform i0 respectively
a)
depends on load impedance Z, does not depends on Z.
b)
depends on Z, depends on Z.
c)
does not depend on Z, does not depend on Z.
d)
does not depend on Z, depends on Z.
 | Pankaj Mehta answered |
In a CSI, load current rather than load voltage is controlled, and the inverter output voltage is dependent upon the load impedance and the output voltage waveform. The load current and hence its waveform is independent of load impedance due to which a CSI has inherent protection against short-circuit across its terminals.
After firing an SCR, the gate pulse is removed. The current in the SCR will- a)rise up.
- b)remain the same.
- c)rise a little and then fall to zero.
- d)immediately fall to zero.
Correct answer is option 'B'. Can you explain this answer?
After firing an SCR, the gate pulse is removed. The current in the SCR will
a)
rise up.
b)
remain the same.
c)
rise a little and then fall to zero.
d)
immediately fall to zero.
 | Ravi Singh answered |
Once the thyristor (SCR) is turned-on, and the anode current is above latching current level, there is no need of gate current, hence, gate signal can be withdrawn i.e. gate looses control over the anode current.
Inverters designed from BJT are preferably used in saturation region than in active region because of- a)high efficiency
- b)high power factor
- c)both (a) and (b)
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Inverters designed from BJT are preferably used in saturation region than in active region because of
a)
high efficiency
b)
high power factor
c)
both (a) and (b)
d)
none of these
 | Rajesh Saha answered |
Introduction:
Inverters are electronic devices that convert DC (Direct Current) power into AC (Alternating Current) power. They are widely used in various applications like power supplies, motor drives, renewable energy systems, etc. Inverters can be designed using different types of electronic components, including BJT (Bipolar Junction Transistor).
Active Region vs Saturation Region:
BJTs operate in two main regions - active region and saturation region. In the active region, the BJT acts as an amplifier, and the output voltage is linearly related to the input voltage. In the saturation region, the BJT acts as a switch, and the output voltage is either fully ON (saturated) or fully OFF (cutoff).
Reasons to Prefer Saturation Region:
In the context of inverters designed from BJTs, the saturation region is preferred over the active region due to the following reasons:
1. High Efficiency: In the saturation region, the BJT operates in a fully ON state, which means it has a low resistance. This leads to lower power dissipation and higher efficiency of the inverter. In contrast, in the active region, the BJT operates as an amplifier, which introduces losses and reduces overall efficiency.
2. High Power Factor: Power factor is an important parameter in AC power systems, and it measures the phase relationship between the voltage and current waveforms. In the saturation region, the BJT acts as a switch, and the output waveform closely follows the input waveform, resulting in a power factor close to unity (1). A high power factor is desirable as it reduces reactive power losses and improves the overall system efficiency. On the other hand, in the active region, the BJT operates as an amplifier, which can introduce phase shifts and distortion in the output waveform, leading to a lower power factor.
Conclusion:
Inverters designed from BJTs are preferably used in the saturation region rather than the active region due to their high efficiency and high power factor. The saturation region allows the BJT to operate as a switch, resulting in lower power dissipation, improved efficiency, and a power factor close to unity.
Inverters are electronic devices that convert DC (Direct Current) power into AC (Alternating Current) power. They are widely used in various applications like power supplies, motor drives, renewable energy systems, etc. Inverters can be designed using different types of electronic components, including BJT (Bipolar Junction Transistor).
Active Region vs Saturation Region:
BJTs operate in two main regions - active region and saturation region. In the active region, the BJT acts as an amplifier, and the output voltage is linearly related to the input voltage. In the saturation region, the BJT acts as a switch, and the output voltage is either fully ON (saturated) or fully OFF (cutoff).
Reasons to Prefer Saturation Region:
In the context of inverters designed from BJTs, the saturation region is preferred over the active region due to the following reasons:
1. High Efficiency: In the saturation region, the BJT operates in a fully ON state, which means it has a low resistance. This leads to lower power dissipation and higher efficiency of the inverter. In contrast, in the active region, the BJT operates as an amplifier, which introduces losses and reduces overall efficiency.
2. High Power Factor: Power factor is an important parameter in AC power systems, and it measures the phase relationship between the voltage and current waveforms. In the saturation region, the BJT acts as a switch, and the output waveform closely follows the input waveform, resulting in a power factor close to unity (1). A high power factor is desirable as it reduces reactive power losses and improves the overall system efficiency. On the other hand, in the active region, the BJT operates as an amplifier, which can introduce phase shifts and distortion in the output waveform, leading to a lower power factor.
Conclusion:
Inverters designed from BJTs are preferably used in the saturation region rather than the active region due to their high efficiency and high power factor. The saturation region allows the BJT to operate as a switch, resulting in lower power dissipation, improved efficiency, and a power factor close to unity.
Assertion (A): The lower order harmonics are reduced by using some technique while the higher order harmonics are reduced by using filter.
Reason (R): The cost of the filter is reduced and at the same time transient response is also improved to a great extent.- a)Both A and R are true and R is the correct explanation of A.
- b)Both A and R are true but R is not the correct explanation of A.
- c)A is true but R is false,
- d)A is false but R is true.
Correct answer is option 'A'. Can you explain this answer?
Assertion (A): The lower order harmonics are reduced by using some technique while the higher order harmonics are reduced by using filter.
Reason (R): The cost of the filter is reduced and at the same time transient response is also improved to a great extent.
Reason (R): The cost of the filter is reduced and at the same time transient response is also improved to a great extent.
a)
Both A and R are true and R is the correct explanation of A.
b)
Both A and R are true but R is not the correct explanation of A.
c)
A is true but R is false,
d)
A is false but R is true.
| | Luminary Institute answered |
The machine operated at 50 Hz.
Synchronous speed Ns
A slip of S = 0.02
Nr = (1 - S) Ns = 980 rpm.
Synchronous speed Ns
A slip of S = 0.02
Nr = (1 - S) Ns = 980 rpm.
In continues gating- a)overlap angle is very high
- b)SCR is heated up
- c)size of the pulse transformer is small
- d)commutation cannot be achieved effectively
Correct answer is option 'B'. Can you explain this answer?
In continues gating
a)
overlap angle is very high
b)
SCR is heated up
c)
size of the pulse transformer is small
d)
commutation cannot be achieved effectively
| | Debanshi Nair answered |
As the gating is applied for a longer duration, the device is heated up.
In power transistor, during turn on process voltage and current change linearly from 400 V and 0 V and 0 A to 300 A respectively in 3 μS. the energy loss during turn on process in given by _____Joule (up to 2 decimal places)
Correct answer is '0.06'. Can you explain this answer?
In power transistor, during turn on process voltage and current change linearly from 400 V and 0 V and 0 A to 300 A respectively in 3 μS. the energy loss during turn on process in given by _____Joule (up to 2 decimal places)
 | Ankita Das answered |
Concept:
Change in V and L are shown below
Energy loss during turn on process =
Calculations:
Change in V and L are shown below
Energy loss during turn on process =
Calculations:
= 20 × 103 × 3 × 10-6
= 0.06 J
In dc choppers, the waveforms for input and output voltages are respectively- a)continuous, discontinuous
- b)both discontinuous
- c)discontinuous, continuous
- d)both continuous
Correct answer is option 'A'. Can you explain this answer?
In dc choppers, the waveforms for input and output voltages are respectively
a)
continuous, discontinuous
b)
both discontinuous
c)
discontinuous, continuous
d)
both continuous
 | Prisha Sengupta answered |
Explanation:
DC choppers are electronic devices that are used to convert a fixed DC voltage into a variable DC voltage. The waveform for input and output voltages in DC choppers is as follows:
Input Voltage Waveform:
The input voltage waveform is continuous in DC choppers. It means that the voltage is present at the input of the chopper throughout the operation.
Output Voltage Waveform:
The output voltage waveform is discontinuous in DC choppers. It means that the output voltage is present only for a certain period of time during the operation.
Reason:
The reason for the continuous input voltage waveform is that the input voltage is connected to a DC source that provides a constant voltage. On the other hand, the output voltage waveform is discontinuous because the chopper switches ON and OFF at a certain frequency. During the ON state, the output voltage is present, and during the OFF state, the output voltage is not present.
Conclusion:
Hence, we can conclude that the waveforms for input and output voltages in DC choppers are respectively continuous and discontinuous.
DC choppers are electronic devices that are used to convert a fixed DC voltage into a variable DC voltage. The waveform for input and output voltages in DC choppers is as follows:
Input Voltage Waveform:
The input voltage waveform is continuous in DC choppers. It means that the voltage is present at the input of the chopper throughout the operation.
Output Voltage Waveform:
The output voltage waveform is discontinuous in DC choppers. It means that the output voltage is present only for a certain period of time during the operation.
Reason:
The reason for the continuous input voltage waveform is that the input voltage is connected to a DC source that provides a constant voltage. On the other hand, the output voltage waveform is discontinuous because the chopper switches ON and OFF at a certain frequency. During the ON state, the output voltage is present, and during the OFF state, the output voltage is not present.
Conclusion:
Hence, we can conclude that the waveforms for input and output voltages in DC choppers are respectively continuous and discontinuous.
The single phase mid-point type cycloconverter uses __________ number of SCRs.- a)4
- b)8
- c)6
- d)none of the mentioned
Correct answer is option 'A'. Can you explain this answer?
The single phase mid-point type cycloconverter uses __________ number of SCRs.
a)
4
b)
8
c)
6
d)
none of the mentioned
| | Ishan Chawla answered |
Introduction:
A cycloconverter is a type of power electronic device that converts the frequency of an AC signal. It is commonly used in applications where variable frequency and voltage control is required. The single-phase mid-point type cycloconverter is a specific configuration of a cycloconverter that can be used to convert single-phase AC power.
Explanation:
In a single-phase mid-point type cycloconverter, the AC input voltage is connected to a pair of SCRs (Silicon Controlled Rectifiers) in an anti-parallel configuration. These SCRs are responsible for controlling the power flow and generating the desired output waveform.
Working of single-phase mid-point type cycloconverter:
The operation of a single-phase mid-point type cycloconverter can be divided into two modes:
1. Positive half-cycle mode: During the positive half-cycle of the input voltage, one of the SCRs is triggered and conducts for the duration of the positive half-cycle. The output voltage is generated during this time and is controlled by the triggering angle of the SCR.
2. Negative half-cycle mode: During the negative half-cycle of the input voltage, the other SCR is triggered and conducts for the duration of the negative half-cycle. The output voltage is generated during this time and is also controlled by the triggering angle of the SCR.
Number of SCRs:
In a single-phase mid-point type cycloconverter, there are two SCRs used in total. Each SCR is responsible for controlling one half-cycle of the input voltage. By triggering the appropriate SCR at the desired angle, the output voltage waveform can be controlled.
Advantages of single-phase mid-point type cycloconverter:
- Provides variable frequency control.
- Can operate with a wide range of input voltages.
- Suitable for applications that require precise control of output voltage and frequency.
Conclusion:
The single-phase mid-point type cycloconverter uses two SCRs in total. These SCRs control the power flow and generate the desired output waveform. This configuration allows for variable frequency and voltage control, making it suitable for various applications.
A cycloconverter is a type of power electronic device that converts the frequency of an AC signal. It is commonly used in applications where variable frequency and voltage control is required. The single-phase mid-point type cycloconverter is a specific configuration of a cycloconverter that can be used to convert single-phase AC power.
Explanation:
In a single-phase mid-point type cycloconverter, the AC input voltage is connected to a pair of SCRs (Silicon Controlled Rectifiers) in an anti-parallel configuration. These SCRs are responsible for controlling the power flow and generating the desired output waveform.
Working of single-phase mid-point type cycloconverter:
The operation of a single-phase mid-point type cycloconverter can be divided into two modes:
1. Positive half-cycle mode: During the positive half-cycle of the input voltage, one of the SCRs is triggered and conducts for the duration of the positive half-cycle. The output voltage is generated during this time and is controlled by the triggering angle of the SCR.
2. Negative half-cycle mode: During the negative half-cycle of the input voltage, the other SCR is triggered and conducts for the duration of the negative half-cycle. The output voltage is generated during this time and is also controlled by the triggering angle of the SCR.
Number of SCRs:
In a single-phase mid-point type cycloconverter, there are two SCRs used in total. Each SCR is responsible for controlling one half-cycle of the input voltage. By triggering the appropriate SCR at the desired angle, the output voltage waveform can be controlled.
Advantages of single-phase mid-point type cycloconverter:
- Provides variable frequency control.
- Can operate with a wide range of input voltages.
- Suitable for applications that require precise control of output voltage and frequency.
Conclusion:
The single-phase mid-point type cycloconverter uses two SCRs in total. These SCRs control the power flow and generate the desired output waveform. This configuration allows for variable frequency and voltage control, making it suitable for various applications.
In case of a three phase full controlled converter with 6 SCRs, commutation occurs every- a)120°
- b)60°
- c)180°
- d)30°
Correct answer is option 'B'. Can you explain this answer?
In case of a three phase full controlled converter with 6 SCRs, commutation occurs every
a)
120°
b)
60°
c)
180°
d)
30°
| | Kunal Sharma answered |
Every SCR conducts for 120°. This means that the SCRs from the positive group are fired 120° among themselves, same is true for SCRs from negative group. For example, if T1 starts conducting at 90° it will conduct till 90+120 = 210°. But while T1 is conducting, half of the time i.e. from 90 to 150, T6 is conducting and another half of the time T2 is conducting. Hence, commutation (change in the SCR which is conducting) takes place every 60 degrees irrespective of the firing angle. Construct the firing sequence table for better understanding.
The single-phase half-bridge inverter has a resistive load of 10Ω and the centre-tap dc input voltage is 96 V. The fundamental power consumed by the load is- a)628.6 Watt
- b)525.0Watt'
- c)746.5 Watt
- d)824.4 Watt
Correct answer is option 'C'. Can you explain this answer?
The single-phase half-bridge inverter has a resistive load of 10Ω and the centre-tap dc input voltage is 96 V. The fundamental power consumed by the load is
a)
628.6 Watt
b)
525.0Watt'
c)
746.5 Watt
d)
824.4 Watt
| | Niharika Basu answered |
The nth harmonic-component of output voltage is
A 220 V dc shunt motor runs at 1500 rpm at no-load. The motor is fed through a type-A chopper. If the armature resistance is 1 Ω and duty cycle of chopper is 60%, the speed of the motor when it draw a current of 20 A would be approximately equal to- a)636 rpm
- b)842 rpm
- c)764 rpm
- d)535 rpm
Correct answer is option 'C'. Can you explain this answer?
A 220 V dc shunt motor runs at 1500 rpm at no-load. The motor is fed through a type-A chopper. If the armature resistance is 1 Ω and duty cycle of chopper is 60%, the speed of the motor when it draw a current of 20 A would be approximately equal to
a)
636 rpm
b)
842 rpm
c)
764 rpm
d)
535 rpm
 | Aashna Dey answered |
The output voltage,
V0 = αVs = 0.6 x 220 - 132 V
∴ Back emf
At no- load, N1, = 1500 rpm and Eb1 = 220 V Now, Eb α N
∴
or,
or,
A three-phase full converter charges a battery from a three-phase supply of 230 V. Find the value of the power delivered to the load if a continues current of 20A is flowing through the battery of emf 200 V and internal resistance of 0.5 Ω.- a)0 W
- b)5600 W
- c)4200 W
- d)1040 W
Correct answer is option 'C'. Can you explain this answer?
A three-phase full converter charges a battery from a three-phase supply of 230 V. Find the value of the power delivered to the load if a continues current of 20A is flowing through the battery of emf 200 V and internal resistance of 0.5 Ω.
a)
0 W
b)
5600 W
c)
4200 W
d)
1040 W
 | Nandita Kulkarni answered |
Iavg = Irms = 20 A
P = E x Iavg + Irms2 x R = 4200 W.
P = E x Iavg + Irms2 x R = 4200 W.
Assertion (A): With increased reverse bias, the reverse blocking safe operating area increases in size.
Reason (R): The reverse blocking safe operating area is a plot of collector current versus collector-emitter voltage.- a)Both A and R are true and R is the correct explanation of A.
- b)Both A and R are true but R is not the correct explanation of A.
- c)A is true but R is false.
- d)A is false but R is true.
Correct answer is option 'D'. Can you explain this answer?
Assertion (A): With increased reverse bias, the reverse blocking safe operating area increases in size.
Reason (R): The reverse blocking safe operating area is a plot of collector current versus collector-emitter voltage.
Reason (R): The reverse blocking safe operating area is a plot of collector current versus collector-emitter voltage.
a)
Both A and R are true and R is the correct explanation of A.
b)
Both A and R are true but R is not the correct explanation of A.
c)
A is true but R is false.
d)
A is false but R is true.
| | Shail Nambiar answered |
The reverse blocking safe operating area (RBSOA) for a power transistor is shown below.
It is clear from above curve that with increased reverse bias, the RBSOA decreases in size.
It is clear from above curve that with increased reverse bias, the RBSOA decreases in size.
A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the input pf.- a)0.6
- b)0.7746
- c)0.855
- d)0.236
Correct answer is option 'B'. Can you explain this answer?
A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the input pf.
a)
0.6
b)
0.7746
c)
0.855
d)
0.236
 | Jatin Mukherjee answered |
k = (6/6+4) = 6/10 = 0.6
input pf = √0.6 = 0.7746.
input pf = √0.6 = 0.7746.
For an SCR, the gate – cathode characteristic has a straight line slope of 150. For trigger source voltage of 20 V and allowable gate power dissipation of 1.2 watts, the gate – source resistance is _______ (in Ω)
Correct answer is between '73,74'. Can you explain this answer?
For an SCR, the gate – cathode characteristic has a straight line slope of 150. For trigger source voltage of 20 V and allowable gate power dissipation of 1.2 watts, the gate – source resistance is _______ (in Ω)
 | Amar Sengupta answered |
VgIg = 1.2 W
Assertion (A): Single phase half-wave converter introduces a dc component into the supply line.
Reason (R) : The supply current taken from the source is unidirectional and is in the form of dc pulses- a)Both A and R are true and R is the correct explanation of A.
- b)Both A and R are true but R is not the correct explanation of A,
- c)A is true but R is false.
- d)A is false but R is true.
Correct answer is option 'A'. Can you explain this answer?
Assertion (A): Single phase half-wave converter introduces a dc component into the supply line.
Reason (R) : The supply current taken from the source is unidirectional and is in the form of dc pulses
Reason (R) : The supply current taken from the source is unidirectional and is in the form of dc pulses
a)
Both A and R are true and R is the correct explanation of A.
b)
Both A and R are true but R is not the correct explanation of A,
c)
A is true but R is false.
d)
A is false but R is true.
| | Gaurav Chauhan answered |
Assertion (A): Single phase half-wave converter introduces a dc component into the supply line.
Reason (R): The supply current taken from the source is unidirectional and is in the form of dc pulses.
The correct answer is option 'A', which means that both the assertion and reason are true, and the reason is the correct explanation of the assertion.
Explanation:
In order to understand the given assertion and reason, let's break it down into two parts and discuss each part separately.
Part 1: Single phase half-wave converter introduces a dc component into the supply line
Single phase half-wave converter is a rectifier circuit that converts alternating current (AC) into pulsating direct current (DC). It consists of a diode connected in series with a load and an AC source. The diode allows current to flow only in one direction, which means it allows only the positive half-cycle of the input AC waveform to pass through. During the negative half-cycle, the diode blocks the current.
As a result, the output current of the half-wave converter is a pulsating DC waveform, where the current flows only during the positive half-cycle and becomes zero during the negative half-cycle. This pulsating DC waveform has a significant ripple component, which means it is not pure DC. However, this pulsating DC waveform does have a constant, non-zero component, known as the DC component.
Therefore, it can be concluded that a single phase half-wave converter introduces a DC component into the supply line.
Part 2: The supply current taken from the source is unidirectional and is in the form of DC pulses
The supply current taken from the source in a single phase half-wave converter is unidirectional, meaning it flows in only one direction. During the positive half-cycle of the input AC waveform, the diode allows the current to flow through the load, creating a positive pulse of current. However, during the negative half-cycle, the diode blocks the current, resulting in zero current flow.
This means that the supply current taken from the source is in the form of DC pulses, where the current flows only during the positive half-cycle and is zero during the negative half-cycle.
Therefore, it can be concluded that the supply current taken from the source in a single phase half-wave converter is unidirectional and in the form of DC pulses.
Conclusion:
The assertion that a single phase half-wave converter introduces a DC component into the supply line is true, and the reason that the supply current taken from the source is unidirectional and in the form of DC pulses is the correct explanation of this assertion. Hence, the correct answer is option 'A'.
Reason (R): The supply current taken from the source is unidirectional and is in the form of dc pulses.
The correct answer is option 'A', which means that both the assertion and reason are true, and the reason is the correct explanation of the assertion.
Explanation:
In order to understand the given assertion and reason, let's break it down into two parts and discuss each part separately.
Part 1: Single phase half-wave converter introduces a dc component into the supply line
Single phase half-wave converter is a rectifier circuit that converts alternating current (AC) into pulsating direct current (DC). It consists of a diode connected in series with a load and an AC source. The diode allows current to flow only in one direction, which means it allows only the positive half-cycle of the input AC waveform to pass through. During the negative half-cycle, the diode blocks the current.
As a result, the output current of the half-wave converter is a pulsating DC waveform, where the current flows only during the positive half-cycle and becomes zero during the negative half-cycle. This pulsating DC waveform has a significant ripple component, which means it is not pure DC. However, this pulsating DC waveform does have a constant, non-zero component, known as the DC component.
Therefore, it can be concluded that a single phase half-wave converter introduces a DC component into the supply line.
Part 2: The supply current taken from the source is unidirectional and is in the form of DC pulses
The supply current taken from the source in a single phase half-wave converter is unidirectional, meaning it flows in only one direction. During the positive half-cycle of the input AC waveform, the diode allows the current to flow through the load, creating a positive pulse of current. However, during the negative half-cycle, the diode blocks the current, resulting in zero current flow.
This means that the supply current taken from the source is in the form of DC pulses, where the current flows only during the positive half-cycle and is zero during the negative half-cycle.
Therefore, it can be concluded that the supply current taken from the source in a single phase half-wave converter is unidirectional and in the form of DC pulses.
Conclusion:
The assertion that a single phase half-wave converter introduces a DC component into the supply line is true, and the reason that the supply current taken from the source is unidirectional and in the form of DC pulses is the correct explanation of this assertion. Hence, the correct answer is option 'A'.
A three-phase M-3 converter is operated from a 3-phase, 230 V, 50 Hz supply with load resistance R = 10 Ω. Find the value of firing angle if an average output voltage of 50% of the maximum possible output voltage is required.Hint: α > 30°.- a)92.7°
- b)67.7°
- c)45°
- d)75.7°
Correct answer is option 'B'. Can you explain this answer?
A three-phase M-3 converter is operated from a 3-phase, 230 V, 50 Hz supply with load resistance R = 10 Ω. Find the value of firing angle if an average output voltage of 50% of the maximum possible output voltage is required.Hint: α > 30°.
a)
92.7°
b)
67.7°
c)
45°
d)
75.7°
| | Aniket Shah answered |
We need , Vo = 0.5 Vom.
α>30° hence we use the equation Vo = (3√3/2π) x Vmp x [1+cos(30+α)] √3Vmp = Vml = √2×230
Therefore, Vo = (3/2π) x √3 Vml x [1+cos(30+α)] = 0.5Vom
(1/√3) x [1+cos(30+α)] = Vo x 2π/3Vml = Vo/Vom = 1/2
α = 67.7°.
α>30° hence we use the equation Vo = (3√3/2π) x Vmp x [1+cos(30+α)] √3Vmp = Vml = √2×230
Therefore, Vo = (3/2π) x √3 Vml x [1+cos(30+α)] = 0.5Vom
(1/√3) x [1+cos(30+α)] = Vo x 2π/3Vml = Vo/Vom = 1/2
α = 67.7°.
Semi-converters are- a)single quadrant converters
- b)double quadrant converters
- c)three quadrant converters
- d)none of the mentioned
Correct answer is option 'A'. Can you explain this answer?
Semi-converters are
a)
single quadrant converters
b)
double quadrant converters
c)
three quadrant converters
d)
none of the mentioned
| | Rajesh Kumar answered |
Semi-converters are single quadrant converters, because the voltage and current can only be both positive due to the diodes connected.
A d.c. chopper circuit connected to a 100 V d.c. source supplies an inductive load having 40 mH in series with a resistance of 5 Ω A freewheeling diode is placed across the load. The load current varies between the limits of 10 A and 12 A. The time ratio (Ton/Toff) of the chopper is- a)1.65
- b)0.98
- c)1.22
- d)0.75
Correct answer is option 'C'. Can you explain this answer?
A d.c. chopper circuit connected to a 100 V d.c. source supplies an inductive load having 40 mH in series with a resistance of 5 Ω A freewheeling diode is placed across the load. The load current varies between the limits of 10 A and 12 A. The time ratio (Ton/Toff) of the chopper is
a)
1.65
b)
0.98
c)
1.22
d)
0.75
 | Kiran Iyer answered |
The average value of load current
The maximum value of load current
Now, the average value of the voltage is
Also,
or,
or,
or,
The maximum value of load current
Now, the average value of the voltage is
Also,
or,
or,
or,
A three-phase half-wave controlled converter is fed from a 3-phase, 400 V source and is connected to a load which takes a constant current of 36 A. Find,the value of average output voltage and average current rating of SCR for a firing angle of 30°.- a)234 V, 36 A
- b)234 V, 12 A
- c)135 V, 36 A
- d)135 V, 12 A
Correct answer is option 'B'. Can you explain this answer?
A three-phase half-wave controlled converter is fed from a 3-phase, 400 V source and is connected to a load which takes a constant current of 36 A. Find,the value of average output voltage and average current rating of SCR for a firing angle of 30°.
a)
234 V, 36 A
b)
234 V, 12 A
c)
135 V, 36 A
d)
135 V, 12 A
| | Sparsh Nambiar answered |
Vo = (3√3/2π) x Vmp x cosα = 233.874 V.
Ia = Io/3 = 12 A.
Ia = Io/3 = 12 A.
A single-phase half bridge inverter has supply voltage of 200 V. For a load resistance of 10 Ω, the output power is equal to- a)2000 W
- b)500 W
- c)225 W
- d)1000W
Correct answer is option 'D'. Can you explain this answer?
A single-phase half bridge inverter has supply voltage of 200 V. For a load resistance of 10 Ω, the output power is equal to
a)
2000 W
b)
500 W
c)
225 W
d)
1000W
| | Prisha Sen answered |
Output is a square wave with an amplitude of 100V.
RMS value of a square wave is equal to its peak value.
So, rms value of output voltage is
RMS value of a square wave is equal to its peak value.
So, rms value of output voltage is
Assertion (A): The commutation overlap is more predominant is semiconverters than in full converters.
Reason (R): The commutation period in seconds, when outgoing and incoming SCRs are conducting, is known as the overlap period.- a)Both A and R are true and R is the correct explanation of A.
- b)Both A and R are true but R is not the correct explanation of A.
- c)A is true but R is false.
- d)A is false but R is true.
Correct answer is option 'D'. Can you explain this answer?
Assertion (A): The commutation overlap is more predominant is semiconverters than in full converters.
Reason (R): The commutation period in seconds, when outgoing and incoming SCRs are conducting, is known as the overlap period.
Reason (R): The commutation period in seconds, when outgoing and incoming SCRs are conducting, is known as the overlap period.
a)
Both A and R are true and R is the correct explanation of A.
b)
Both A and R are true but R is not the correct explanation of A.
c)
A is true but R is false.
d)
A is false but R is true.
| | Poulomi Chopra answered |
The commutation overlap is more predominant in full converter than in semi converter. Hence, assertion is a false statement.
Which of the following converter circuits would require a neutral point?- a)3-phase semi-converter
- b)3-phase full converter
- c)3-phase full converter with freewheeling diode
- d)3-phase half wave converter
Correct answer is option 'D'. Can you explain this answer?
Which of the following converter circuits would require a neutral point?
a)
3-phase semi-converter
b)
3-phase full converter
c)
3-phase full converter with freewheeling diode
d)
3-phase half wave converter
| | Arya Mukherjee answered |
Answer:
In a converter circuit, a neutral point is required when there is a need to connect the common or neutral terminal of the load or source to a reference potential. This reference potential is usually the earth or ground potential.
Among the given options, the converter circuit that requires a neutral point is the 3-phase half wave converter (option D).
Explanation:
A half wave converter is a type of converter circuit that converts AC voltage to DC voltage. It is commonly used in various applications such as motor control, power supplies, and battery charging.
In a 3-phase half wave converter, the AC input is a 3-phase supply with three phases: Phase A, Phase B, and Phase C. The AC input is connected to a diode bridge rectifier circuit consisting of six diodes. The output of the rectifier circuit is a pulsating DC voltage.
Reason for requiring a neutral point:
The reason why a neutral point is required in a 3-phase half wave converter is because the load or source may have a common or neutral terminal that needs to be connected to a reference potential. In many practical applications, the neutral point is connected to the ground potential.
Importance of neutral point:
1. Ground reference: The neutral point provides a reference potential for the load or source. By connecting the neutral point to the ground potential, it ensures that the voltage levels are measured and controlled with respect to the ground.
2. Safety: The neutral point helps in ensuring the safety of the system. By grounding the neutral point, any leakage current or fault current can be effectively diverted to the ground, preventing potential hazards such as electric shock.
3. Stability: The neutral point helps in stabilizing the system. By connecting the neutral point to the ground potential, it provides a stable reference point for voltage measurements and control, minimizing fluctuations and ensuring stable operation.
4. Common mode voltage control: The neutral point helps in controlling the common mode voltage in the system. By connecting the neutral point to the ground potential, it helps in reducing common mode noise and interference.
Therefore, in a 3-phase half wave converter, a neutral point is required to provide a reference potential for the load or source and to ensure safety, stability, and common mode voltage control.
In a converter circuit, a neutral point is required when there is a need to connect the common or neutral terminal of the load or source to a reference potential. This reference potential is usually the earth or ground potential.
Among the given options, the converter circuit that requires a neutral point is the 3-phase half wave converter (option D).
Explanation:
A half wave converter is a type of converter circuit that converts AC voltage to DC voltage. It is commonly used in various applications such as motor control, power supplies, and battery charging.
In a 3-phase half wave converter, the AC input is a 3-phase supply with three phases: Phase A, Phase B, and Phase C. The AC input is connected to a diode bridge rectifier circuit consisting of six diodes. The output of the rectifier circuit is a pulsating DC voltage.
Reason for requiring a neutral point:
The reason why a neutral point is required in a 3-phase half wave converter is because the load or source may have a common or neutral terminal that needs to be connected to a reference potential. In many practical applications, the neutral point is connected to the ground potential.
Importance of neutral point:
1. Ground reference: The neutral point provides a reference potential for the load or source. By connecting the neutral point to the ground potential, it ensures that the voltage levels are measured and controlled with respect to the ground.
2. Safety: The neutral point helps in ensuring the safety of the system. By grounding the neutral point, any leakage current or fault current can be effectively diverted to the ground, preventing potential hazards such as electric shock.
3. Stability: The neutral point helps in stabilizing the system. By connecting the neutral point to the ground potential, it provides a stable reference point for voltage measurements and control, minimizing fluctuations and ensuring stable operation.
4. Common mode voltage control: The neutral point helps in controlling the common mode voltage in the system. By connecting the neutral point to the ground potential, it helps in reducing common mode noise and interference.
Therefore, in a 3-phase half wave converter, a neutral point is required to provide a reference potential for the load or source and to ensure safety, stability, and common mode voltage control.
A three-phase, three-pulse, M-3 type controlled converter uses ____________ number of SCRs.- a)1
- b)2
- c)3
- d)4
Correct answer is option 'C'. Can you explain this answer?
A three-phase, three-pulse, M-3 type controlled converter uses ____________ number of SCRs.
a)
1
b)
2
c)
3
d)
4
| | Suyash Joshi answered |
It uses three SCRs with a three-phase transformer. M-3 type 3-pulse converters are not practically used.
Assertion (A): For high power applications, inverters are used instead of transistors.
Reason (R): For high power applications, inverter is operated in active region.- a)Both A and R are true and R is the correct explanation of A.
- b)Both A and R are true but R is not the correct explanation of A.
- c)A is true but R is false.
- d)A is false but R is true.
Correct answer is option 'C'. Can you explain this answer?
Assertion (A): For high power applications, inverters are used instead of transistors.
Reason (R): For high power applications, inverter is operated in active region.
Reason (R): For high power applications, inverter is operated in active region.
a)
Both A and R are true and R is the correct explanation of A.
b)
Both A and R are true but R is not the correct explanation of A.
c)
A is true but R is false.
d)
A is false but R is true.
| | Nayanika Singh answered |
In low-power electronic circuits oscillators are used for converting dc power into ac power. These oscillator use transistors for converting dc voltage into sinusoidal ac voltage. Since transistor is used in active region, therefore there is substantial loss of power which decreases efficiency. In high power applications inverters are used instead of transistors and the inverters operate in saturation region or cut-off region. Thus, assertion is true but reason is false.
Assertion (A): The principal of step-up chopper can be employed for the regenerative braking of dc motors.
Reason (R) : The step-up chopper can be employed in regenerative braking only for increasing motor speeds- a)Both A and R are true and R is the correct explanation of A.
- b)Both A and R are true but R is not the correct explanation of A
- c)A is true but R is false
- d)A is false but R is true
Correct answer is option 'C'. Can you explain this answer?
Assertion (A): The principal of step-up chopper can be employed for the regenerative braking of dc motors.
Reason (R) : The step-up chopper can be employed in regenerative braking only for increasing motor speeds
Reason (R) : The step-up chopper can be employed in regenerative braking only for increasing motor speeds
a)
Both A and R are true and R is the correct explanation of A.
b)
Both A and R are true but R is not the correct explanation of A
c)
A is true but R is false
d)
A is false but R is true
| | Abhishek Chauhan answered |
The step-up chopper can be employed in regenerative braking both for increasing and decreasing motor speed provided duty cycle a is so adjusted that Vs/( 1 - a) exceeds the fixed source voiiage V0. Hence, reason is a false statement.
A 3-phase full converter feeds power to an R load of 10 Ω. For a firing angle delay of 30° the load takes 5 kW. An inductor of large value is also connected to the load to make the current ripple free. Find the value of per phase input voltage.- a)133 V
- b)230/√3 V
- c)191/√3 V
- d)298/√3 V
Correct answer is option 'C'. Can you explain this answer?
A 3-phase full converter feeds power to an R load of 10 Ω. For a firing angle delay of 30° the load takes 5 kW. An inductor of large value is also connected to the load to make the current ripple free. Find the value of per phase input voltage.
a)
133 V
b)
230/√3 V
c)
191/√3 V
d)
298/√3 V
| | Arpita Banerjee answered |
Ior = Vo/R = (3Vml/Rπ) cos α
P = 5 kW = Ior2 x R = [(3Vml/π) cos α]2 x 1/R] Therefore, Vs (line) = √50000 x (π/√2 x 3 x cos30) = 191.22 V
Vs (phase) = 191/√3 V.
P = 5 kW = Ior2 x R = [(3Vml/π) cos α]2 x 1/R] Therefore, Vs (line) = √50000 x (π/√2 x 3 x cos30) = 191.22 V
Vs (phase) = 191/√3 V.
The drawback of fi-firing circuit for SCR is- a)firing angle limitation to
- b)reduced response time π/2
- c)high power loss
- d)requirement of high on-time of SCRs
Correct answer is option 'A'. Can you explain this answer?
The drawback of fi-firing circuit for SCR is
a)
firing angle limitation to
b)
reduced response time π/2
c)
high power loss
d)
requirement of high on-time of SCRs
| | Nilesh Joshi answered |
The R-firing circuit suffers from several disadvantages. One of them is that the trigger angle can be varied only upto an approximate value of 90°.
SCR with a rating of 1200 V and 250 A are available to be used in a string to handle 8 kV and 1kA. The number of parallel and series units required are respectively _____ (Take derating factor is 0.2)- a)8, 5
- b)9, 5
- c)5, 9
- d)5, 8
Correct answer is option 'C'. Can you explain this answer?
SCR with a rating of 1200 V and 250 A are available to be used in a string to handle 8 kV and 1kA. The number of parallel and series units required are respectively _____ (Take derating factor is 0.2)
a)
8, 5
b)
9, 5
c)
5, 9
d)
5, 8
| | Arya Mukherjee answered |
Derating factor = 1 – string efficiency
In dc choppers, per unit ripple is maximum when duty cycle α is- a)0.9
- b)0.7
- c)0.5
- d)0.2
Correct answer is option 'C'. Can you explain this answer?
In dc choppers, per unit ripple is maximum when duty cycle α is
a)
0.9
b)
0.7
c)
0.5
d)
0.2
 | EduRev GATE answered |
The minimum turn-ON time of the SCR is = 
For step down chopper, Vo = δ VS
(Vo )min = 50 π V
For step down chopper, Vo = δ VS
(Vo )min = 50 π V
Assertion (A) : If the one-state anode current drops below a minimum level, designated as the holding current IH, the thyristor reverts to the forward blocking state.
Reason (R) : The loop gain of the equivalent pnp-npn transistors fall below unity and the regenerative hold-on action ceases.- a)Both A and R are true and R is the correct explanation of A.
- b)Both A and R are true but R is not the correct explanation of A.
- c)A is true but R is false.
- d)A is false but R is true.
Correct answer is option 'A'. Can you explain this answer?
Assertion (A) : If the one-state anode current drops below a minimum level, designated as the holding current IH, the thyristor reverts to the forward blocking state.
Reason (R) : The loop gain of the equivalent pnp-npn transistors fall below unity and the regenerative hold-on action ceases.
Reason (R) : The loop gain of the equivalent pnp-npn transistors fall below unity and the regenerative hold-on action ceases.
a)
Both A and R are true and R is the correct explanation of A.
b)
Both A and R are true but R is not the correct explanation of A.
c)
A is true but R is false.
d)
A is false but R is true.
| | Luminary Institute answered |
For successful turn ON i1 + i2 ≥ 100 mA
I = i1 + i2 = (8 + 7.9) (1 - e-20t ) = 100 mA
1 – e-20t = 6.30μ
t = 3.15 × 10-4 s
I = i1 + i2 = (8 + 7.9) (1 - e-20t ) = 100 mA
1 – e-20t = 6.30μ
t = 3.15 × 10-4 s
The principle of three phase cycloconverter is to- a)add and remove number of SCRs
- b)vary progressively the firing angle of the devices
- c)keep the firing angle as 0° for all the devices
- d)none of the mentioned
Correct answer is option 'B'. Can you explain this answer?
The principle of three phase cycloconverter is to
a)
add and remove number of SCRs
b)
vary progressively the firing angle of the devices
c)
keep the firing angle as 0° for all the devices
d)
none of the mentioned
| | Charvi Reddy answered |
B) vary progressively the firing angle of the devices
Consider the following statements associated with single phase full converters:
1. Mid-point converter configuration is used in case the terminals on dc side have to be grounded.
2. The transformer rating in mid-point converter is double the load rating.
3. SCRs are subjected to a peak inverse voltage of 2 Vm in single phase mid-point converter.
4. Bridge converter is preferred over mid-point converter.Which of these statements are correct?- a)1,3 and 4
- b)2, 3 and 4
- c)1,2 and 3
- d)1, 2, 3 and 4
Correct answer is option 'A'. Can you explain this answer?
Consider the following statements associated with single phase full converters:
1. Mid-point converter configuration is used in case the terminals on dc side have to be grounded.
2. The transformer rating in mid-point converter is double the load rating.
3. SCRs are subjected to a peak inverse voltage of 2 Vm in single phase mid-point converter.
4. Bridge converter is preferred over mid-point converter.
1. Mid-point converter configuration is used in case the terminals on dc side have to be grounded.
2. The transformer rating in mid-point converter is double the load rating.
3. SCRs are subjected to a peak inverse voltage of 2 Vm in single phase mid-point converter.
4. Bridge converter is preferred over mid-point converter.
Which of these statements are correct?
a)
1,3 and 4
b)
2, 3 and 4
c)
1,2 and 3
d)
1, 2, 3 and 4
| | Snehal Rane answered |
In mid-point converter, each secondary should be able to supply the load power. Due to this reason, the transformer rating in mid-point converter is double the load rating. Thus, statement-2 is not correct.
A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the average value of SCR current.- a)21.68 A
- b)200 mA
- c)4.14 A
- d)2.07 A
Correct answer is option 'C'. Can you explain this answer?
A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the average value of SCR current.
a)
21.68 A
b)
200 mA
c)
4.14 A
d)
2.07 A
 | Manoj Chaudhary answered |
Average value of SCR current can be determined using the formula:
\(\text{Average current} = \frac{{\text{On time current} \times \text{On time duration} + \text{Off time current} \times \text{Off time duration}}}{{\text{Total duration}}}\)
Given data:
- Input voltage (V) = 230 V
- Load resistance (R) = 15 Ω
- On time duration = 6 cycles
- Off time duration = 4 cycles
1. Calculate On time current:
The on time current can be calculated using Ohm's law:
\(I_{\text{on}} = \frac{V}{R}\)
Substituting the given values:
\(I_{\text{on}} = \frac{230}{15}\)
2. Calculate Off time current:
During the off time, the SCR is off, so the current is zero.
\(I_{\text{off}} = 0\)
3. Calculate Total duration:
The total duration is the sum of on time and off time durations:
\(\text{Total duration} = \text{On time duration} + \text{Off time duration}\)
4. Calculate the average current:
Substituting the values into the formula:
\(\text{Average current} = \frac{{I_{\text{on}} \times \text{On time duration} + I_{\text{off}} \times \text{Off time duration}}}{{\text{Total duration}}}\)
Simplifying the equation:
\(\text{Average current} = \frac{{I_{\text{on}} \times \text{On time duration}}}{{\text{Total duration}}}\)
Substituting the given values:
\(\text{Average current} = \frac{{\frac{230}{15} \times 6}}{{6 + 4}}\)
\(\text{Average current} = \frac{{230 \times 6}}{{15 \times 10}}\)
\(\text{Average current} = \frac{{1380}}{{150}}\)
\(\text{Average current} = 9.2\) A
Therefore, the average value of SCR current is 9.2 A, which is not one of the given options. It seems that there might be an error in the options provided.
\(\text{Average current} = \frac{{\text{On time current} \times \text{On time duration} + \text{Off time current} \times \text{Off time duration}}}{{\text{Total duration}}}\)
Given data:
- Input voltage (V) = 230 V
- Load resistance (R) = 15 Ω
- On time duration = 6 cycles
- Off time duration = 4 cycles
1. Calculate On time current:
The on time current can be calculated using Ohm's law:
\(I_{\text{on}} = \frac{V}{R}\)
Substituting the given values:
\(I_{\text{on}} = \frac{230}{15}\)
2. Calculate Off time current:
During the off time, the SCR is off, so the current is zero.
\(I_{\text{off}} = 0\)
3. Calculate Total duration:
The total duration is the sum of on time and off time durations:
\(\text{Total duration} = \text{On time duration} + \text{Off time duration}\)
4. Calculate the average current:
Substituting the values into the formula:
\(\text{Average current} = \frac{{I_{\text{on}} \times \text{On time duration} + I_{\text{off}} \times \text{Off time duration}}}{{\text{Total duration}}}\)
Simplifying the equation:
\(\text{Average current} = \frac{{I_{\text{on}} \times \text{On time duration}}}{{\text{Total duration}}}\)
Substituting the given values:
\(\text{Average current} = \frac{{\frac{230}{15} \times 6}}{{6 + 4}}\)
\(\text{Average current} = \frac{{230 \times 6}}{{15 \times 10}}\)
\(\text{Average current} = \frac{{1380}}{{150}}\)
\(\text{Average current} = 9.2\) A
Therefore, the average value of SCR current is 9.2 A, which is not one of the given options. It seems that there might be an error in the options provided.
IA class C chopper is operated from a 220 V battery. The load Is a dc motor with R = 0.1Ω, L= 10 mH and Eb= 100 V. The duty cycle of the chopper to achieve regenerative braking at the rated current of 10 Ampere would be equal to- a)25%
- b)45%
- c)66%
- d)72%
Correct answer is option 'B'. Can you explain this answer?
IA class C chopper is operated from a 220 V battery. The load Is a dc motor with R = 0.1Ω, L= 10 mH and Eb= 100 V. The duty cycle of the chopper to achieve regenerative braking at the rated current of 10 Ampere would be equal to
a)
25%
b)
45%
c)
66%
d)
72%
| | Bhavana Reddy answered |
Thyristor is a semiconductor switch which is- a)unilateral and bistable
- b)bilateral and bistable
- c)unilateral and astable
- d)bistable and astable
Correct answer is option 'A'. Can you explain this answer?
Thyristor is a semiconductor switch which is
a)
unilateral and bistable
b)
bilateral and bistable
c)
unilateral and astable
d)
bistable and astable
| | Preethi Banerjee answered |
Like the diode, an SCR is an unidirectional device that blocks the current flow from cathode to anode i.e. unilateral. An SCR or a thyristor has three modes of operation namely forward conduction mode, forward blocking mode and reverse blocking mode. Out of these three modes of operation the first two are stable states due to which an SCR is bistable.
The nature of load current, l.e., whether load is continuous or discontinuous in controlled rectifiers- a)depends only on the type of load
- b)does not depend on type of load and firing angle delay
- c)depends only on the firing angle delay
- d)depends both on the type of load and firing angle delay
Correct answer is option 'D'. Can you explain this answer?
The nature of load current, l.e., whether load is continuous or discontinuous in controlled rectifiers
a)
depends only on the type of load
b)
does not depend on type of load and firing angle delay
c)
depends only on the firing angle delay
d)
depends both on the type of load and firing angle delay
| | Debanshi Iyer answered |
The term continuous means that load current never ceases but continues to flow through SCR or their combination. The term discontinuous is applied to the condition when load current reaches zero during each half cycle before the next SCR in sequence is fired. In practice, the output current may become discontinuous at high values of firing angle or at low values of load current.
The range of firing angle for a 3-phase, 3-pulse converter feeding a resistive load is __________ (in degrees).- a)0 to 180
- b)0 to 150
- c)30 to 150
- d)30 to 180
Correct answer is option 'B'. Can you explain this answer?
The range of firing angle for a 3-phase, 3-pulse converter feeding a resistive load is __________ (in degrees).
a)
0 to 180
b)
0 to 150
c)
30 to 150
d)
30 to 180
 | Mainak Roy answered |
Range of Firing Angle for a 3-phase, 3-pulse Converter
To determine the range of firing angles for a 3-phase, 3-pulse converter feeding a resistive load, we need to consider the operation of the converter and the limitations imposed by the system.
Operation of a 3-phase, 3-pulse Converter
- A 3-phase, 3-pulse converter is a type of rectifier that converts AC power into DC power.
- It consists of three diodes connected in a bridge configuration, with each diode connected to one phase of the AC supply.
- The converter operates by turning on the diodes in a specific sequence to rectify the AC voltage into a pulsating DC voltage.
- The firing angle determines the delay between the AC voltage waveform and the turning on of the diodes.
Limitations and Constraints
- For a resistive load, the converter should operate in the continuous conduction mode to ensure a smooth DC output voltage.
- In the continuous conduction mode, the diodes should not turn off during the conduction period of the AC waveforms.
- The firing angle should be chosen such that the diodes remain conducting until the next phase is turned on.
- The maximum firing angle is limited by the requirement that the diodes should not turn off before the next phase is turned on.
Determining the Range of Firing Angle
- In a 3-pulse converter, each phase is turned on for 120 degrees of the AC cycle.
- The firing angle for each phase can be defined as the delay in turning on the diodes with respect to the AC voltage waveform.
- The minimum firing angle is 0 degrees, where the diodes turn on immediately at the start of each phase.
- The maximum firing angle is limited by the condition that the diodes should not turn off before the next phase is turned on.
- Since each phase is turned on for 120 degrees, the diodes should remain conducting for at least 120 degrees to ensure continuous conduction.
- Therefore, the maximum firing angle is 150 degrees, allowing a 30-degree safety margin to ensure continuous conduction.
Conclusion
The range of firing angle for a 3-phase, 3-pulse converter feeding a resistive load is 0 to 150 degrees. This range ensures continuous conduction and prevents the diodes from turning off before the next phase is turned on.
To determine the range of firing angles for a 3-phase, 3-pulse converter feeding a resistive load, we need to consider the operation of the converter and the limitations imposed by the system.
Operation of a 3-phase, 3-pulse Converter
- A 3-phase, 3-pulse converter is a type of rectifier that converts AC power into DC power.
- It consists of three diodes connected in a bridge configuration, with each diode connected to one phase of the AC supply.
- The converter operates by turning on the diodes in a specific sequence to rectify the AC voltage into a pulsating DC voltage.
- The firing angle determines the delay between the AC voltage waveform and the turning on of the diodes.
Limitations and Constraints
- For a resistive load, the converter should operate in the continuous conduction mode to ensure a smooth DC output voltage.
- In the continuous conduction mode, the diodes should not turn off during the conduction period of the AC waveforms.
- The firing angle should be chosen such that the diodes remain conducting until the next phase is turned on.
- The maximum firing angle is limited by the requirement that the diodes should not turn off before the next phase is turned on.
Determining the Range of Firing Angle
- In a 3-pulse converter, each phase is turned on for 120 degrees of the AC cycle.
- The firing angle for each phase can be defined as the delay in turning on the diodes with respect to the AC voltage waveform.
- The minimum firing angle is 0 degrees, where the diodes turn on immediately at the start of each phase.
- The maximum firing angle is limited by the condition that the diodes should not turn off before the next phase is turned on.
- Since each phase is turned on for 120 degrees, the diodes should remain conducting for at least 120 degrees to ensure continuous conduction.
- Therefore, the maximum firing angle is 150 degrees, allowing a 30-degree safety margin to ensure continuous conduction.
Conclusion
The range of firing angle for a 3-phase, 3-pulse converter feeding a resistive load is 0 to 150 degrees. This range ensures continuous conduction and prevents the diodes from turning off before the next phase is turned on.
A single-phase voltage controller, using one SCR in anti parallel with a diode, feeds a load R and Vs = 230 V. For a firing angel of 90° for the SCR, the PMMC voltage connected across R would read- a)0
- b)51.8 V
- c)–51.8 V
- d)–36.82 V
Correct answer is option 'C'. Can you explain this answer?
A single-phase voltage controller, using one SCR in anti parallel with a diode, feeds a load R and Vs = 230 V. For a firing angel of 90° for the SCR, the PMMC voltage connected across R would read
a)
0
b)
51.8 V
c)
–51.8 V
d)
–36.82 V
| | Nandita Bajaj answered |
As firing angle is 90, there is ideally be no conduction in the positive half. Hence, the average value will be zero.
Vo = (√2 Vs)/2π x (cos90 – 1) = – 51.8 V.
Vo = (√2 Vs)/2π x (cos90 – 1) = – 51.8 V.
A three-phase three-pulse converter would operate as a line commutated inverter when- a)30° < α < 60°
- b)90° < α <180°
- c)90° > α
- d)it can never operate as a line commutated inverter
Correct answer is option 'B'. Can you explain this answer?
A three-phase three-pulse converter would operate as a line commutated inverter when
a)
30° < α < 60°
b)
90° < α <180°
c)
90° > α
d)
it can never operate as a line commutated inverter
| | Harshitha Pillai answered |
A) The converter is supplied with three-phase AC voltage and the thyristors are triggered in a specific sequence to convert the input AC voltage into a variable frequency AC output voltage.
To turn-off or commutate a thyristor- a)forced commutation is used
- b)reverse blocking voltage is applied
- c)gate current is made zero
- d)none of the above
Correct answer is option 'A'. Can you explain this answer?
To turn-off or commutate a thyristor
a)
forced commutation is used
b)
reverse blocking voltage is applied
c)
gate current is made zero
d)
none of the above
| | Saranya Mishra answered |
To turn-off a thyristor, anode current must be reduced below holding current and a reverse bias must be applied across thyristor for a finite period of time. If all these two conditions are not met simultaneously then forced commutation can be used to turn-off the thyristor.
A full-bridge bipolar PWM inverter is fed from a 240 V battery and is driving an RL load. The fundamental output voltage for a modulation index of 0.8 is- a)135V
- b)215V
- c)205 V
- d)165 V
Correct answer is option 'A'. Can you explain this answer?
A full-bridge bipolar PWM inverter is fed from a 240 V battery and is driving an RL load. The fundamental output voltage for a modulation index of 0.8 is
a)
135V
b)
215V
c)
205 V
d)
165 V
| | Sanchita Sharma answered |
The fundamental output voltage for a modulation index of M is
A 3-phase bridge converter is given a three-phase supply in the phase sequence R-Y-B. Let the neutral to R phase voltage be Vm sinωt. The first SCR (connected to R phase) is fired at an angle of 15°. What is the maximum value at the output terminals at this instant?- a)Vm/√2
- b)Vm
- c)1.5 Vm
- d)3 Vm
Correct answer is option 'C'. Can you explain this answer?
A 3-phase bridge converter is given a three-phase supply in the phase sequence R-Y-B. Let the neutral to R phase voltage be Vm sinωt. The first SCR (connected to R phase) is fired at an angle of 15°. What is the maximum value at the output terminals at this instant?
a)
Vm/√2
b)
Vm
c)
1.5 Vm
d)
3 Vm
| | Alok Khanna answered |
In case of a 3 phase bridge converter, the maximum value of voltage at the output terminal is always 1.5 Vm.
Assertion (A): The terminal voltage of a voltage source inverter remains substantially constant with variations in load.
Reason (R): Any short-circuit across the terminals of a voltage source inverter causes current to rise very fast.- a)Both A and R are true and R is the correct explanation of A
- b)Both A and R are true but R is not the correct explanation of A
- c)A is true but R is false
- d)A is false but R is true
Correct answer is option 'B'. Can you explain this answer?
Assertion (A): The terminal voltage of a voltage source inverter remains substantially constant with variations in load.
Reason (R): Any short-circuit across the terminals of a voltage source inverter causes current to rise very fast.
Reason (R): Any short-circuit across the terminals of a voltage source inverter causes current to rise very fast.
a)
Both A and R are true and R is the correct explanation of A
b)
Both A and R are true but R is not the correct explanation of A
c)
A is true but R is false
d)
A is false but R is true
| | Sharmila Bajaj answered |
Explanation:
The correct answer is option 'B': Both A and R are true but R is not the correct explanation of A.
Assertion (A): The terminal voltage of a voltage source inverter remains substantially constant with variations in load.
Reason (R): Any short-circuit across the terminals of a voltage source inverter causes current to rise very fast.
Explanation:
Terminal Voltage of a Voltage Source Inverter:
- A voltage source inverter (VSI) is an electronic device that converts a DC voltage source into an AC voltage source.
- The terminal voltage of a VSI refers to the voltage across the output terminals of the inverter.
- Ideally, the terminal voltage of a VSI should remain constant with variations in load.
Reason Explanation:
- The reason states that any short-circuit across the terminals of a VSI causes current to rise very fast.
- This statement is true because in a short-circuit condition, the impedance across the terminals becomes very low, resulting in a high current flow.
- However, this reason does not directly explain why the terminal voltage of a VSI remains constant with load variations.
Explanation of Assertion:
- The assertion states that the terminal voltage of a VSI remains substantially constant with variations in load.
- This assertion is true because VSIs are designed to regulate the output voltage regardless of the load variations.
- VSIs achieve this regulation by using control techniques such as pulse width modulation (PWM).
- PWM adjusts the width of the output pulses based on the load requirements, ensuring that the average output voltage remains constant.
Conclusion:
- Both the assertion and reason are true.
- However, the reason does not provide a correct explanation for the assertion.
- The terminal voltage of a VSI remains constant with variations in load due to the control techniques used, not solely because of the potential of a short-circuit.
The correct answer is option 'B': Both A and R are true but R is not the correct explanation of A.
Assertion (A): The terminal voltage of a voltage source inverter remains substantially constant with variations in load.
Reason (R): Any short-circuit across the terminals of a voltage source inverter causes current to rise very fast.
Explanation:
Terminal Voltage of a Voltage Source Inverter:
- A voltage source inverter (VSI) is an electronic device that converts a DC voltage source into an AC voltage source.
- The terminal voltage of a VSI refers to the voltage across the output terminals of the inverter.
- Ideally, the terminal voltage of a VSI should remain constant with variations in load.
Reason Explanation:
- The reason states that any short-circuit across the terminals of a VSI causes current to rise very fast.
- This statement is true because in a short-circuit condition, the impedance across the terminals becomes very low, resulting in a high current flow.
- However, this reason does not directly explain why the terminal voltage of a VSI remains constant with load variations.
Explanation of Assertion:
- The assertion states that the terminal voltage of a VSI remains substantially constant with variations in load.
- This assertion is true because VSIs are designed to regulate the output voltage regardless of the load variations.
- VSIs achieve this regulation by using control techniques such as pulse width modulation (PWM).
- PWM adjusts the width of the output pulses based on the load requirements, ensuring that the average output voltage remains constant.
Conclusion:
- Both the assertion and reason are true.
- However, the reason does not provide a correct explanation for the assertion.
- The terminal voltage of a VSI remains constant with variations in load due to the control techniques used, not solely because of the potential of a short-circuit.
In the integral cycle control of ac voltage controller, is the load is on for n cycles and off for m cycles, then the periodicity is given by? Consider the output is sinusoidal.- a)m/2π(m+n)
- b)n/2π(m+n)
- c)m/π(m+n)
- d)n/π(m+n)
Correct answer is option 'B'. Can you explain this answer?
In the integral cycle control of ac voltage controller, is the load is on for n cycles and off for m cycles, then the periodicity is given by? Consider the output is sinusoidal.
a)
m/2π(m+n)
b)
n/2π(m+n)
c)
m/π(m+n)
d)
n/π(m+n)
| | Debanshi Iyer answered |
Over a complete cycle of 2π x (on cycles + off cycles) the power is delivered for n cycles.
Hence, P = n/2π(m+n).
Hence, P = n/2π(m+n).
Assertion (A) : In 3-phase converters, the ripple frequency of the converter output voltage is higher than in single phase converter.
Reason (R) : The load current is mostly discontinuous in 3-phase converters.- a)Both A and R are true and R is the correct explanation of A.
- b)Both A and R are true but R is not the correct explanation of A.
- c)A is true but R is false.
- d)A is false but R is true.
Correct answer is option 'C'. Can you explain this answer?
Assertion (A) : In 3-phase converters, the ripple frequency of the converter output voltage is higher than in single phase converter.
Reason (R) : The load current is mostly discontinuous in 3-phase converters.
Reason (R) : The load current is mostly discontinuous in 3-phase converters.
a)
Both A and R are true and R is the correct explanation of A.
b)
Both A and R are true but R is not the correct explanation of A.
c)
A is true but R is false.
d)
A is false but R is true.
| | Avik Iyer answered |
The load current is mostly continuous in 3-phase converters. Hence reason is a false statement.
A four-quadrant chopper is driving a separately excited dc motor load. The motor parameters are R = 0.1 Ω, L = 10 mH. The supply voltage is 200 V d.c. If the rated current of the motor is 10 A with Eb = 150 V and motor is driving the rated torque, then it is operating under- a)forward motoring mode
- b)reverse braking mode
- c)reverse motoring mode
- d)forward brakina mode
Correct answer is option 'A'. Can you explain this answer?
A four-quadrant chopper is driving a separately excited dc motor load. The motor parameters are R = 0.1 Ω, L = 10 mH. The supply voltage is 200 V d.c. If the rated current of the motor is 10 A with Eb = 150 V and motor is driving the rated torque, then it is operating under
a)
forward motoring mode
b)
reverse braking mode
c)
reverse motoring mode
d)
forward brakina mode
| | Hiral Kulkarni answered |
For a four-quadrant chopper, the average voltage in all the four-modes is given by
V0 = 2 Vs (α - 0.5)
The average curreni,
or,
or, α = 0.877
Since α > 0.5, the motor is operating undi
forward motorina mode.
V0 = 2 Vs (α - 0.5)
The average curreni,
or,
or, α = 0.877
Since α > 0.5, the motor is operating undi
forward motorina mode.
In a single-pulse modulation of PWM inverters if pulse width is 120° then- a)5th harmonic will be eliminated
- b)3rd harmonic will be eliminated
- c)7th harmonic will be eliminated
- d)none of the above
Correct answer is option 'B'. Can you explain this answer?
In a single-pulse modulation of PWM inverters if pulse width is 120° then
a)
5th harmonic will be eliminated
b)
3rd harmonic will be eliminated
c)
7th harmonic will be eliminated
d)
none of the above
| | Manoj Chaudhary answered |
The rms value of amplitude of harmonic voltage of a single, pulse modulated wave is given by
(where, p = width of pulse an Vdc = supply dc voltage)
If the 3rd harmonic Is to be eliminated, then
EL3 = 0
i.e.
or,
or,
= Required pulse width
(where, p = width of pulse an Vdc = supply dc voltage)
If the 3rd harmonic Is to be eliminated, then
EL3 = 0
i.e.
or,
or,
= Required pulse width
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