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All questions of AC Voltage Controllers & Cyclo Converters for Electrical Engineering (EE) Exam

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?

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 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?

Amar Sengupta answered
Given:
- Input voltage (V) = 230 V
- Load resistance (R) = 15 Ω
- On time (Ton) = 6 cycles
- Off time (Toff) = 4 cycles

To find:
- Power factor (pf)

Solution:
Step 1: Calculate the time period (T) of each cycle.
The time period can be calculated using the formula:
T = 1/frequency

Since it is not mentioned, let's assume the frequency as 50 Hz (standard frequency in most countries).
So, T = 1/50 = 0.02 seconds

Step 2: Calculate the total time for one cycle (Ttotal).
Ttotal = Ton + Toff

Given:
Ton = 6 cycles
Toff = 4 cycles

Ttotal = 6T + 4T = 10T
Ttotal = 10 * 0.02 = 0.2 seconds

Step 3: Calculate the average power (Pavg) delivered to the load.
The average power can be calculated as:
Pavg = V^2/R

Given:
V = 230 V
R = 15 Ω

Pavg = (230^2)/15 = 52900/15 = 3526.67 W

Step 4: Calculate the apparent power (S) delivered to the load.
The apparent power can be calculated as:
S = V * I

In a resistive load, the current can be calculated using Ohm's Law:
I = V/R

Substituting the values:
I = 230/15 = 15.33 A

S = 230 * 15.33 = 3524.9 VA

Step 5: Calculate the power factor (pf).
The power factor can be calculated as:
pf = Pavg / S

Substituting the values:
pf = 3526.67 / 3524.9
pf ≈ 0.9995

Step 6: Round off the power factor to the nearest decimal place.
Since the given options are rounded to three decimal places, the power factor is approximately 0.775.

Therefore, the correct answer is option B: 0.7746.

From the below given statements regarding sequence control of ac voltage, which of them are true?
i) It improves system power factor
ii) It reduces the harmonic content at the output
iii) Wider control of output voltage is possible
  • a)
    only (i)
  • b)
    only (i) and (ii)
  • c)
     only (i) and (iii)
  • d)
    all of them are true
Correct answer is option 'D'. Can you explain this answer?

Shivani Saha answered
Improving System Power Factor

The statement (i) states that sequence control of AC voltage improves system power factor. This statement is true.

- Sequence control of AC voltage refers to the control of individual phase voltages in a three-phase system. By controlling the magnitude and phase angle of each phase voltage, it is possible to improve the power factor of the system.
- Power factor is a measure of how effectively electrical power is being used in a system. It is the ratio of real power to apparent power. A higher power factor indicates a more efficient use of electrical power.
- By adjusting the sequence of the phase voltages, it is possible to optimize the power factor of the system. This can be achieved by balancing the reactive power between the three phases, minimizing the reactive power flow in the system, and reducing the overall power losses.
- Therefore, by implementing sequence control of AC voltage, it is possible to improve the system power factor and achieve a more efficient use of electrical power.

Reducing Harmonic Content at the Output

The statement (ii) states that sequence control of AC voltage reduces the harmonic content at the output. This statement is true.

- Harmonics are unwanted frequencies that can distort the waveform of an electrical signal. They are typically caused by nonlinear loads in the system, such as power electronics devices, and can lead to various issues such as increased power losses, heating of equipment, and interference with other sensitive equipment.
- By implementing sequence control of AC voltage, it is possible to reduce the harmonic content at the output. This can be achieved by controlling the phase angle and magnitude of each phase voltage, which helps to minimize the distortion in the waveform and reduce the harmonic content.
- By reducing the harmonic content at the output, it is possible to improve the quality of the electrical signal and ensure that it meets the required standards and specifications.

Wider Control of Output Voltage

The statement (iii) states that wider control of output voltage is possible with sequence control of AC voltage. This statement is true.

- Sequence control of AC voltage allows for individual control of each phase voltage in a three-phase system. This enables a wider range of control over the output voltage.
- By adjusting the magnitude and phase angle of each phase voltage, it is possible to achieve a wide range of output voltage levels. This can be beneficial in various applications where precise control of voltage is required, such as in motor control, power supply regulation, and voltage conditioning.
- The ability to have wider control over the output voltage allows for greater flexibility and versatility in system design and operation.

Overall, all the given statements are true. Sequence control of AC voltage improves system power factor, reduces the harmonic content at the output, and enables wider control of the output voltage.

In three phase cycloconverters, the reduction factor is given by
  • a)
    input frequency/output frequency
  • b)
    (input frequency/output frequency) -1
  • c)
    (input frequency/output frequency) -1/2
  • d)
    (input frequency/output frequency) 1/2
Correct answer is option 'B'. Can you explain this answer?

Ameya Gupta answered
Reduction factor in three-phase cycloconverters

The reduction factor in a three-phase cycloconverter refers to the ratio of the input frequency to the output frequency. It is an important parameter that determines the level of frequency conversion achieved by the cycloconverter.

Working principle of three-phase cycloconverters

Three-phase cycloconverters are power electronic devices used to convert the frequency of three-phase AC power. They consist of a set of thyristor switches that control the flow of power to the output. The cycloconverter operates by switching the thyristors in a controlled manner to create the desired output frequency.

The input to the cycloconverter is a three-phase AC supply with a fixed frequency, typically 50 Hz or 60 Hz. The output frequency can be lower or higher than the input frequency, depending on the application requirements.

Reduction factor formula

The reduction factor is given by the formula:

Reduction factor = (Input frequency / Output frequency) - 1

Explanation of the formula

The reduction factor formula can be derived by considering the operating principle of the cycloconverter.

- In a three-phase cycloconverter, each cycle of the output frequency is composed of a certain number of cycles of the input frequency.
- The number of input cycles required to complete one output cycle is determined by the reduction factor.
- For example, if the reduction factor is 2, it means that two cycles of the input frequency are required to complete one cycle of the output frequency.
- Therefore, the reduction factor can be calculated by dividing the input frequency by the output frequency and subtracting 1.

Significance of the reduction factor

The reduction factor determines the level of frequency conversion achieved by the cycloconverter. It indicates how many cycles of the input frequency are required to generate one cycle of the output frequency.

- A reduction factor of 1 means that the output frequency is the same as the input frequency, resulting in no frequency conversion.
- A reduction factor less than 1 means that the output frequency is higher than the input frequency, resulting in up-conversion.
- A reduction factor greater than 1 means that the output frequency is lower than the input frequency, resulting in down-conversion.

Conclusion

In three-phase cycloconverters, the reduction factor is given by the formula (input frequency / output frequency) - 1. This formula provides a quantitative measure of the frequency conversion achieved by the cycloconverter. The reduction factor determines the ratio of input cycles to output cycles and indicates whether the output frequency is higher or lower than the input frequency.

The below shown controller circuit is a
  • a)
    half wave controller
  • b)
    full wave controller
  • c)
    none of the mentioned
  • d)
    will depend upon the firing angle
Correct answer is option 'A'. Can you explain this answer?

Sneha Bose answered
As it consists one diode and one SCR only, the control is only in one cycle (the positive half in this case), hence it is a half wave controller.

Applications of cycloconverters include
  • a)
    speed control of ac drives
  • b)
    induction heating
  • c)
    static VAr compensation
  • d)
    all of the mentioned
Correct answer is option 'D'. Can you explain this answer?

Palak Verma answered
Applications of Cycloconverters

Cycloconverters are power electronic devices that convert AC power at one frequency to AC power at another frequency. They are widely used in various industrial applications due to their versatility and ability to provide variable frequency and voltage output. Some of the common applications of cycloconverters include:

1. Speed Control of AC Drives:
Cycloconverters are extensively used in speed control systems for AC drives, especially in large industrial applications. By using cycloconverters, the speed of AC motors can be controlled efficiently and smoothly. This is achieved by varying the output frequency and voltage of the cycloconverter, which in turn controls the speed of the motor. This application is particularly useful in industries where precise control of motor speed is required, such as in conveyor systems, elevators, and machine tools.

2. Induction Heating:
Cycloconverters are also used in induction heating applications where high-frequency power is required. Induction heating is a process in which an electrically conductive material is heated by electromagnetic induction. Cycloconverters provide the necessary high-frequency power to generate the electromagnetic field required for induction heating. This application is commonly used in industries such as metal processing, forging, and heat treatment.

3. Static VAr Compensation:
Cycloconverters are utilized in static VAr compensation systems to regulate the reactive power in power transmission and distribution networks. Reactive power is required to maintain voltage stability in the grid and compensate for the reactive power demand of inductive loads. By using cycloconverters, the reactive power can be controlled and adjusted according to the system requirements. This helps in reducing voltage fluctuations, improving power factor, and enhancing the overall efficiency of the power system.

4. Other Applications:
Apart from the above-mentioned applications, cycloconverters find use in various other industrial and commercial applications. These include frequency conversion for testing electrical equipment, voltage conversion for specialized machinery, variable speed control for fans and pumps, and AC power generation for aerospace and marine applications.

In conclusion, cycloconverters have a wide range of applications in the field of power electronics. They are extensively used for speed control of AC drives, induction heating, static VAr compensation, and various other industrial applications. The ability to provide variable frequency and voltage output makes them highly versatile and suitable for a diverse range of applications.

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 power delivered to the load.
  • a)
    2.1 W
  • b)
    2.1 kW
  • c)
    516 W
  • d)
    5.16 kW
Correct answer is option 'B'. Can you explain this answer?

Bibek Saha answered
Calculation:

Given:
Input voltage = 230 V
Load resistance = 15 Ω
Cycles on = 6
Cycles off = 4

To find the power delivered to the load, we need to calculate the average power.

Average Power:
The average power delivered to a resistive load can be calculated using the formula:

Average Power = (1/T) * ∫(0 to T) [v(t) * i(t)] dt

Where:
T = Time period of the waveform
v(t) = Instantaneous voltage
i(t) = Instantaneous current

Time Period:
The time period of the waveform is the sum of the on and off cycles:

T = (cycles on + cycles off) * Time period of one cycle

Since one cycle consists of both on and off periods, the time period of one cycle can be found using the frequency:

f = 1 / Time period of one cycle

Frequency:
The frequency can be calculated using the formula:

f = 1 / Time period

Time Period of One Cycle:
The time period of one cycle can be calculated by dividing the time period by the sum of on and off cycles:

Time period of one cycle = Time period / (cycles on + cycles off)

Calculating Time period:
The time period can be calculated using the formula:

Time period = 1 / Frequency

Calculating Frequency:
The frequency can be calculated using the formula:

Frequency = 1 / Time period

Calculating Time period of One Cycle:
Substituting the values, we get:

Time period of one cycle = (1 / Frequency) / (cycles on + cycles off)

Calculating Average Power:
Substituting the values of voltage and current into the formula for average power, we get:

Average Power = (1 / Time period) * ∫(0 to Time period) [(v(t))^2 / R] dt

Since the load is resistive, the instantaneous current is given by Ohm's law:

i(t) = v(t) / R

Substituting this into the formula for average power, we get:

Average Power = (1 / Time period) * ∫(0 to Time period) [(v(t))^2 / R] dt
= (1 / Time period) * ∫(0 to Time period) [(v(t))^2 / R] dt
= (1 / Time period) * (1 / R) * ∫(0 to Time period) [(v(t))^2] dt

Since the voltage waveform is a sinusoidal waveform, the integral of the square of the voltage waveform over one time period is equal to the square of the RMS voltage:

∫(0 to Time period) [(v(t))^2] dt = (Vrms)^2 * Time period

Substituting this back into the formula for average power, we get:

Average Power = (1 / Time period) * (1 / R) * (Vrms)^2 * Time period
= (Vrms)^2 / R

Calculating RMS Voltage:
The RMS voltage can be calculated using the formula:

Vrms = Vm / √2

Where:
Vm = Maximum voltage

Calculating Maximum

In AC voltage controllers the
  • a)
    variable ac with fixed frequency is obtained
  • b)
    variable ac with variable frequency is obtained
  • c)
    variable dc with fixed frequency is obtained
  • d)
    variable dc with variable frequency is obtained
Correct answer is option 'A'. Can you explain this answer?

Partho Saha answered
AC Voltage Controllers

AC voltage controllers are devices that allow for the control of AC voltage by varying the amplitude of the voltage waveform. They are commonly used in applications where precise control of AC voltage is required, such as in heating, lighting, and motor speed control.

Variable AC with Fixed Frequency

The main function of an AC voltage controller is to provide variable AC voltage with a fixed frequency. This is achieved by controlling the amplitude of the AC voltage waveform. The AC voltage controller uses a thyristor or a silicon-controlled rectifier (SCR) to control the voltage.

The thyristor or SCR is a solid-state device that can be turned on and off by a small control signal. When the thyristor is turned on, it conducts current in only one direction, allowing the AC voltage waveform to pass through. When the thyristor is turned off, it blocks the current flow and stops the AC voltage waveform.

The AC voltage controller uses a control circuit to turn the thyristor on and off at specific times during the AC voltage waveform. By varying the timing of the thyristor, the amplitude of the AC voltage waveform can be controlled, providing variable AC voltage with a fixed frequency.

Advantages of AC Voltage Controllers

AC voltage controllers have several advantages over other voltage control methods, including:

- High efficiency
- Precise control of AC voltage
- Low maintenance
- Compact size
- Low cost

Applications of AC Voltage Controllers

AC voltage controllers are commonly used in applications where precise control of AC voltage is required, such as:

- Heating systems
- Lighting systems
- Motor speed control
- Power supplies
- Welding equipment

Conclusion

In conclusion, AC voltage controllers provide variable AC voltage with a fixed frequency by controlling the amplitude of the AC voltage waveform using a thyristor or SCR. They have several advantages over other voltage control methods and are commonly used in applications where precise control of AC voltage is required.

In a N-stage sequence controller, each secondary is rated for __________
  • a)
    n x Vs
  • b)
    Vs
  • c)
    Vs/n
  • d)
    Vs x (n-1)
Correct answer is option 'C'. Can you explain this answer?

Surbhi Chopra answered
In a N-stage sequence controller, each secondary is rated for Vs/n.

Explanation:
A sequence controller is a device used in power systems to control the sequential operation of various electrical equipment. It ensures that the equipment starts and stops in a specific order to prevent overloading or damage.

In a N-stage sequence controller, there are N secondary windings, each connected to a separate electrical load. The primary winding is connected to the power supply. The secondary windings are designed to have a specific voltage rating, Vs.

The voltage rating of each secondary winding in a N-stage sequence controller is given by Vs/n, where n represents the number of stages or secondary windings.

The reason for this is that the voltage across each secondary winding should be reduced to ensure proper operation of the connected electrical load. When multiple secondary windings are connected in a sequence controller, the total voltage is divided equally among them.

To achieve this division of voltage, the voltage rating of each secondary winding is set to Vs/n. This ensures that the voltage across each secondary winding is Vs/n, and the total voltage across all the secondary windings is Vs.

By providing the secondary windings with a voltage rating of Vs/n, the sequence controller ensures that each connected load receives the appropriate voltage level for its operation. This prevents overloading or damage to the electrical equipment.

Therefore, the correct answer is option c) Vs/n, where each secondary winding in a N-stage sequence controller is rated for Vs/n.

Pulse gating is suitable for
  • a)
    R loads only
  • b)
    R and RL loads
  • c)
    RL loads only
  • d)
    all types of loads
Correct answer is option 'A'. Can you explain this answer?

Malavika Nair answered
In RL loads with pulse gating, the incoming SCR may be fired during the interval when it is reversed biased by the outgoing SCR, thus it won’t get turn on even after the outgoing SCR is not reveres basing the incoming SCR is the pulse width is over before forward biasing the SCR.

AC voltage controllers convert
  • a)
    fixed ac to fixed dc
  • b)
    variable ac to variable dc
  • c)
    fixed ac to variable ac
  • d)
    variable ac to fixed ac
Correct answer is option 'C'. Can you explain this answer?

Hiral Kulkarni answered
AC voltage controllers are electronic devices used to control the voltage of an alternating current (AC) power supply. They are commonly used in applications where precise control of the output voltage is required, such as in industrial machinery, motor speed control, and power supplies for electronic devices.

The correct answer is option 'C', which states that AC voltage controllers convert fixed AC to variable AC. This means that they take a fixed AC input voltage and provide a variable AC output voltage, allowing for control and adjustment of the voltage level.

Here is a detailed explanation of why option 'C' is the correct answer:

1. AC to AC Conversion:
AC voltage controllers are designed to perform AC to AC conversion. They take an AC input voltage, which is typically fixed at a specific frequency and amplitude, and convert it into a variable AC output voltage.

2. Fixed AC input:
The input voltage to an AC voltage controller is usually fixed at a specific frequency and amplitude. This means that the input voltage remains constant throughout the operation of the controller. It could be sourced from the mains power supply or from a generator.

3. Variable AC output:
The primary function of an AC voltage controller is to provide a variable AC output voltage. This means that the controller can adjust the output voltage level based on the desired requirements. The output voltage can be increased or decreased as needed, allowing for precise control of the voltage.

4. Control methods:
AC voltage controllers use various control methods to adjust the output voltage. One common method is phase control, where the firing angle of the thyristors or other solid-state switching devices is adjusted to control the portion of the input waveform that is transferred to the output. This allows for smooth and continuous control of the output voltage.

5. Applications:
AC voltage controllers find applications in various industries and systems. They are commonly used in motor speed control, where the output voltage is adjusted to control the speed of AC motors. They are also used in power supplies for electronic devices, where the output voltage needs to be precisely regulated.

In conclusion, AC voltage controllers convert fixed AC input voltage to variable AC output voltage. They are used in applications where precise control of the voltage level is required.

In the integral cycle control method with duty cycle = k and maximum load current = Im. Find the value of average SCR current.
  • a)
    Im/k.π
  • b)
    Im
  • c)
    k.Im
  • d)
    k.Im/π
Correct answer is option 'D'. Can you explain this answer?

Gargi Mishra answered
As each SCR conducts for π radians during each cycle of n on cycles, the average value of SCR current is
I = (k/2π) ∫Im sin ωt d(ωt). Where the integration is from 0 to π.

 In the positive half cycle from ωt = 0 to π
  • a)
    P1 and P2 are forward biased
  • b)
    N1 and P2 are forward biased
  • c)
    P1 and N2 are forward biased
  • d)
    None of the mentioned
Correct answer is option 'C'. Can you explain this answer?

Pooja Patel answered
In the positive half cycle, upper terminal is positive of the upper secondary and the lower terminal is negative for the lower secondary. Hence, P1 and N2 are forward biased.

In the below given cycloconverter circuit, _________ and _________ conduct in one cycle together.
  • a)
    P1, P2 and N1, N2
  • b)
    P1, N2 and N1, P2
  • c)
    N1, P1 and N1, P2
  • d)
    None of the mentioned
Correct answer is option 'B'. Can you explain this answer?

Pooja Patel answered
In the positive half cycle, upper terminal is positive of the upper secondary and the lower terminal is negative for the lower secondary. Hence, P1 and N2 are forward biased. And likewise, N1 and P2 will conduct in the next half cycle.

In the below shown circuit, the diode conducts for
  • a)
    90°
  • b)
    > 90°
  • c)
    < 90°
  • d)
Correct answer is option 'A'. Can you explain this answer?

Poulomi Chopra answered
The diode conducts from π to 2π in the negative half cycle. If it were a non resistive load, the diode would conduct for less than 90°, as the inductor would force conduct the SCR for some time.

The below given circuit has Vs = 230V and R = 20 Ω. Find the value of the average output voltage at the R load for a firing angle of 45°.
  • a)
    224 V
  • b)
    -15.17 V
  • c)
    15.17 V
  • d)
    –224 V
Correct answer is option 'B'. Can you explain this answer?

Poulomi Chopra answered
Vo = [(√2 x 230) x (cos45 – 1)]/2π = -15.17 V.
Negative value is due to the fact that the average value in the positive half cycle is less than that in the negative half cycle.

The below given output voltage waveform can be obtained by a
  • a)
    half wave ac voltage controller
  • b)
    full wave ac voltage controller
  • c)
    half wave controller with firing angle = 0° for T1
  • d)
    full wave controller with firing angle = 0° for both T1 and T2
Correct answer is option 'A'. Can you explain this answer?

Gargi Mishra answered
As the positive half is chopped off due to some value of firing angle, the firing angle is not equal to zero. As the negative half is a half sine wave, either it is a full wave controller with firing angle for T2 set to zero or it is a half wave controller with a Thyristor (T1) and a diode.

The below given circuit has Vs = 230V and R = 20 Ω. Find the value of the average output load current at the R load for a firing angle of 45°.
  • a)
    – 0.7585
  • b)
    0.7585
  • c)
    -0.6396
  • d)
    -0.5
Correct answer is option 'A'. Can you explain this answer?

Poulomi Ahuja answered
Vo = [(√2 x 230) x (cos45 – 1)]/2π = -15.17 V.
Io = (Vo)/R = -0.7585
Negative value is due to the fact that the average value in the positive half cycle is less than that in the negative half cycle.

 If T1, T2 are kept off and T3, T4 are having a firing angle of 180° each, then the output voltage is ____ if the turns ratio is 1:1
  • a)
    Vs
  • b)
    √2Vs
  • c)
    0
  • d)
    2Vs
Correct answer is option 'C'. Can you explain this answer?

 If T3 and T4 are fired at an angle of 180°, they are equivalent to an off switch, as they are fired and naturally commutated at the same time, hence they are never on.
Hence, as all the four SCRs are always off, the output voltage is zero.

For obtaining the output voltage control from V to 2V, the firing angle must be
  • a)
    always 0 for T3, T4
  • b)
    always 180 for T3, T4
  • c)
    0 to 180 from T3, T4
  • d)
    none of the mentioned
Correct answer is option 'A'. Can you explain this answer?

Sanskriti Kaur answered
For obtaining the voltage control from V to 2V, the lower controller i.e. T3, T4 pair must always be on hence the firing angle from them must always be 0° ideally.

For obtaining the output voltage control from V to 2V, the firing angle must be
  • a)
    always 0 for T1, T2
  • b)
    always 180 for T1, T2
  • c)
    0 to 180 from T1, T2
  • d)
    none of the mentioned
Correct answer is option 'C'. Can you explain this answer?

 For obtaining the voltage control from V to 2V, the upper controller must be controlled by varying its firing angle from 0 to 180° whereas the T3, T4 pair must always be on hence the firing angle from them must always be 0° ideally.

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