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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?
Verified Answer
A four-quadrant chopper is driving a separately excited dc motor load....
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.
Most Upvoted Answer
A four-quadrant chopper is driving a separately excited dc motor load....
Ohms, L = 0.5 mH, J = 0.01 kgm^2, and B = 0.1 Nm/s. The chopper is operating with a switching frequency of 10 kHz and a duty cycle of 0.5. The input voltage to the chopper is 100 V.
To solve this problem, we need to use the equations of motion for the separately excited dc motor. These equations relate the voltage applied to the motor (which is controlled by the chopper) to the motor speed and current.
The equations of motion for the motor are:
V = R*i + L*di/dt + e
T = J*dω/dt + B*ω
where V is the voltage applied to the motor, R is the motor resistance, L is the motor inductance, i is the motor current, e is the back EMF (which is proportional to motor speed ω), T is the motor torque, J is the motor inertia, and B is the motor damping coefficient.
To solve for the motor speed and current, we need to solve these equations simultaneously. First, we can simplify the equations by assuming that the motor is in steady state (i.e. di/dt = dω/dt = 0). Then, we can solve for the current i:
i = (V - e)/R
To solve for the back EMF e, we can use the equation:
e = kω
where k is a constant that depends on the motor parameters. For a separately excited dc motor, k = φ*ω_m, where φ is the motor flux and ω_m is the angular speed of the motor. Assuming that the motor flux is constant, we can write:
e = k*ω = const.
Therefore, we can solve for the motor speed ω:
ω = e/k = (V - Ri)/k
Now we can calculate the torque produced by the motor:
T = J*B*ω/(R^2 + L^2*ω^2)
Plugging in the values given in the problem, we get:
i = (100 - 0.1*ω)/(0.1)
e = const.
ω = (100 - 0.1*i)/(0.01*π*10^4) = 318.3 rad/s
T = 0.01*0.1*318.3/(0.1^2 + 0.5*10^-3*(318.3)^2) = 0.32 Nm
Therefore, the motor is producing a torque of 0.32 Nm.
To solve this problem, we need to use the equations of motion for the separately excited dc motor. These equations relate the voltage applied to the motor (which is controlled by the chopper) to the motor speed and current.
The equations of motion for the motor are:
V = R*i + L*di/dt + e
T = J*dω/dt + B*ω
where V is the voltage applied to the motor, R is the motor resistance, L is the motor inductance, i is the motor current, e is the back EMF (which is proportional to motor speed ω), T is the motor torque, J is the motor inertia, and B is the motor damping coefficient.
To solve for the motor speed and current, we need to solve these equations simultaneously. First, we can simplify the equations by assuming that the motor is in steady state (i.e. di/dt = dω/dt = 0). Then, we can solve for the current i:
i = (V - e)/R
To solve for the back EMF e, we can use the equation:
e = kω
where k is a constant that depends on the motor parameters. For a separately excited dc motor, k = φ*ω_m, where φ is the motor flux and ω_m is the angular speed of the motor. Assuming that the motor flux is constant, we can write:
e = k*ω = const.
Therefore, we can solve for the motor speed ω:
ω = e/k = (V - Ri)/k
Now we can calculate the torque produced by the motor:
T = J*B*ω/(R^2 + L^2*ω^2)
Plugging in the values given in the problem, we get:
i = (100 - 0.1*ω)/(0.1)
e = const.
ω = (100 - 0.1*i)/(0.01*π*10^4) = 318.3 rad/s
T = 0.01*0.1*318.3/(0.1^2 + 0.5*10^-3*(318.3)^2) = 0.32 Nm
Therefore, the motor is producing a torque of 0.32 Nm.
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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 undera)forward motoring modeb)reverse braking modec)reverse motoring moded)forward brakina modeCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 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 undera)forward motoring modeb)reverse braking modec)reverse motoring moded)forward brakina modeCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 undera)forward motoring modeb)reverse braking modec)reverse motoring moded)forward brakina modeCorrect 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 undera)forward motoring modeb)reverse braking modec)reverse motoring moded)forward brakina modeCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about 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 undera)forward motoring modeb)reverse braking modec)reverse motoring moded)forward brakina modeCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 undera)forward motoring modeb)reverse braking modec)reverse motoring moded)forward brakina modeCorrect answer is option 'A'. Can you explain this answer?.
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