GATE Exam > GATE Questions > A 200 V DC series motor, when operating from ...
Start Learning for Free
A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 Ω. The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .
Correct answer is '823 to 827'. Can you explain this answer?
Most Upvoted Answer
A 200 V DC series motor, when operating from rated voltage while drivi...
To determine the new speed of the motor when the load torque is increased by 44%, we need to analyze the motor's characteristics and use the torque-speed relationship.
1. Initial Parameters:
- Supply voltage (V) = 200 V
- Current (I) = 10 A
- Speed (N) = 1000 RPM
- Total series resistance (R) = 1 Ω
2. Calculating Initial Torque:
The torque (T) of a DC series motor is given by the equation:
T = K * Φ * I
where K is a constant, Φ is the flux, and I is the current.
Since the magnetic circuit is assumed to be linear, the flux Φ is directly proportional to the supply voltage V:
Φ ∝ V
Therefore, the torque T is directly proportional to the square of the supply voltage:
T ∝ V^2
We can calculate the initial torque by substituting the values:
T_initial = K * V^2 * I
3. Calculating New Torque:
When the load torque is increased by 44%, the new torque can be expressed as:
T_new = 1.44 * T_initial
4. Calculating New Speed:
The torque-speed relationship of a DC series motor is given by the equation:
T = K * (N - No)
where No is the no-load speed.
From the torque equation, we can rearrange it to solve for the speed:
N = T/K + No
Substituting the values, we have:
N_new = (1.44 * T_initial) / K + No
5. Finding the No-Load Speed:
To find the no-load speed, we need the torque at no-load conditions. At no-load, the torque is zero, so we can rewrite the torque equation as:
0 = K * (N_no-load - No)
N_no-load = No
Therefore, the no-load speed is equal to the initial speed:
N_no-load = N_initial = 1000 RPM
6. Final Calculation:
Substituting the values into the new speed equation:
N_new = (1.44 * T_initial) / K + N_no-load
Since N_no-load = N_initial, we have:
N_new = (1.44 * T_initial) / K + N_initial
Now, we know that the motor is operating from the same supply voltage, so the constant K remains the same. Therefore, the new speed can be calculated as:
N_new = (1.44 * T_initial) / K + N_initial
Substituting the values, we get:
N_new = (1.44 * (K * V^2 * I)) / K + N_initial
Simplifying the equation:
N_new = 1.44 * V^2 * I + N_initial
Plugging in the given values:
N_new = 1.44 * 200^2 * 10 + 1000
N_new ≈ 823 to 827 RPM
Hence, the speed of the motor, when the load torque is increased by 44%, is approximately 823 to 827 RPM.
1. Initial Parameters:
- Supply voltage (V) = 200 V
- Current (I) = 10 A
- Speed (N) = 1000 RPM
- Total series resistance (R) = 1 Ω
2. Calculating Initial Torque:
The torque (T) of a DC series motor is given by the equation:
T = K * Φ * I
where K is a constant, Φ is the flux, and I is the current.
Since the magnetic circuit is assumed to be linear, the flux Φ is directly proportional to the supply voltage V:
Φ ∝ V
Therefore, the torque T is directly proportional to the square of the supply voltage:
T ∝ V^2
We can calculate the initial torque by substituting the values:
T_initial = K * V^2 * I
3. Calculating New Torque:
When the load torque is increased by 44%, the new torque can be expressed as:
T_new = 1.44 * T_initial
4. Calculating New Speed:
The torque-speed relationship of a DC series motor is given by the equation:
T = K * (N - No)
where No is the no-load speed.
From the torque equation, we can rearrange it to solve for the speed:
N = T/K + No
Substituting the values, we have:
N_new = (1.44 * T_initial) / K + No
5. Finding the No-Load Speed:
To find the no-load speed, we need the torque at no-load conditions. At no-load, the torque is zero, so we can rewrite the torque equation as:
0 = K * (N_no-load - No)
N_no-load = No
Therefore, the no-load speed is equal to the initial speed:
N_no-load = N_initial = 1000 RPM
6. Final Calculation:
Substituting the values into the new speed equation:
N_new = (1.44 * T_initial) / K + N_no-load
Since N_no-load = N_initial, we have:
N_new = (1.44 * T_initial) / K + N_initial
Now, we know that the motor is operating from the same supply voltage, so the constant K remains the same. Therefore, the new speed can be calculated as:
N_new = (1.44 * T_initial) / K + N_initial
Substituting the values, we get:
N_new = (1.44 * (K * V^2 * I)) / K + N_initial
Simplifying the equation:
N_new = 1.44 * V^2 * I + N_initial
Plugging in the given values:
N_new = 1.44 * 200^2 * 10 + 1000
N_new ≈ 823 to 827 RPM
Hence, the speed of the motor, when the load torque is increased by 44%, is approximately 823 to 827 RPM.
|
Explore Courses for GATE exam
|
|
Similar GATE Doubts
A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer?
Question Description
A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer? for GATE 2025 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer? covers all topics & solutions for GATE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer?.
A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer? for GATE 2025 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer? covers all topics & solutions for GATE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer?.
Solutions for A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free.
Here you can find the meaning of A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer?, a detailed solution for A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer? has been provided alongside types of A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .Correct answer is '823 to 827'. Can you explain this answer? tests, examples and also practice GATE tests.
|
|
Explore Courses for GATE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup