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A 230 V d.c. series motor develops its rated output at 1500 rpm while taking 20 A. Armature and series field resistance are 0.6 Q and 0.4 Q respectively. To obtain rated torque at 1000 rpm, external resistance mustbe added is____________
- a)4.5 Q
- b)3.5 Q
- c)2.5 Q
- d)2.0 Q
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A 230 V d.c. series motor develops its rated output at 1500 rpm while ...
- Initial voltage drop:
Armature + Series field = 0.6 Ω + 0.4 Ω = 1 Ω. At 20 A, voltage drop = 20 A 1 x Ω = 20 V. - Back EMF at 1500 rpm:
230 V - 20 V = 210 V. - Back EMF at 1000 rpm for same torque (current):
Back EMF ∝ Speed, so at 1000 rpm, Back EMF = (1000/1500) x 210 V = 140 V. - Total voltage drop needed:
230 V - 140 V = 90 V. - Additional resistance calculation:
Required voltage drop = 90 V - 20 V = 70 V. Resistance = 70 V / 20 A = 3.5 Ω. - Thus, the external resistance required is 3.5 Ω.
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Most Upvoted Answer
A 230 V d.c. series motor develops its rated output at 1500 rpm while ...
Solution:
Given data:
Voltage rating (V) = 230 V
Rated speed (N1) = 1500 rpm
Rated current (I1) = 20 A
Armature resistance (Ra) = 0.6 Ω
Series field resistance (Rs) = 0.4 Ω
Let's calculate the rated power of the motor.
P = V*I
= 230*20
= 4600 W
Let's calculate the rated torque of the motor.
ωm = 2πN/60
= 2π*1500/60
= 157.1 rad/s
T1 = P/ωm
= 4600/157.1
= 29.3 Nm
Now, let's calculate the external resistance required to obtain rated torque at 1000 rpm.
At 1000 rpm,
ωm = 2πN/60
= 2π*1000/60
= 104.7 rad/s
From the torque-speed characteristic of the series motor, we know that the torque is proportional to the square of the armature current.
T2/T1 = (I2/I1)^2
T2 = T1*(I2/I1)^2
At rated torque, I1 = I2
T2 = T1
At 1000 rpm, back EMF (Eb) can be calculated as follows:
Eb = V - Ia*Ra - Ia*Rs
At rated torque, Ia = I1
Eb = 230 - 20*0.6 - 20*0.4
= 220 V
At 1000 rpm,
Eb = K*Φ*N
220 = K*Φ*1000
Let's assume that the flux is constant.
K*Φ = 0.22
At 1000 rpm and rated torque, the armature current is given by:
Ia = (Eb - V)/Rs
= (220 - 230)/0.4
= -25 A
The negative sign indicates that the direction of the current is opposite to that of the voltage. This is due to the fact that the external resistance is added in series with the motor.
Let's calculate the external resistance required to obtain rated torque at 1000 rpm.
T2 = T1*(I2/I1)^2
T2/T1 = (N2/N1)
(I2/I1)^2 = (N2/N1)
(I2/I1)^2 = (1000/1500)
I2/I1 = 0.82
I2 = 0.82*20
= 16.4 A
Ia = (Eb - V)/Rs
-25 = (220 - 230 - I2*Ra)/Rs
Rs = 3.5 Ω
Therefore, the external resistance required to obtain rated torque at 1000 rpm is 3.5 Ω.
Given data:
Voltage rating (V) = 230 V
Rated speed (N1) = 1500 rpm
Rated current (I1) = 20 A
Armature resistance (Ra) = 0.6 Ω
Series field resistance (Rs) = 0.4 Ω
Let's calculate the rated power of the motor.
P = V*I
= 230*20
= 4600 W
Let's calculate the rated torque of the motor.
ωm = 2πN/60
= 2π*1500/60
= 157.1 rad/s
T1 = P/ωm
= 4600/157.1
= 29.3 Nm
Now, let's calculate the external resistance required to obtain rated torque at 1000 rpm.
At 1000 rpm,
ωm = 2πN/60
= 2π*1000/60
= 104.7 rad/s
From the torque-speed characteristic of the series motor, we know that the torque is proportional to the square of the armature current.
T2/T1 = (I2/I1)^2
T2 = T1*(I2/I1)^2
At rated torque, I1 = I2
T2 = T1
At 1000 rpm, back EMF (Eb) can be calculated as follows:
Eb = V - Ia*Ra - Ia*Rs
At rated torque, Ia = I1
Eb = 230 - 20*0.6 - 20*0.4
= 220 V
At 1000 rpm,
Eb = K*Φ*N
220 = K*Φ*1000
Let's assume that the flux is constant.
K*Φ = 0.22
At 1000 rpm and rated torque, the armature current is given by:
Ia = (Eb - V)/Rs
= (220 - 230)/0.4
= -25 A
The negative sign indicates that the direction of the current is opposite to that of the voltage. This is due to the fact that the external resistance is added in series with the motor.
Let's calculate the external resistance required to obtain rated torque at 1000 rpm.
T2 = T1*(I2/I1)^2
T2/T1 = (N2/N1)
(I2/I1)^2 = (N2/N1)
(I2/I1)^2 = (1000/1500)
I2/I1 = 0.82
I2 = 0.82*20
= 16.4 A
Ia = (Eb - V)/Rs
-25 = (220 - 230 - I2*Ra)/Rs
Rs = 3.5 Ω
Therefore, the external resistance required to obtain rated torque at 1000 rpm is 3.5 Ω.
Community Answer
A 230 V d.c. series motor develops its rated output at 1500 rpm while ...
E1=230-(20*0.6)
e1=218
extrnl resistance added
e2=230-20(r+0.6)
e directly proportional to n
230-20(r+0.6)/218=(1000/1500)
on solving we get r=3.5
e1=218
extrnl resistance added
e2=230-20(r+0.6)
e directly proportional to n
230-20(r+0.6)/218=(1000/1500)
on solving we get r=3.5
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A 230 V d.c. series motor develops its rated output at 1500 rpm while taking 20 A. Armature and series field resistance are 0.6 Q and 0.4 Q respectively. To obtain rated torque at 1000 rpm, external resistance mustbe added is____________a)4.5 Qb)3.5 Qc)2.5 Qd)2.0 QCorrect answer is option 'B'. 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 230 V d.c. series motor develops its rated output at 1500 rpm while taking 20 A. Armature and series field resistance are 0.6 Q and 0.4 Q respectively. To obtain rated torque at 1000 rpm, external resistance mustbe added is____________a)4.5 Qb)3.5 Qc)2.5 Qd)2.0 QCorrect answer is option 'B'. 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 230 V d.c. series motor develops its rated output at 1500 rpm while taking 20 A. Armature and series field resistance are 0.6 Q and 0.4 Q respectively. To obtain rated torque at 1000 rpm, external resistance mustbe added is____________a)4.5 Qb)3.5 Qc)2.5 Qd)2.0 QCorrect answer is option 'B'. Can you explain this answer?.
A 230 V d.c. series motor develops its rated output at 1500 rpm while taking 20 A. Armature and series field resistance are 0.6 Q and 0.4 Q respectively. To obtain rated torque at 1000 rpm, external resistance mustbe added is____________a)4.5 Qb)3.5 Qc)2.5 Qd)2.0 QCorrect answer is option 'B'. 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 230 V d.c. series motor develops its rated output at 1500 rpm while taking 20 A. Armature and series field resistance are 0.6 Q and 0.4 Q respectively. To obtain rated torque at 1000 rpm, external resistance mustbe added is____________a)4.5 Qb)3.5 Qc)2.5 Qd)2.0 QCorrect answer is option 'B'. 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 230 V d.c. series motor develops its rated output at 1500 rpm while taking 20 A. Armature and series field resistance are 0.6 Q and 0.4 Q respectively. To obtain rated torque at 1000 rpm, external resistance mustbe added is____________a)4.5 Qb)3.5 Qc)2.5 Qd)2.0 QCorrect answer is option 'B'. Can you explain this answer?.
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A 230 V d.c. series motor develops its rated output at 1500 rpm while taking 20 A. Armature and series field resistance are 0.6 Q and 0.4 Q respectively. To obtain rated torque at 1000 rpm, external resistance mustbe added is____________a)4.5 Qb)3.5 Qc)2.5 Qd)2.0 QCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A 230 V d.c. series motor develops its rated output at 1500 rpm while taking 20 A. Armature and series field resistance are 0.6 Q and 0.4 Q respectively. To obtain rated torque at 1000 rpm, external resistance mustbe added is____________a)4.5 Qb)3.5 Qc)2.5 Qd)2.0 QCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A 230 V d.c. series motor develops its rated output at 1500 rpm while taking 20 A. Armature and series field resistance are 0.6 Q and 0.4 Q respectively. To obtain rated torque at 1000 rpm, external resistance mustbe added is____________a)4.5 Qb)3.5 Qc)2.5 Qd)2.0 QCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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