A 100hp, 250 V, 1200 rpm shunt DC motor with an armature resistance of 0.03 Ω and a field resistance of 41.67 Ω. The motor has compensating windings, so armature reaction can be ignored. Mechanical and core losses may be assumed to be negligible for the purposes of this problem. The motor is assumed to be driving a load with a line current of 126 A and an initial speed of 1103 rpm. To simplify the problem, assume that the amount of armature current drawn by the motor remains constant. what is the motor's speed if the field resistance is raised to 50 Ω? can you help me to find N2?
A 100hp, 250 V, 1200 rpm shunt DC motor with an armature resistance of 0.03 Ω and a field resistance of 41.67 Ω. The motor has compensating windings, so armature reaction can be ignored. Mechanical and core losses may be assumed to be negligible for the purposes of this problem. The motor is assumed to be driving a load with a line current of 126 A and an initial speed of 1103 rpm. To simplify the problem, assume that the amount of armature current drawn by the motor remains constant. what is the motor's speed if the field resistance is raised to 50 Ω? can you help me to find N2?
Introductory Circuit Analysis (13th Edition)
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A 100hp, 250 V, 1200 rpm shunt DC motor with an armature resistance of 0.03 Ω and a field resistance of 41.67 Ω. The motor has compensating windings, so armature reaction can be ignored.
what is the motor's speed if the field resistance is raised to 50 Ω?
can you help me to find N2?
![Pour= 100hp
V = 250V
N = 1200rpm
RA = 0.032
Rei = 41.672
ILI = 126A
N = 1103 rpm
R2 = 50 2
I= 120A
I2= 125A
Ip = V + Re1 = 250 + 41.67 = 6A
In = ILI -IF= 126 - 6 = 120A (constant)
En = Vr - I(RA) = 250 - 120(0.03) = 246.4V
I2 = Vr + Re2 = 250 - 50 = 5A|
I2=I, + I2 = 120 + 5 = 125A
E12= Vr - I(R) = 250 - 120(0.03) = 246.4V
EAI = kON = 246.4 = ko = 1103.
-(1)
E12= KON2 = 246.4 = ko = (N2).(2)
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F287e156b-7981-40d0-8aab-01f55065120d%2Fcb84aa78-6743-4993-b80e-e6d1a7424b5d%2Foocudbl_processed.png&w=3840&q=75)
Transcribed Image Text:Pour= 100hp
V = 250V
N = 1200rpm
RA = 0.032
Rei = 41.672
ILI = 126A
N = 1103 rpm
R2 = 50 2
I= 120A
I2= 125A
Ip = V + Re1 = 250 + 41.67 = 6A
In = ILI -IF= 126 - 6 = 120A (constant)
En = Vr - I(RA) = 250 - 120(0.03) = 246.4V
I2 = Vr + Re2 = 250 - 50 = 5A|
I2=I, + I2 = 120 + 5 = 125A
E12= Vr - I(R) = 250 - 120(0.03) = 246.4V
EAI = kON = 246.4 = ko = 1103.
-(1)
E12= KON2 = 246.4 = ko = (N2).(2)
%3D
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