I just need answers (H) In a shunt wound DC motor you can increase the speed by
ID: 2087387 • Letter: I
Question
I just need answers
(H) In a shunt wound DC motor you can increase the speed by increasing the resistance of shunt resistor Rf. What happens to the motor torque when you do this and why? (1) Torque increases. Increasing RF increases current in the field windings and increases magnetic flux. Because torque is proportional to magnetic flux, it increases (2) Torque decreases. Increasing Rf reduces current in the field windings and reduces magnetic flux. Because torque is proportional to magnetic flux, it decreases (3) Torque increases. Increasing RF increases current in the field windings and increases magnetic flux. Because torque is proportional to magnetic flux, it increases (4) Torque decreases. Increasing Rf increases current in the fiel windings and reduces magnetic flux. Because torque is proportional to magnetic flux, it decreases (5) Torque decreases. Increasing Rf increases current in the fiel windings and increases magnetic flux. Because torque is proportional to magnetic flux, it decrease:s (I) Refer to the diagram on "Speed Control of a Shunt Wound DC Motor". Suppose the no load voltage V - 90V, Ra R1 5 ohms, RsRF- 140 ohms, and the motor runs at a rated speed of 1500 RPM when 4 amps. What is the armature current lA in amps when the motor runs at 6 amps? Note: Type a 1-decimal number, rounded off to the nearest 0.1Explanation / Answer
H. 1 is the correct option because shunt winding has very high resistance , therefore it draws very less current . On increasing the shunt resistance, the current in the field winding furthur increases due to decrease of current in shunt fields, thus magnetic field increases , consequently increasing the Torque as the torque produced is proportional to magnetic flux.
2. No load back emf = 90 - current* armature resistance
= 90- 4*5 = 70V
I = I (armature)+ I (shunt)
6 = I (armature)+ 90/140
I(armature) = 5.357A