0
\$\begingroup\$

I have a problem regarding DC motors that I am trying to solve. I think that I might not understand the assumptions that are given. The question given is:

A DC series motor is connected to a load. The torque varies as the square of the speed. With the diverter circuit open, the motor takes 20 A and runs at 500 rpm. Determine the motor current and speed when the diverter circuit resistance is made equal to the series field resistance. Neglect saturation and the voltage drops across the series field resistance, as well as the armature resistance.

My solution can be seen here: http://imgur.com/zOaHxfo.

I am not sure if I am simply misinterpreting the given assumptions (bolded above) or if they are nonsensical. For instance, if there is no voltage drop, doesn't that mean that there is no current, and no power is being supplied?

Intuitively, it seems to me that increasing the resistance in parallel with the series field resistance should increase the current and speed while decreasing the back (induced) EMF.

EDIT: It seems the correct answer is given by Bruce Abbott. The speed becomes 1000 rpm and the current becomes 160 A.

\$\endgroup\$
  • 1
    \$\begingroup\$ I believe the assumptions mean that connecting the diverter circuit makes the field current 50% of the armature current but otherwise consider the armature and field resistances to be zero. The field flux is directly proportional to the field current. I don't have time to look at your solution. \$\endgroup\$ – Charles Cowie Dec 10 '15 at 6:04
  • \$\begingroup\$ Under these assumptions, it would appear that the solution becomes: imgur.com/TC3cqGN for anyone curious. Any input as to the validity of this result is welcome. \$\endgroup\$ – Alex Smith Dec 10 '15 at 16:45
  • \$\begingroup\$ I have not been able to understand your solution or develop one of my own. The fact that I am sick with a cold has a lot to do with that. My treatment of the assumptions implies that this is a lossless motor that has only insignificant resistance in the field and diverter to form a 50/50% current divider so that only 50% of the armature current flows through the field coil under the second set of conditions. What isn't stated in the problem, but must be assumed for the problem to make sense, is that the supply voltage must be adjusted in order to reach the second operating point. \$\endgroup\$ – Charles Cowie Dec 11 '15 at 1:32
1
\$\begingroup\$

Neglecting voltage drops across field and armature resistances, and assuming supply voltage is not changed:-

The diverter circuit cuts field coil current in half, thereby halving magnetic field strength, doubling Kv, and halving Kt.

Doubling Kv causes speed to increase to 1000rpm. As load torque is proportional to speed squared, torque will increase by 4 times. Since Kt is half, 8 times more current (160A) is required to deliver this torque.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your response Bruce, I think this is the correct answer. \$\endgroup\$ – Alex Smith Dec 11 '15 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.