# DC motor question

#### electrical_ck

Joined Dec 7, 2009
10
Exam tomorrow, professor didn't go over previous exam and Im trying to figure out where I messed up......

A 100V, 50A and 1200 rpm permanent magnet DC motor has an armature resistance of 0.2 ohms.

a) determine starting resistance= 100V/0.2 = 500 A
b) determine the external resistance that needs to be inserted in the armature to limit the starting current to twice the rated current

Rst = ((VL-EC)/(IR*M))- Ra= ((100-0)/(50*2))-.2= .8ohm (Don't know if I did this right)

c) determine the counter emf when the motor speed reaches 500 rpm

so at 1200 rpm the voltage is 100 V and 50 A. So at 500 rpm the voltage is 41.6 V and 20.8 A. Is this the correct way for finding my values for my formula?

Ec= VL-IA(Ra+Rs) = 41.6V-20.8(.2)= 33.28 V

Any help would be great

#### nuckollsr

Joined Dec 17, 2009
16
A 100V, 50A and 1200 rpm permanent magnet DC motor has an armature resistance of 0.2 ohms.

a) determine starting resistance= 100V/0.2 = 500 A

Correct Amps, actually this is "locked rotor" or "start-up" or "inrush" current.

b) determine the external resistance that needs to be inserted in the armature to limit the starting current to twice the rated current

Okay, 50A rated calls for 100A max or 1.0 ohms total resistance.
You already have 0.2 ohms in the armature windings, so 0.8 ohms
added outside will get you the 100A max inrush.

Rst = ((VL-EC)/(IR*M))- Ra= ((100-0)/(50*2))-.2= .8ohm (Don't know if I did this right)

Yup, you did.

c) determine the counter emf when the motor speed reaches 500 rpm

so at 1200 rpm the voltage is 100 V and 50 A. So at 500 rpm the voltage is 41.6 V and 20.8 A. Is this the correct way for finding my values for my formula?

Hmmmm. Rated current 50A, 1200 rpm at 100v applied is "rated" load then
I*R drops in armature wires add up to 50 X 0.2 or 10 v. This means that
CEMF for the motor is (100-10)/1.2KRPM = 75 volts/KRPM which is
a constant for the motor no matter what. So at 500 RPM, CEMF is
37.5 volts

Ec= VL-IA(Ra+Rs) = 41.6V-20.8(.2)= 33.28 V

I think adding the resistor is a "ringer" . . . Adding 0.8 ohms
in series means that at full rated torque, the motor draws 50A
and a total of 50V is tossed off in the total loop resistance.
This means the motor is turning 50/75 = 666 RPM at 50A and 100v Supply