Unit 7-Part 3

Question

The armature windings of a dc motor have a resistance of 5.0 Ω. The motor is connected to a 120 V line and when the motor reaches full speed against its normal load, the counter emf is 108 V. Calculate:
a) The current into the motor when it is just starting up. Be sure to draw the equivalent circuit showing your motor as a resistor, the battery, and the back emf (another battery) if it exists.



b) The current when the motor reaches full speed. Be sure to draw the equivalent circuit showing your motor as a resistor, the battery, and the back emf (another battery) if it exists.

Answer 1

(a)

 

When the motor first starts up, the back emf is zero.  You can use Ohm's law to get the initial current I₀. 

 

I₀ = E/R = (120 V) / (5.0 Ω) = 24.0 A

 

(b)

 

When the motor reaches full speed, the voltage of the circuit becomes the difference between the line voltage and the back emf.  Use Ohm's law to get the final, lower current I_f based on the reduced voltage E_f.  The resistance is the same.

 

E_f = (120 V) - (108 V) = 12 V

 

I_f = E_f/R = (12 V) / (5.0 Ω) = 2.4 A

 

The circuit diagram will be a single closed loop with a battery of 120 V in opposition to a battery of 108 volts, in series with a 5.0 Ω resistor.