In a shunt-wound dc motor with the field coils and rotor connected in parallel (
ID: 2143514 • Letter: I
Question
In a shunt-wound dc motor with the field coils and rotor connected in parallel (see the figure (Figure 1) ), the resistance Rf of the field coils is Rf = 107? and the resistance Rr of the rotor is Rr = 6.0? . When a potential difference of V = 120V is applied to the brushes and the motor is running at full speed delivering mechanical power, the current supplied to it is 4.85A .
Explanation / Answer
The field current is: V/R = 120/107 = 1.12 A
The only other current flowing is the rotor current: Ir = Itot - Ifld
Ir = 4.85 - 1.12 = 3.73
The emf: The rotor current multiplied by the rotor resistance is the difference between the line voltage and the counter emf:
emf = Vline - Irotor*Rrotor = 120 - 3.73*6 = 97.62 V
"rate of development of thermal energy" is assumed to be the dissipative losses, i.e. the ohmic power. For the field, it's: Ifld^2*Rfld = 1.12^2*107 = 134.22W
For the rotor electrical losses: Irot^2*Rrot = 3.73^2*6 = 83.47 W
The power output is the power input minus the losses:
120*4.85 - 134.22 - 83.47 = 364.31 W