Problem 20 In a shunt-wound dc motor with the field coils and rotor connected in

Question

Problem 20 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 = 110 and the resistance R, of the rotor is R,-57 . When a potential difference of V= 120 V is applied to the brushes and the motor is running at full speed delivering mechanical power, the current supplied to it is 4.87 A Submit My Answers Give Up Part B What is the current in the rotor? Submit My Answers Give Up Part C What is the induced emf developed by the motor? Express your answer using two significant figures Figure 1 of! Submit My Answers Give Up Part D 120V R How much mechanical power is developed by this motor? Express your answer using two significant figures Pmech

Explanation / Answer

The field current is: V/R = 120/110 = 1.1 A

The only other current flowing is the rotor current: Ir = Itot - Ifld
Ir = 4.87 – 1.1 = 3.77 A

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 – 3.77*5.7 = 98.5 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 = 1.1^2*110 = 133.1 W

For the rotor electrical losses: Irot^2*Rrot = 3.77^2*5.7 = 81 W

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