In a shunt-wound dc motor with the field coils and rotor connected in parallel (

Question

In a shunt-wound dc motor with the field coils and rotor connected in parallel (see the figure (Figure 1)), the resistance Rt of the field coils is R 110 S2 and the resistance Rr of the rotor is RT 36.2 2 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 4.83A Figure 1 v of 1 120 V R E, R Part A What is the current in the field coils? Submit My Answers Give U Part B What is the current in the rotor? Submit My Answers Give U

Explanation / Answer

V = 120 V ; Rf = 110 Ohm ; Rr = 6.2 Ohm ; I(t) = 4.83 A

A)The current in the field coild will be:

If = Vf/Rf

If = 120/110 = 1.1 A

Hence, If = 1.1 A

B)The current in the rotor will be:

Ir = 4.83 - 1.1 = 3.73 A

Hence, Ir = 3.73 A

C)emf will be

emf = 120 - 3.73 x 6.2 = 96.87 Volts

Hence, emf = 96.87 V

D)Mechanical power

P = If^2Rf = 1.1^2 x 110 = 133.1 W

P(rot) = 3.73^2 x 6.2 = 86.26 W

P(out) = 120 x 4.83 - 133.1 - 86.26 = 360.24 W

Hence, P(out) = 360.24 W

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