A shunt motor rotating at 1500 r/min is fed by a 120 V source (Fig. 9a). The line current is 51 A and the shunt-field resistance is 120 Ω. If the armature resistance is 0.1 Ω, calculate the following:
a. The current in the armature
b. The counter-emf
c. The mechanical power developed by the motor

Question

Accepted Answer

a. The field current (Fig. 9b) is

[latex]I_{ x }=120 V / 120 \Omega=1 A[/latex]

The armature current is

[latex]I=51-1=50 A[/latex]

b. The voltage across the armature is

[latex]E=120 V[/latex]

Voltage drop due to armature resistance is

[latex]I R=50 imes 0.1=5 V[/latex]

The cemf generated by the armature is

[latex]E_{ o }=120-5=115 V[/latex]

c. The total power supplied to the motor is

[latex]P_{ i }=E I=120 imes 51=6120 W[/latex]

Power absorbed by the armature is

[latex]P_{ a }=E I=120 imes 50=6000 W[/latex]

Power dissipated in the armature is

[latex]P=I R^{2}=50^{2} imes 0.1=250 W[/latex]

Mechanical power developed by the armature is

P = 6000 - 250 = 5750 W

(equivalent to 5750/746 = 7.7 hp)

The actual mechanical output is slightly less than 5750 W because some of the mechanical power is dissipated in bearing friction losses, in windage losses, and in armature iron losses.

Question 5.4: A shunt motor rotating at 1500 r/min is fed by a 120 V sourc...

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