A shunt-wound DC motor with the field coils and rotor connected in parallel (see

Question

A shunt-wound DC motor with the field coils and rotor connected in parallel (see the figure) (Figure 1) operates from a 130 V DC power line. The resistance of the field windings, Rf, is 248 . The resistance of the rotor, Rr, is 5.80 . When the motor is running, the rotor develops an emf E. The motor draws a current of 4.64 A from the line. Friction losses amount to 43.0 W

Q1) Compute the field current If.

Q2) Compute the rotor current Ir.

Q3) Compute the emf E.

Q4) Compute the rate of development of thermal energy in the field windings.

Q5) Compute the rate Pth,rotor of development of thermal energy in the rotor.

Q6) Compute the power input to the motor Pin.

Q7) Compute the efficiency of the motor.

Explanation / Answer

field current=V/R=130/248

=0.524A

rotor current=ir=4.64-0.524

=4.116.

emf=Vline-irotor*Rrotor

=130-4.116*5.80

=106.12v

rate of development of thermal energy

=ifld^2*Rfld=0.524^2*248

=68.09W.

rotor of development of thermal energy

irot^2*Rrot

=4.116^2*5.80

=98.26W.

power input to the motor pin

130*4.64-68.09-98.26-43.0

=393.85W.

the efficiency of the motor

=393.85/(130*4.64)

=65.2%.

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