The armature of a permanent-magnet dc generator has a resistance of 1 Ω and generates a voltage of 50 V when the speed is 500 r/min. If the armature is connected to a source of 150 V, calculate the following:

a. The starting current
b. The counter-emf when the motor runs at 1000 r/min. At 1460 r/min.
c. The armature current at 1000 r/min. At 1460 r/min.

Question

The armature of a permanent-magnet dc generator has a resistance of 1 Ω and generates a voltage of 50 V when the speed is 500 r/min. If the armature is connected to a source of 150 V, calculate the following:

a. The starting current
b. The counter-emf when the motor runs at 1000 r/min. At 1460 r/min.
c. The armature current at 1000 r/min. At 1460 r/min.

Accepted Answer

a. At the moment of start-up, the armature is
stationary, so [latex]E_{ o }=0 V[/latex] (Fig. 3a). The starting current is limited only by the armature
resistance:

[latex]I=E_{ s } / R=150 V / 1 \Omega=150 A[/latex]

b. Because the generator voltage is 50 V at 500 r/min, the cemf of the motor will be 100 V at 1000 r/min and 146 V at 1460 r/min.

c. The net voltage in the armature circuit at 1000 r/min is

[latex]E_{ s }-E_{ o }=150-100=50 V[/latex]

The corresponding armature current is

[latex]\begin{aligned}I &=\left(E_{ s }-E_{ o } ight) / R \&=50 / 1=50 A ( ext { Fig. } 3 b )\end{aligned}[/latex]

When the motor speed reaches 1460 r/min, the cemf will be 146 V, almost equal to the source voltage. Under these conditions, the armature current is only

[latex]\begin{aligned}I &=\left(E_{ s }-E_{ o } ight) / R=(150-146) / 1 \&=4 A\end{aligned}[/latex]

and the corresponding motor torque is much smaller than before (Fig. 3c).

Question 5.1: The armature of a permanent-magnet dc generator has a resist...

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