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Question

A DC shunt motor takes an rated armature current of 50 A at its rated voltage of 240 V. Its armature circuit resistance is 0.2 Ω. If an external resistance of 1 Ω is inserted in series with the armature and field flux remains unchanged. The percentage increase or decrease in speed for half of the load torque is

A
8.7% increase
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B
21. 739% decrease
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C
8.7% decrease
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D
21.739% increase
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Solution

The correct option is C 8.7% decrease
Ia=50A, Vt=240 V
Ra=0.2 Ω
E1=VtIaRa
=24050×0.2
=230 V
Now as torque is reduced to half the rated load torque
T ϕIa
as flux is constant
T Ia

So, T1T2=Ia1Ia2

10.5=50Ia2

Ia2=25 A

1 Ω resistor is added to armature circuit as external resistor
E2=VtIa2(Ra+Rext)
E2=24025(1.2)
E2=210 V
E ϕN
as flux is constant, EN

E1E2=N1N2

230210=N1N2

N2=0.913 N1

% change in speed =N2N1N1×100%=0.913N1N1N1×100%=8.7%decrease

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