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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=Vt−IaRa =240−50×0.2 =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=Vt−Ia2(Ra+Rext) E2=240−25(1.2) E2=210 V E∝ ϕN as flux is constant, E∝N E1E2=N1N2 230210=N1N2 N2=0.913 N1 % change in speed =N2−N1N1×100%=0.913N1−N1N1×100%=8.7%decrease

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