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Question

A 240 V, series motor takes 40 A when giving its rated output at 1500 rpm. Its resistance is 0.3Ω. The value of resistance which must be added to obtain rated torque at starting will be

A
3Ω
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B
5.7Ω
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C
2.7Ω
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D
6Ω
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Solution

The correct option is B 5.7ΩV=240V,Ia=40A N1=1500rpmRa=0.3Ω T∝I2a, since the torque is constant, ∴ I2a1=I2a2 ⇒ Ia1=Ia2=40A During starting the induced emf is zero, hence the current is limited only by the resistance in the armature circuit. ∴ Total resistance =24040=6Ω Extra resistance to be added in series with armature =6−0.3=5.7Ω

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