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The armature of a `DC` motor has `20Omega` resistance. It draws a current of `1.5 A` when run by `200 V DC` supply The value of back emf induced in it will beA. 150 VB. 170 VC. 180 VD. 190V

Answer» Correct Answer - D
For DC motor, `I=(V_(S) - V_(B))/R` and `V_(B) propto omega`
`1.5 = (220-V_(R))/(20) rArr V_(B) = 190 V`


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