The armature of a DC motor has $20 Ω$ resistance. It draws a current of $1.5$ amp when run by $200 V$ DC supply. The value of back emf induced in it will be

150 V

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170 V

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180 V

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190 V

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Solution

The correct option is B $170 V$
Back emf $=$ Applied voltage potential drop across armature coil
$= 200 - i R$
$= 200 - 1.5 \times 20$
$= 170 V$

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