Question

Q A SIMPLE ELECTRIC MOTOR HAS AN ARMATURE RESISTANCE OF 1 ohm  & RUNS FROM A D.C SOURCE OF 12V . IT DRAWS A CURRENT OF 2A WHEN UNLOADED . WHEN A CERTAIN LOAD IS CONNECTED TO IT ,IT SPEED REDUCES BY 10% OF ITS INITIAL VALUE ,THE CURRENT DRAWN BY THE UNLOADED MOTOR IS??

ANS 3A

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 SIR , ISTHE FOLLOWG METH. IS CORRECT

R=1ohm , E=12V , I=2A

I=(E-e)/R =>  e=10  ON PUTTING D VALUES

NOW AS IT IS REDUCES BY 10% MEANS  NOW e' = 9V ( 10/100 X 10 = 1 CHANGE , THUS FINAL = 10-1=9)

I= E-e'/R  = ( 12-9)/ 1 = 3A (RESISTANCE IS KEPT CONST.)  RESISTANCE WILL CHANGE OR NOT??

SARIKA SHARMA 11 year ago is this solution helpfull: 26 7

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