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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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 12 year ago
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