A direct current of 5 A is superimposed on an alternating current .I=10 sin ωtflowing through a wire. The effective value of the resulting current will be

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(15/2) A

53A

55A

15 A

answer is B.

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Detailed Solution

Given l=5+10 sinωtIeff=∫0TI2dt∫0Tdt1/2=1T∫0T(5+10sinωt)2dt1/2=1T∫0T(25+100sinωt+100sin2ωt)1/2But as 1T∫0Tsinωtdt=0and1T∫0Tsin2ωtdt=12so Ieff=25+12×1001/2=53A

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