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

Given l=5+10sinωt Ieff=[∫0TI2dt∫0Tdt]1/2 =[1T∫0T(5+10sinωt)2dt]1/2 =[1T∫0T(25+100sinωt+100sin2ωt)]1/2 But as 1T∫0Tsinωtdt=0 And 1T∫0Tsinωtdt=12 So leff=[25+12×100]1/2=53A