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

Given: I=5+10sinωtIeff=∫0T I2dt∫0T dt1/2=1T∫0T (5 + 10 sinωt)2dt1/2=1T∫0T 25 + 100 sinωt  + 100 sin2 ωt1/2 But as  1T∫0T sin ωt dt=0 and  1T∫0T sin2 ωt dt=12 so  Ieff=25 + 12 × 1001/2=53A