45 id lb 3 10 0 Va 12 240 V 180 tc 18 4 SolutionApply KVL fo

45 id lb 3 10 0% Va 12 240 V 180 tc 18 4

Apply KVL for loop-1

R*Ie +180*Ic =240 ----1

Apply kvl for loop 2

10*Ib+12*Ia +18*Ia-180*Ic =0 ----2

Apply kvl for loop 3

45*Id -10*Ib – R*Ie =0 -----3

Ie =Ib+Ic----4

Ia=Ib+Id

Given condition:

12*Ia=60

Ia=60/12=5 amp