In the circuit shown below E1 = 4.0 V,R1=2Ω ,E2= 6.0 V, R2=4Ω and R3 = 2Ω. The current I1is
Applying Kirchhoff's law for the loops (1) and (2) as shown in figure For loop(1)
−2i1−2(i1−i2)+4=0⇒ 2i1−i2=2 ....(i)
For loop (2)
−2(i1−i2)+4i2−6=0 ⇒ −i1+3i2=3 ...(ii)
On solving equation (i) and (ii) i1=1.8A.