Applying KVL in loop ABDA, we get 2I1+4−I2‌=0 I2‌=2I1+4.......(i) Applying KVL in loop BCDB, we get ‌1(I1+I3)−4(I2−I3)−4‌=0 ⇒I1+I3−4I2+4I3−4‌=0 ⇒I1+5I3−4(2I1+4)−4‌=0 ⇒−7I1+5I3‌=20 ⇒‌I3‌=‌
20+7I1
5
.......(ii) By applying KVL in loop ADCA, we get ‌I2+4(I2−I3)−10‌=0 ⇒5I2−4I3‌=10 ⇒‌5(2I1+4)−4(‌
20+7I1
5
)‌=10 [From Eqs. (i) and (ii)] ⇒‌‌I1=1.364A From Eq. (i), I2=2×1.364+4=6.278A From Eq (ii), I3=‌