Solve 1112 and 13 a at point C I1 I2 I3 b Loop rule bcfeb

Solve 11,12 and 13 (a) at point C: I1 + I2 = I3 (b) Loop rule bcfeb ()10 ()6I1 ()14 ()4I2 = 0 (c) Loop rule bcdab ()10 ()6I1 ()2I3 = 0

at Junction C

I₁+I₂=I₃

I₁+I₂-I₃=0------------------------1

For Loop bcfeb

10+6I₁-14+4I₂=0

6I₁+4I₂=4----------------------2

For loop bcdab

14-6I₁-2I₃=0

6I₁+2I₃=14--------------------3

Solving 1 ,2 and 3 we get

I₁=2.4 A

I₂=-2.6 A

I₃=-0.2 A