Using Kirchhoff's laws, determine the currents . . .

Question : Using Kirchhoff's laws, determine the currents I₁, I₂ and I₃ for the network shown in figure.

Solution :

Apllying Kirchhoff's Junction Rule

I₁=I₂+I₃ — (1)

Apllying Krichhoff's Loop Rule in Loop 1

2I₁+6I₂=24-27

2I₁+6I₂=-3
2(I₂+I₃)+6I₂=-3
2I₂+2I₃+6I₂=-3
8I₂+2I₃=-3 — (2)

Apllying Krichhoff's Loop Rule in Loop 2

6I₂-4I₃=-27 — (3)

Solving Equation (2) and (3)

[8I₂+2I₃=-3]×2

6I₂-4I₃=-27

16I₂+4I₃=-6
6I₂-4I₃=-27
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22I₂+0 = -33

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22I₂=-33
I₂=-33/22
I₂=-3/2
I₂=-1.5 A

-ve sign shows that the direction of current is actually opposite to what has been marked in the circuit diagram.

Putting I₂=-1.5 A in equation (2)

8(-1.5)+2I₃=-3
-12+2I₃=-3
2I₃=-3+12
2I₃=9
I₃=9/2
I₃=4.5 A

Putting the value of I₂ and I₃ in equation (1)

I₁=-1.5+4.5
I₁=3A

Hence, I₁= 3A, I₂=-1.5 A, I₃=4.5 A

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