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)
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
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
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
-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
I1=3A
Hence, I1= 3A, I2=-1.5 A, I3=4.5 A
