application of Kirchoff's law.

simple network

There are two unknown  quantity is this circuit that is i1 and i2 .in this circuit there are two closed loop one is loop ABCFA and another is FCDEF.

Now using Kirchhoff’s law find out the unknowns quantity.

Consider the node C. we know by Kirchhoff’s current law incoming current equal to outgoing current

For this reason in this node (i1) + (i2) = (i1+i2)

Now Loop ABCFA 

30-2(i1)-10+5(i2) = 0

Or 2(i1)-5(i2) =20………………………….. (1)

Now Loop FCDEF

-5(i2) +10-3(i1+i2)-5-4(i1+i2) =0

Or 7(i1) +12(i2) =5…………………………… (2)

Multiplying equation (1) by 7 and equation (2) by 2, we get,

14(i1)-35(i2) =140…………………………… (3)

14(i1) +24(i2) =10…………………………. (4)

Subtraction equation (4) from equation (3) we get

-59(i2) =130

 i2= -130/59

    = -2.2 A

This negative sign means the direction of i2 current is opposite that direction we gave.

For that i2=2.2 from C to F

Substituting the value of i2= -2.2 in equation (1) we get

 i1=4.5 A

Current in branch CDEF= (i1+i2)

                                          = (4.5) + (-2.2)

                                          =2.3 A

And now the figure is.

simple network