Question

I have to calculate the resultant current in R2 on my circuit using kirchhoff's current laws. Please help

Answer 1

Let the current in the left hand mesh be a, and the direction be clockwise.
Let the current in the right hand mesh be b, and the direction clockwise.
Then using Kirchhoff's second law we can write the following equations for each mesh:

17 = 35a + 6a -6b which simplifies to:
17 = 41a - 6b ............(1)
and
34 = 4b + 6b - 6a which simplifies to:
34 = 10b - 6a ............(2)

Multiplying equation (1) by 5 we get:
85 = 205a - 30b .........(3)
Multiplying equation (2) by 3 we get:
102 = 30b - 18a .........(4)
Adding equations (3) and (4) gives:
187 = 187a, which gives the value of current a as 1 amp.
Substituting the value of 1 for a in equation (1) we find the value of current b to be 4 amps.
Therefore the resultant current in R2 is 4 - 1 = 3 amps.