Question

Find all branch currents in the network shown in Fig. 3-14(a).

(My question is at the end.)

Figures 3-14(a) and 3-14(b): http://i.imgur.com/JFg3M.jpg

Solution:
The equivalent resistances to the left and right of nodes a and b are
R_[eq(left)] = [5 + 12(8)]/20 = 9.8 ohms
Req(right = (6)(3)/9 = 2 ohms

Now referring to the reduced network of Fig. 3-14(b),

I_3 = 2/11.8(13.7) = 2.32 A

I_4 = 9.8/11.8(13.7) = 11.38 A

Then referring to the original network,
I_1 = 8/20(2.32) = 0.93 A
I_2 = 2.32 - 0.93 = 1.39 A
I_5 = 3/9(11.38) = 3.79 A
I_6 = 11.38 - 3.79 = 7.59 A

Answer 1

My question:
Could someone please elaborate as to what is being done for finding I_3 and I_4?

Answer 2

In your equations, R_eq left = 4.8, you have 9.8.

Answer 3

Because it is a parallel circuit you know that the voltage in one element is the same in all of the elements. So once you find the voltage in one, you know the voltage in them all. V=IR, I=V/R.

Answer 4

i can explain ur problem when u r online

Answer 5

R_eq left = 5 + 12(8)/20 = 9.8 ohms is what I meant.

Also, I know that and I found V_eq = (2ohms)(13.7 A) = 27.4 V but then when I try to find the current in the left and right branches, I get:

27.4/9.8 = 2.7959 (left)

27.4/2 = 13.7 (right)

What am I doing wrong?