# Find the currents in the circuit branches

1. Jun 19, 2006

### endeavor

http://img335.imageshack.us/img335/8383/physics4a8ua.th.png [Broken]http://img335.imageshack.us/img335/8000/physics4b6vo.th.png [Broken]
"Find the currents in the circuit branches in [the first figure on the left above]."

V1 = 20V, V2 = 10V, R1 = 5.0 ohms, R2 = 4.0 ohms, R3 = 6.0 ohms, R4=R5=R6= 2.0 ohms

Rp-1 = 3/2 (from R4,R5,R6)
Rp = 2/3 ohms

I assigned loops in the second figure above. There are 6 different currents.
Using Kirchhoff's first rule:
(C1) I1 = I2 + I3
(C2) I3 = I4 + I5
(C3) I1 = 3I6
Then using the Kirchhoff's second rule, I get a bunch of equations (L1 = LOOP 1):
(L1) -V1 +I2R1 + I1Rp = 0
(L2) +V2 +I5R2 - I2R1 = 0
(L3) +I4R3 - I5R2 = 0
(L7) -V1 + V2 +I5R2 +I1Rp = 0
(L8) +V2 - I2R1 +I4R3 = 0
(L9) -V1 +I1Rp +I4R3 + V2 = 0
Then substituting values for the V's and R's:
(L1) 5*I2 + (2/3)*I1 = 20
(L2) 4*I5 - 5*I2 = -10
(L3) 6*I4 - 4*I5 = 0
(L7) 4*I5 + (2/3)*I1 = 10
(L8) -5*I2 + 6*I4 = -10
(L9) (2/3)*I1 + 6*I4 = 10
I then found that I3 = (5/2)*I4. Then using this equation, and equations C1 and L9, I finally got:
I1 = - 90/462 A = -0.19 A

I think the fact that my answer is negative only shows that I chose the wrong loop direction. But the absolute value of my answer is still wrong... help

Last edited by a moderator: May 2, 2017
2. Jun 19, 2006

### lightgrav

You should simplify a bit more before assigning loops ... R2 || R3.
Then you have a loop on the left and a loop on the right.

- I_L R_P - (I_L - I_R) R_1 - V_1 = 0 ... because I_1(upward) = I_L - I_R
- I_R R_{2||3} + V_2 - (I_R - I_L) R_1 = 0 .

3. Jun 20, 2006

### endeavor

thanks! Simplifying really worked. After I simplified the parallel parts, I only had 3 loops, which was a lot easier to work with.