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"Find the currents in the circuit branches in [the first figure on the left above]."

V_{1}= 20V, V_{2}= 10V, R_{1}= 5.0 ohms, R_{2}= 4.0 ohms, R_{3}= 6.0 ohms, R_{4}=R_{5}=R_{6}= 2.0 ohms

R_{p}^{-1}= 3/2 (from R_{4},R_{5},R_{6})

R_{p}= 2/3 ohms

I assigned loops in the second figure above. There are 6 different currents.

Using Kirchhoff's first rule:

(C1) I_{1}= I_{2}+ I_{3}

(C2) I_{3}= I_{4}+ I_{5}

(C3) I_{1}= 3I_{6}

Then using the Kirchhoff's second rule, I get a bunch of equations (L1 = LOOP 1):

(L1) -V_{1}+I_{2}R_{1}+ I_{1}R_{p}= 0

(L2) +V_{2}+I_{5}R_{2}- I_{2}R_{1}= 0

(L3) +I_{4}R_{3}- I_{5}R_{2}= 0

(L7) -V_{1}+ V_{2}+I_{5}R_{2}+I_{1}R_{p}= 0

(L8) +V_{2}- I_{2}R_{1}+I_{4}R_{3}= 0

(L9) -V_{1}+I_{1}R_{p}+I_{4}R_{3}+ V_{2}= 0

Then substituting values for the V's and R's:

(L1) 5*I_{2}+ (2/3)*I_{1}= 20

(L2) 4*I_{5}- 5*I_{2}= -10

(L3) 6*I_{4}- 4*I_{5}= 0

(L7) 4*I_{5}+ (2/3)*I_{1}= 10

(L8) -5*I_{2}+ 6*I_{4}= -10

(L9) (2/3)*I_{1}+ 6*I_{4}= 10

I then found that I_{3}= (5/2)*I_{4}. Then using this equation, and equations C1 and L9, I finally got:

I_{1}= - 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

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# Find the currents in the circuit branches

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