Calculating i_1, i_2, v_A, and v_H in a Ladder Network | Kirchhoff's Laws

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The discussion revolves around calculating the currents i_1, i_2, and voltages v_A and v_H in a ladder network using Kirchhoff's Laws. The user attempts to simplify the circuit to one loop and apply Ohm's law but struggles with the calculations and the resulting values seem incorrect. They express confusion over their system of equations and seek clarification on their approach. The importance of maintaining clear steps in calculations is emphasized, particularly for applying the potential divider rule to find voltages. The conversation highlights common challenges in circuit analysis and the need for methodical problem-solving.
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Homework Statement



Find i_1,i_2,v_A,\text{ and }v_H in the ladder network. Calculate the power dissipated in the 2\;\Omega resistor.

http://i.snag.gy/PIhxl.jpg


Homework Equations



Kirchhoff's Laws (i.e. current law, voltage law).


The Attempt at a Solution



Ok, I know that I can simplify the circuit down to one loop then use Ohm's law to calculate i_1, but then I have to do other things to get the other stuff.

I tried to make a system to solve this problem using the laws.
Can somebody tell me where I'm doing wrong here?

<br /> \begin{align}<br /> i_b&amp;=i_c+i_a\\<br /> i_c&amp;=i_e+i_d\\<br /> i_e&amp;=i_g+i_f\\<br /> 12i_a-28+8i_b&amp;=0\\<br /> 12i_a+6i_c+15i_d&amp;=0\\<br /> 4i_e+10i_f+15i_d&amp;=0\\<br /> 2i_g+13i_h+10i_f&amp;=0\\<br /> 4i_e+2i_g+13i_h+15i_d&amp;=0<br /> \end{align}

Getting me something like:
-\frac{7}{39},\frac{49}{13},\frac{154}{39},\dots

which I know is completely off.
 
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V_a is the one w/ the 12 ohms.

V_h is the 13 ohms one.
 
It is because of the direction of the currents?
 
I'd probably just simplify it. If you keep your working for each step it will help when it comes to working out the voltages using the potential divider rule.
 
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