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

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SUMMARY

The discussion focuses on 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 into a single loop and apply Ohm's Law but struggles with the calculations and system of equations. Key equations provided include current relationships and voltage equations, but the user reports incorrect results. The discussion emphasizes the importance of maintaining clear calculations for accurate voltage determination using the potential divider rule.

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  • Familiarity with potential divider rule for voltage calculations.
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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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