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

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Homework Help Overview

The problem involves calculating the currents i_1, i_2, and the voltages v_A and v_H in a ladder network using Kirchhoff's Laws. The context includes a circuit diagram with specific resistances and a request to calculate power dissipated in a resistor.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to simplify the circuit and apply Kirchhoff's Laws to set up a system of equations. They express uncertainty about their calculations and seek clarification on potential errors. Other participants question whether the direction of the currents might be affecting the results and suggest simplifying the circuit further.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the circuit and the implications of current direction. Some guidance has been offered regarding simplification and maintaining clarity in calculations, but no consensus has been reached on the correct approach or solution.

Contextual Notes

The original poster indicates potential confusion due to handwriting and specifies which resistors correspond to v_A and v_H. There may be assumptions about the circuit configuration that are being questioned.

vaio911
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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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If you can't understand my handwriting,

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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