Finding potential difference between two points in a ladder circuit

AI Thread Summary
The discussion focuses on finding the potential difference between two points in a ladder circuit using Kirchhoff's laws. The user has set potential at point A to zero and established relationships between the currents flowing through the resistances. Suggestions include using Kirchhoff's Current Law (KCL) and creating node equations for supernodes to simplify the analysis. A recommendation is made to modify the circuit by replacing a vertical resistor with a parallel combination to exploit symmetry, making it easier to analyze. The conversation emphasizes the importance of establishing correct relationships between currents to solve for the potential difference effectively.
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Homework Statement


Find the potential difference ##V_A– V_B## for the circuit shown in the figure.
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Homework Equations


Kirchhoff's laws.

The Attempt at a Solution


If we assign potentials to junctions starting by setting ##A=0## Current ##i_1## flow through each vertical resistances and ##i_2 ## through slant resistances. B has potential ##i_1+2## and there exist a relation ## 1+i_2+i_1=0## between ##i_1## and ##i_2##. I need one more equation. In which loop should I apply KVL. Is there a better way to approach this ?
 
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Welcome to PF!

You might try working with some junctions (KCL).
 
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How exactly ? Could you please elaborate ?
 
Choose A as the reference node. You have two supernodes, one associated with B and the other A (the reference node). Write a node equation for supernode B.
 
Have I proceeded correctly ?
 
arutor said:
How exactly ? Could you please elaborate ?
Consider the 4 emf's along the bottom. In terms of ##i_1## and ##i_2##, how much current is in the emf next to point A? Then how much current is in the next emf as you move to the left along the bottom of the circuit. Keep going for all 4 emf's along the bottom. This should get you another relation between the two currents.

Your relation ##1+i_1+i_2 = 0## looks correct for a particular choice of current directions.

I don't know what a supernode is, but gneill's suggestion might get you to the answer quicker.
 
KCL will surely give you the answer. But you need not consider all the 8 sources.
Replace the vertical 1Ω below point B with a parallel combination of two 2Ω resistors and cut the circuit into two halves at point B such that each 2Ω resistor is on either half. Now, you have two circuits with 4 sources in each. Use the one containing both A and B(i.e. discard the left circuit). It will be easier. This is possible because of the symmetry of the circuit w.r.t. point B.
 
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