How can I use KVL and KCL to find the potential differences in this circuit?

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To find the potential differences in the given circuit using Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL), the equations ε - i1R1 - i2R2 = 0 and additional KVL applications at loop ACB are essential. R3 affects the circuit by creating a current flow that must be accounted for, as it bridges the parallel paths and influences the potential differences. The left path has a lower resistance, leading to a lower potential at point C compared to A and D, while the right path has a higher potential at point B. By establishing two equations with the unknown currents i1 and i2, one can solve for these currents and subsequently determine the potential differences VA - VB, VB - VC, VC - VD, and VA - VC. Understanding the symmetry in the circuit will aid in simplifying the calculations.
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Homework Statement
Finding potential difference across a fancy circuit
Relevant Equations
as attached
In the figure ε = 9.89 V, R1 = 1150 Ω, R2 = 2890 Ω, and R3 = 4940 Ω. What are the potential differences (in V) (a) VA - VB, (b) VB - VC, (c) VC - VD, and (d) VA - VC?What I've tried
I have derived the equation ε - i1R1 - i2R2 = 0

where i1 is the current running through R1 and vice versaQuestions
How does R3 affect the circuit? I am quit confused as it bridges the parallel lines and I'm now not sure how the current runs

Because I want to find out i1 and i2 such that I can then use:

V = IR

to calculate the Voltage on R1 aka point B

and find VA - VB etc
 

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Pochen Liu said:
I have derived the equation ε - i1R1 - i2R2 = 0

where i1 is the current running through R1 and vice versa
The current through which R1?
There is a symmetry you can use but you'll have to discover it first.
Pochen Liu said:
How does R3 affect the circuit? I am quit confused as it bridges the parallel lines and I'm now not sure how the current runs
The two sides are not identical. In the left path you have the smaller resistance at the bottom (C will have a lower potential than the average of A and D), in the right path you have the smaller resistance at the top (B will have a higher potential than the average of A and D). There will be a current flow through R3.

You don't need to discover this directly. You can calculate VB-VC assuming that no current flows through R3. If it is non-zero (and it will be) then that assumption is wrong.
 
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you have already made 1 correct equation by applying KVL at the loop ABD+voltage source that contains two unknowns ##I_1,I_2##

Now apply KVL at the loop ACB. The symmetry that @mfb talks about, if you identify it correctly will help you make this equation also with the same unknowns ##I_1, I_2## (and also with the help of applying KCL at node B or node C.

So you ll have two equations with two uknowns, solve the system of equations and you ll know ##I_1,I_2## and everything else can be derived by knowing those 2 values.
 
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