How Do You Calculate the Potential Difference Between Two Points in a Circuit?

AI Thread Summary
To calculate the potential difference between points a and b in the circuit, the current through the resistors must be determined using the junction and loop rules. The user initially calculated the current I(3) as 0.76 A but made an error by not accounting for the potential differences across the two 1Ω resistors correctly. It was noted that the currents through these resistors should be added since they flow in the same direction. Additionally, two loop equations are typically needed for a two-loop circuit, although symmetry can simplify the problem. Clarification on the voltage of the batteries was also suggested to ensure accurate calculations.
abcdmichelle
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Homework Statement


For the circuit in the figure below, V = 3.56 V and R = 2.04 Ω. Find the potential difference between points a and b.

This is the link to the figure: http://i749.photobucket.com/albums/xx137/abcdmichelle/physics/26-62alt1.gif


Homework Equations


V=IR


The Attempt at a Solution


So I needed to find the current passing through R.
I said that the resistor at the top left had current I, the resistor at the top right was I(2), the resistor at the bottom left was I, the resistor at the bottom right was I(2), and the resistor R in the middle had a current of I(3).
I applied the junction rule and the loop rule to the left hand side loop, and found the current to be:

2-I-2.04I(3)-3.56+I=0
Current I(3)=0.76

So then to find the potential difference I did:
V= -(0.76)(2.04)-3.56

This is wrong, I'm not sure if I did anything right here!
Please help!
 
Last edited:
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welcome to pf!

hi abcdmichelle! welcome to pf! :smile:

(try using the X2 tag just above the Reply box :wink:)
abcdmichelle said:
2-I-2.04I(3)-3.56+I=0
Current I(3)=0.76

no, you've canceled the potential differences for the two 1Ω resistors … the current is flowing the "same" way through them, so they will add

also, you normally need two loop-equations to solve a two-loop problem (though in this case you can get a relationship between I1 and I2 from symmetry) :wink:

(btw, are the 2V batteries 2 volts, or 2 times 3.56 V ? :confused:)
 
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