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

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SUMMARY

The discussion focuses on calculating the potential difference between two points in a circuit with a voltage of 3.56 V and a resistor value of 2.04 Ω. The user attempted to apply Kirchhoff's laws but made errors in accounting for the current through the resistors. The correct approach involves recognizing that the potential differences across identical resistors add up, and two loop equations are typically necessary for solving a two-loop circuit problem. The user was advised to clarify the battery voltage configuration for accurate calculations.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Familiarity with Ohm's Law (V=IR)
  • Basic circuit analysis techniques, including loop and junction rules
  • Knowledge of series and parallel resistor configurations
NEXT STEPS
  • Review Kirchhoff's Laws for circuit analysis
  • Practice solving two-loop circuits using simultaneous equations
  • Learn about series and parallel resistor combinations
  • Explore advanced circuit simulation tools like LTspice or Multisim
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Students studying electrical engineering, physics enthusiasts, and anyone looking to improve their circuit analysis skills.

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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