Solving a Wheatstone Bridge with 5 Resistors

AI Thread Summary
To solve a Wheatstone Bridge with five resistors, Kirchhoff's laws are essential for determining the current at each junction and loop. The resistor in the middle complicates the analysis, as it represents the sum of the currents from the two loops. By choosing clockwise and counterclockwise directions for the loop currents, one can formulate two equations based on these currents. Solving these equations will yield the bottom current, which can then be used to calculate power using the formula P = I²R. This approach effectively addresses the complexities of the circuit configuration.
katrascythe
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Homework Statement



A Wheatstone Bridge with 5 resistors

http://img529.imageshack.us/my.php?image=physicsproblemmo7.png

Homework Equations



Use Kirchoff's Law

The Attempt at a Solution



I know that I have to use Kirchoff's laws to find the current at each junction and then at the loops. The only deal is that I'm not sure how to handle the resistor in the middle.
 
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Hi katrascythe! :smile:

Your link isn't working. :cry:
 
katrascythe said:

Homework Statement



A Wheatstone Bridge with 5 resistors

http://img529.imageshack.us/my.php?image=physicsproblemmo7.png

Homework Equations



Use Kirchoff's Law

The Attempt at a Solution



I know that I have to use Kirchoff's laws to find the current at each junction and then at the loops. The only deal is that I'm not sure how to handle the resistor in the middle.

Choose directions for your loop currents in each of the 2 loops. (I would choose clockwise for the top and counterclockwise for the bottom.) The voltage then of the resister in the middle represents the current of the sum of the 2 currents, top and bottom current, and becomes a term in each of your loop equations.

You have 2 equations, 2 unknowns, solve to the bottom current and that current2 times R is your power by P = I2R.
 

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