How to find current of resistor in complex circuit?

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Homework Help Overview

The discussion revolves around finding the current through a resistor in a complex circuit involving multiple resistors and branches. The original poster presents a circuit diagram and equations related to current and resistance, expressing difficulty in solving part C of the problem.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of voltage drops and equivalent resistances in the circuit. The original poster questions the rules regarding equivalent resistors at a junction, while others suggest calculating voltage drops across specific resistors to find currents.

Discussion Status

The discussion includes attempts to clarify the process of calculating current and voltage drops. Some participants provide guidance on how to approach the problem, while others express confusion about specific rules and calculations. There is no explicit consensus, but productive suggestions have been made.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available or the methods they can use. The original poster and others express uncertainty about the application of certain electrical principles in the context of the circuit presented.

Schoomy
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Homework Statement



http://dl.getdropbox.com/u/119186/Picture%201.png

Homework Equations



I = I2+I3
I= e/R

The Attempt at a Solution



http://dl.getdropbox.com/u/119186/snapshot-1253898419.920882.jpg
As you can see I got all of it, but part C. I can't figure out what the process is. I tried making equivalent resistors, but there must be some rule I don't understand about resistors and being equivalent or not when there's a junction?

If someone could point me in the right direction I'd appreciate it.
 
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Oh, you've done all the hard work!
Just to get you started, work out the voltage drop across the 1 ohm resistor due to the 5.8 A current. That leaves about 24.2 V across the rest of the circuit. You got 3.25 ohms for the parallel combo of the 4 and 17.5; add the 3.3 and you've got 24.2 V across the 6.55 ohm leg so you can find that current easily. Use that to find the drop in the 3.3 ohm resister and you'll know the potential remaining across the 4 and 17.5 combo. Knowing the V across the 17.5 ohm resistor, calc the current.
 
I'm still stumped. Voltage drop is given as V=-IR yes?

Voltage drop from 1ohm resistor is 5.8 so I have 24.2V left to go into the lower branch. I take:

24.2/3.3 = 7.33A current through the 3.3ohm resistor, right?

Now I do 7.33A*3.25ohm = 23.8225, which I subtract from 24.2, giving me 0.3775A left for R2?

Doesn't work...
 
Nevermind, got it! Didn't follow your instructions properly :P
 
Why or what rule let you subtract 5.8Amps from 30 Volts? I don't get that...
 
Nevermind again. It's Voltage Drop = (5.8Amps)(1ohm) = 5.8volts drop so 24.2 volts left for the rest.

Divide that by 6.55 and you have the total lower leg current of 3.694.

Now find the voltage drop due to the 3.3ohm resistor:

Vdrop = (3.694)(3.3) = 12.19.

24.2-12.19 = 12.0098V left for the 3.25combo resistors.

12.0098/17.5 = 0.68627Amps.

Thanks!
 

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