Finding Current in a Circuit (to get current going through an amp meter)

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Homework Help Overview

The discussion revolves around finding the current in a circuit involving multiple resistors and ammeters. The original poster is attempting to analyze a circuit diagram to determine the current flowing through an ammeter.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of equivalent resistances and the configuration of resistors in series and parallel. There are questions about the current distribution through different branches of the circuit and the impact of ammeters on the circuit analysis.

Discussion Status

Some participants are exploring different interpretations of the circuit setup, particularly regarding the arrangement of resistors and the effect of the ammeters. Guidance has been offered to redraw the circuit and reconsider the equivalent resistance calculations.

Contextual Notes

There is confusion regarding the representation of the battery voltage and the arrangement of resistors, with some participants noting discrepancies in their calculations of equivalent resistance. The original poster has acknowledged a potential oversight in their understanding of the circuit layout.

Schoomy
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Homework Statement



http://dl.getdropbox.com/u/119186/Picture%20122.png

Homework Equations



Voltage Drop = -I*R
I1=I2+I3

Reqiv= 1/R1 + 1/R2 etc.

The Attempt at a Solution



I really have no idea how to approach this. A1 and A2 won't have the same current will they?

I tried getting an equivalent resistor to get A1, but that didn't work:
1/Req = 1/R1+1/R2 = 1/2.2 + 1/2.2 = 2.2/2 = 1.1=Req

But still that can only get me the voltage drop and that hasn't worked...
 
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Any ideas? I'm still stumped on this...nothing comes out right the way I'm trying this...
 
Bump.
 
Redraw the circuit, but leave out the ammeters since they act as shorts as far as the circuit is concerned.

Then look for parallel and/or series resistors, and come up with a single equivalent resistance for the entire circuit. That will help figure out the total current out of the battery.
 
I got the entire Req as 4.6357 ohm, does that seem right? Then total current was 2.0493 A. Not sure if those are right or what to do from there...
 
I did not get a total resistance of 4.6357 ohms. Can you show the work that you did to get to that point?

Also, once you get the total current through the circuit, you can get the voltage drop across Req. So, will the voltage across Req be the same as the voltage across the 3 ohms? What about the 6 ohms? Why/Why Not, and show us where your thinking is.
 
ravioli said:
I did not get a total resistance of 4.6357 ohms. Can you show the work that you did to get to that point?

Also, once you get the total current through the circuit, you can get the voltage drop across Req. So, will the voltage across Req be the same as the voltage across the 3 ohms? What about the 6 ohms? Why/Why Not, and show us where your thinking is.

I found the problem! a) didn't notice that voltage is double what's given (see how the picture shows two batteries next to each other? oops!)
b) i got an Req for the parallel 2.2ohm resistors but apparently that's not necessary?
 
a) It's one battery at 9.5V. We have multiple lines like that to represent multiple cells of a battery (one cell usually cannot do 9.5V)
b) The 2.2ohm resistors are not in parallel. One way to think about parallel versus series is as follows. Every single straight line is going to represent a node. If two resistors are connected to the same two nodes, then they are in parallel. For instance, an R, 3, and 6 share the same two nodes, thus they are parallel. Let's collapse those three resistors down to an Requivalent at this moment. Are the resulting (equivalent) resistors in series/parallel and why?
 

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