Engineering Circuit Analysis - Resistors and a battery

AI Thread Summary
The discussion revolves around calculating the current I4 in a circuit with resistors and a battery. The original poster attempted to use the voltage divider equation but encountered discrepancies in their results. Participants suggested using loop analysis instead of the voltage divider method, as it may yield more accurate results. The final calculation using the voltage divider approach led to a current of 1.18mA, which was confirmed as correct. Overall, the conversation emphasizes the importance of choosing the right analysis method in circuit problems.
jendrix
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Circuit Analysis -- Resistors and a battery

Homework Statement



Hello, if you see the attached I am trying to find the current I4.My plan was to find Voltage across R4 and use Ohm's law.

Homework Equations



V=IR

The Attempt at a Solution



I have simplified the circuit down(attached pic), I thought I could use the voltage divider equation to find voltage across R4||R6 which would be equal to the voltage across R4 however I'm getting a different answer

I got 50*(3.197/(3.197+4.7)) =20V making I4= 2mA

http://i.imgur.com/vjVbcEi.png
http://i.imgur.com/6FwW7QG.png
 
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So you believe that absolutely no current is flowing through R3? How do you come to that conclusion?
 
Hmm that makes sense, I was thinking that going by Kirchoff's voltage law that the sum of the loop on the right would=0 in which case I could use the potential divider equation.
 
Though it makes sense as Voltage across R3 would be the same as R2 yet if you did the voltage divider equation then you would get a different solution.

So the only way would be to put R2||R3 then do the voltage divider equation? Or is there another way?

Thanks
 
jendrix said:
Though it makes sense as Voltage across R3 would be the same as R2 yet if you did the voltage divider equation then you would get a different solution.

So the only way would be to put R2||R3 then do the voltage divider equation? Or is there another way?

Thanks

This whole "voltage divider" thing is, to me, a brain-dead way of going about it. Don't get me wrong, I know people use it but I just find it a distraction that leads to the kind of error that you got. Just do a full loop analysis. It will take a lot more steps than what you were trying to do but that's because what you were trying to do was wrong.
 
It was the way the question was worded, it guided you towards doing it that way.Personally I prefer loop analysis although my university will only let us use nodal, it gets marked incorrect if we do a loop analysis.
 
jendrix said:
It was the way the question was worded, it guided you towards doing it that way.Personally I prefer loop analysis although my university will only let us use nodal, it gets marked incorrect if we do a loop analysis.

Gads, I HATE that kind of crap. I had a prof once who counted my solution to a calculus problem as wrong because I used a simple method out of Thomas and he wanted us to use a more complicated method out of the book he made us use.

Well, node analysis in this case is about the same as loop analysis, so do that :smile:
 
Thanks, I think you're spot on, I've done it the way he wanted but I'll double check it with loop analysis to see which way is quicker, I think everybody has "their" way of doingthings and that should be acceptable.
 
You can use the potential divider approach, basing it on resistor R3 and all the other resistors combined into a single resistance.

What answer do you get?
 
  • #10
NascentOxygen said:
You can use the potential divider approach, basing it on resistor R3 and all the other resistors combined into a single resistance.

What answer do you get?


I did it using the divider approach by combining R2*R3/(R2+R3) This gives 1.45kΩ

Then 50*(1.45/(1.45+4.7)

=11.79V

Then 11.79/10kΩ

=1.18mA


Thanks for the help and suggestions
 
  • #11
1.18mA ✔[/size][/color]
Looks right.
 

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