Current in multi-battery circuits

  • #1
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Homework Statement



Find the current through R2 (see attached figure)


Homework Equations





The Attempt at a Solution


Not really sure if I did this right, would like if someone would double check my work:

First I tried Using Kirchhoff equations for a node and I got:

I1-I2+I3=0

Next I broke it down into two loops using the fact that the voltage drops across a closed loop is zero:

Loop 1: V1-3*I1-5*I2=0
Loop 2: V2-8*I3-5*I2=0

Using the second equation I got:
8*I3=V2-5*I2
I3=(V2-5*I2)/8

Next, plugging that into the Kirchhoff equation, I can solve for I1 in terms of I2:

I1=I2+(5*I2-v2)/8=(13*I2)/8

Lastly, plugging the values for I1 and I3 into the loop 1 equation, I get:

V1=3((13*I2-V2)/8)+5*I2

V1=(39*I2-3*V2)/8 + (40*I2)/8= (79*I2-45)/8

Using V1=5:

40=79*I2-45
85=79*I2
I2=1.08A

I am not confident in that answer though :/

Any help would be great!
 

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Answers and Replies

  • #2
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Sorry last attachment didn't work!
 

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  • #3
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Whoops, you made a mistake in your currents. The two equations should be.

5-3I1-5(I2+I1)=0
15-8I2-5(I2+I1)=0

When you draw your voltage loops, label the center of them with I1, I2, etc. and then add or subtract currents where loops share common nodes depending on whether the loops both point in the same direction (add) or opposite directions (subtract). You may find more about this if you search mesh current analysis.

Also, I think you should just use your calculator to solve the matrix, but if you want to plug and chug go for it.

Remember, the current through R2 is just I2+I1.
 
  • #4
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wouldn't the current through R2 be I2-I1 (based on my Kirchhoff equations?)
 
  • #5
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Oh wait, you are only using 2 currents, got it!

Thanks for your help :)
 
  • #6
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So you are looking at voltage gains, and I'm more used to looking at voltage drops because that's how current is defined. Anyway, for this problem it doesn't matter because the diagram has few components.

Your equations show that loop one should be going clockwise, indicating a clockwise current flow, and loop two should be going counter clockwise, indicating a counter clockwise current flow. They will add together at R2.

Edit: Cool, glad you got it. Also, sorry, I see what you did now. Yes, our work should be identical for the setup at least. I didn't check your numerics, but you've done the important part right. :)
 
  • #7
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Solving for I1 using the first equation I got I1=(5-5I2)/8

I plugged that back into the second equation and got I2=.19A

Plugging that .19 back into the first equation i got I1=.506A

So, the current through R2 would be .506+.19=.696 A?
 
  • #8
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You should get

8I1+5I2=5
5I1+13I2=15

I think you did it right to begin with because plugging the system into my calculator I get

I1=-.12
I2=1.2

I1+I2=IR2=1.08
 
  • #9
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Oh so I was right in the first place, but your way is much easier! Thanks for showing me a new way to solve the problem!
 
  • #10
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No problem, hope it made sense. :)
 
  • #11
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It makes a lot more sense than the way I did it haha!
 

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