3 battery resistor circuit

  • #1
ReidMerrill
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Homework Statement


Find the current through and power dissipated by each resistor
20160412_181936_zps2kuodzwp.jpg


Homework Equations


V=I/R
ΣIin=∑Iout

The Attempt at a Solution


20160412_183323_zps0by4tlu9.jpg

Did I set up the loop laws correctly and where do I go from here?
 

Answers and Replies

  • #2
gneill
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Your first loop equation looks fine. I can't picture what your second loop is from the equation that you wrote, but I can say that it is incorrect. Your third loop equation has an error in the last term: current I2 does not pass through the 10 Ω resistor.

Can you label the nodes in your diagram and state your loop paths?

Hint: You only need enough loops so that each component is included in at least one of the loops. In this problem you can get away with two loop equations.
 
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  • #3
ReidMerrill
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Since the 10Ω resistor is already in the first loop do I even need to include it in the second?
 
  • #4
gneill
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Since the 10Ω resistor is already in the first loop do I even need to include it in the second?
Every loop must complete a circuit (be a closed path) and some components will be shared by more than one loop; that's expected. You must include a term for every component in the "KVL walk" that you take around given loop in order for KVL to hold.
 
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  • #5
ReidMerrill
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20160412_191653_zpsd6mcpiqa.jpg

I'm not sure if the current going through the 10Ω resistor is different/needs a different name
I'm also going to see if starting from the 8V battery makes more sense to me.
 
  • #6
gneill
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Once you've labeled a current though a component, that's the current that is associated with it. It doesn't change if you apply KVL to different loops that happen to share that component. So ##I_3## flows through the 10 Ω resistor, in the same direction as you chose it, for each KVL equation that contains that resistor. So just be careful of the signs of the terms: if you "KVL walk" over a resistor against the flow of the current, then you will "see" a potential rise instead of a drop.

In your second equation above, you are "walking" clockwise around the loop, so when you traverse the 10 Ω resistor you are proceeding upwards, against the flow of ##I_3##. Will you see a potential drop or a potential rise as a result?
 
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