Finding current using Kirchhoff's circuit rules

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The discussion focuses on applying Kirchhoff's circuit rules to solve for current in a circuit. The user initially set up a system of linear equations but struggled with cancellation issues due to incorrect assumptions about current directions. A suggestion to check signs and use nodal analysis helped clarify the problem, leading to a successful resolution. The importance of correctly defining current directions is emphasized, noting that negative values can indicate the opposite flow. Ultimately, the user expressed gratitude for the guidance received in solving the problem.
slaufer
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Homework Statement



Nt0Cm.png



Homework Equations



Kirchhoff's loop rule and (maybe) Kirchhoff's current law

The Attempt at a Solution



So I attempted to apply Kirchhoff's loop rule to the circuit, and ended up with a system of linear equations:

24 - 6I_{1} - 3I_{3} = 0 \rightarrow 2I_{1} - I_{3} = 8
36 - 6I_{1} - 6I_{2} = 0 \rightarrow I_{1} + I_{2} = 6
12 - 3I_{3} - 6I_{2} = 0 \rightarrow 2I_{2} + I_{3} = 4

but every time I try to solve it, I end up with everything dropping out. I'm not sure what I'm missing here, but I've been banging my head against this one for about 4 hours now, so any help would be appreciated!
 
Last edited:
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slaufer said:

Homework Statement



Nt0Cm.png

Homework Equations



Kirchhoff's loop rule and (maybe) Kirchhoff's current law

The Attempt at a Solution



So I attempted to apply Kirchhoff's loop rule to the circuit, and ended up with a system of linear equations:

24 - 6I_{1} - 3I_{3} = 0 \rightarrow 2I_{1} - I_{3} = 8 wrong
36 - 6I_{1} - 6I_{2} = 0 \rightarrow I_{1} + I_{2} = 6
12 - 3I_{3} - 6I_{2} = 0 wrong\rightarrow 2I_{2} + I_{3} = 4

but every time I try to solve it, I end up with everything dropping out. I'm not sure what I'm missing here, but I've been banging my head against this one for about 4 hours now, so any help would be appreciated!

Check the signs.
You also have a relation between the currents. (Nodal Law) Use that and two loop equations.

ehild
 
Last edited:
Thanks, that makes more sense now.

I might have eventually realized that my system was cancelling out because it was using mutually derived equations, but I probably never would have realized that I was treating I_{3} as if it were flowing in two directions at once.

Using two of the equations from the original system in the nodal rule equation I_{1} = I_{2}+ I_{3} made it all fall together though. I'm still not sure how I'd figure out what direction I_{3} was flowing in if it weren't given in the problem, but at least I'm FINALLY done with this problem.

tl;dr you are the best, thanks
 
You need not find out the direction of the currents in advance. If you have not chosen the proper direction you would get negative value for the current.

ehild
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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