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- Homework Statement
- Calculate ##I## and ##V## in the circuit below
- Relevant Equations
- Kirchhoff's laws
##I=\frac{V}{R}##
greg_rack said:Homework Statement:: Calculate ##I## and ##V## in the circuit below
Relevant Equations:: Kirchhoff's laws
##I=\frac{V}{R}##
View attachment 271797
For this one circuit, I've applied Kirchhoff's laws as following:
$$\left\{\begin{matrix}
V-6\Omega I_{6}-24V-2\Omega I_{2}=0\\
24V-6\Omega I_{6}-12V-24V=0\\
I_{2}+I_{6}=6A
\end{matrix}\right.$$
but the values I get for ##I## and ##V## aren't correct... what's wrong?
I got it from second Kirchhoff's law: current entering a node is equal to the current exiting that node.George Jones said:How did you get the last equation?
That is actually wrong, in accordance with what my choices are, the third equation should be: ##I_{4}+I_{6}+I_{8}=I_{2}##gneill said:Where did the 6 Amps come from in your third equation?
Have I done something wrong? In which way would potential drops in resistors depend upon the current flow?gneill said:On your diagram, label your assumed current directions for each current you will be using in your equations. Apply those directions consistently for each equation that you write. Potential drops across resistors depend upon the direction of current flow!
The same current flows through both the 4 and 8 Ohm resistors. They are in series so there is only one current in that branch.greg_rack said:That is actually wrong, in accordance with what my choices are, the third equation should be: ##I_{4}+I_{6}+I_{8}=I_{2}##
By the way, with 6 amps I meant ##I_{4}+I_{8}##.
You need to draw in and clearly label the currents you are using before writing your equations so that you don't mistakenly change their directions (and hence potential drop directions) midway.Have I done something wrong? In which way would potential drops in resistors depend upon the current flow?
God, I'm finding the application of these rules deceptively easy... the procedure hasn't been explained in class and I had to learn it on my own, but I'm apparently really confused about the method, in spite of its simplicity and algorithmic-nature.gneill said:The same current flows through both the 4 and 8 Ohm resistors. They are in series so there is only one current in that branch.
You need to draw in and clearly label the currents you are using before writing your equations so that you don't mistakenly change their directions (and hence potential drop directions) midway.
As for potential drops in resistors,
View attachment 271817
Note how the direction of the potential drop changes with the direction of the current.
Got it, thank you very much! I'm confident I will soon master this topic!gneill said:Off hand I can't think of a particular article or video, but I do know that they are plentiful on the web if you just do a search on something like "Kirchhoff circuit analysis".
Rather than assign individual currents to each component, you want to take advantage of the fact that series-connected components will always share one current. This will minimize the number of variables that end up in your equations. For your circuit, something like this:
View attachment 271846
So you work with "branch currents" rather than "component currents".