Solving the Node Equation for I_2 in a Circuit

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To solve for the outward-going current I_2 in the circuit, it is essential to apply Kirchhoff's current law, which states that the sum of currents entering a node equals the sum of currents leaving. The correct approach involves choosing a consistent convention for current direction, ensuring that all currents are accounted for properly. After applying this method, I_2 is determined to be 7 A. It is important to note that different methods can yield the same result, as long as the chosen convention is consistently followed. Understanding these principles is crucial for accurately solving circuit problems.
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Homework Statement


Write the node equation for the circuit in the figure. If ##I_1 = 6 A,## ##I_4 = 5 A,## and ##I_3 = 4 A,## what is the value of the outward-going current ##I_2##?

current_conservation3a.gif


Homework Equations



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The Attempt at a Solution



I know that in the steady state, the electron current entering a node in a circuit is equal to the electron current leaving that node, but I am not sure how to do this problem.
 
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Try it from the different way of saying that: all the currents coming into a node have to add up to zero
 
So will ##I_2## equal ##-15##?
 
That's the right idea, but not the right answer.

The arrows on the diagram show you which directions of current flow are positive.
 
So it should be positive ##15## since the direction is in the positive direction?
 
Lee33 said:
So it should be positive ##15## since the direction is in the positive direction?
That's not quite right either.

You get to pick which direction (into or out of the node) is positive or negative. It's your choice. But once you pick it you must be consistent with all currents involved. (Your answer of "15" is incorrect because you didn't follow your chosen convention consistently. Consistency is what is important here.)

Notice that I1 and I4 have different directions than I2 and I3.

Once you pick a convention, if you follow the advice that phinds gives you can't go wrong. All the currents going into a node must add to zero. Or using the other convention, all currents leaving a node must add to zero. Pick either one of the two, then stick with it consistently.

There's also another way to do this: put all currents entering a node on one side of the equation and all currents leaving the node on the other side of the equation. Even though this method works, I still suggest using one of the above methods instead: putting all the currents on one side of the equation and summing them to zero (using your chosen convention of what is positive and what is negative, consistently). You can't go wrong with that.
 
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Collinsmark - Following your advice will ##I_2## be ##7##?
 
Lee33 said:
Collinsmark - Following your advice will ##I_2## be ##7##?
Yes, that's right. Good job. :approve: Keep in mind though, there are several different ways that same answer could have been obtained, depending on your convention. Don't be afraid to explore the different methods! :smile:
 
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Thank you very much!
 
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