Nodal Circuit Analysis using KCL

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Canaldigger
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Homework Statement




I(a)= 11A, I(b)= 2.5A, I(c)= 3A. Apply KCL at node X to find I(x).

This question seems really simple as it tells me what to do and where to do it, but am I missing something or not remembering to apply some concept?


The Attempt at a Solution


My attempt: I assigned current flowing towards node x as positive and away as negative.
Ʃi=0; I(a)+I(b)-I(c)-I(x) = 0
After plugging in, I(x) = 10.5A
Is that all there is to the question?
I don't think this answer is correct and rather I should redraw the circuit as a three loop one and use Current divider equations.
 
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For the given circuit component values the given current values are nonsense.

If you take the current values at face value and ignore the component values, then your answer (Ix = 10.5A) is fine. Otherwise, for the circuit as specified, the given current values are utter nonsense.

Shame on the one(s) who created the problem; they should not be teaching.
 
gneill said:
For the given circuit component values the given current values are nonsense.

If you take the current values at face value and ignore the component values, then your answer (Ix = 10.5A) is fine. Otherwise, for the circuit as specified, the given current values are utter nonsense.

Shame on the one(s) who created the problem; they should not be teaching.

I will email them and see if there was a typo or something. I was wondering why my values didn't match up when using KVL or any other equations that could apply to a circuit. I had an inkling that the figures might be wrong.

Thanks for help! This problem was driving me crazy.
 
Canaldigger said:
I will email them and see if there was a typo or something. I was wondering why my values didn't match up when using KVL or any other equations that could apply to a circuit. I had an inkling that the figures might be wrong.

Thanks for help! This problem was driving me crazy.

No problem. That's why we're here :smile: