# Kirchhoff's Law circuit question

1. May 20, 2014

### RyanTG

I've attached the circuit layout.

Q) Find the current flowing through each of the resistors.

What am I doing wrong? I thoroughly do not understand kirchhoff laws. I follow the sign rules, I try to do the maxwell loop thing and I'm still getting wrong answers.

If I do the maxwell loop on the circuit, is there no need for a I0? If I don't do the maxwell loop then I never get the right answer for any of the questions.

I don't understand why I can't do this.

Thanks.

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2. May 20, 2014

### phinds

You're asking for help and you post the work SIDEWAYS? Doesn't work for me. No interest in getting a kink in my neck.

3. May 20, 2014

### BiGyElLoWhAt

Maybe it could stem from the fact that you have current flowing both to the top of the page and to the bottom of the page, at the same time, through the middle section with the 4 ohm wire... Mark your potentials on the circuit to see which way the current's flowing.

4. May 20, 2014

### BiGyElLoWhAt

What's the potential at the junction above the 4 ohm resistor? What does that tell you about the potential difference across the 1ohm resistor? What does that tell you about the current through the 1 ohm resistor remember V = IR.

5. May 20, 2014

### RyanTG

I'm sorry man, it is saved vertically on my computer, I didn't purposefully upload it to be awkward.

There is no harm in just saving it to your computer and rotating it. I don't understand why I would be denied help because of something as petty as that.

Here you go, rotated.

Having a second go at it, I'm getting I1 as -4.5A, but a different value for I2 depending on which way I do my maxwell loop thing in the second half of the circuit...

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6. May 20, 2014

### RyanTG

Oh I think I get it now.

Is the correct answer: I1 = -5A, I2 = 5.5A?

Should the maxwell loop always go in the direction that the current would flow in that part of the circuit? So for the second half, it should go positive to negative, so my loop is the wrong way around?

And I've been plugging I1 back into the first simultaneous equation that governs the emf for the first loop, instead of plugging it into the second one, to find I2.

7. May 20, 2014

### RyanTG

You know what I don't even know anymore....

8. May 20, 2014

### RyanTG

figured it out finally i think... my answers were correct in that image, my method is wrong but just happened to be correct in that instance.

9. May 20, 2014

### BiGyElLoWhAt

I got those numbers as well, but i think you have to add them together through the 4 ohm resistor, so .5 through the 1 ohm and 1.5A through the 1 ohm.