# Finding Current with Kirchoffs Laws

Discussion in 'Homework Help' started by hogesjzz30, Sep 3, 2013.

1. ### hogesjzz30 Thread Starter New Member

Sep 1, 2013
12
0
I am really stuck on a question for my first year engineering physics course involving a circuit with 2 loops and 3 voltage sources. I have tried 6 times and gotten 6 wrong answers! Can anybody offer me any advice for where I am going wrong?

Here is the question, my latest attempt, and the circuit in PHeT simulator

Aug 17, 2013
858
39
In your first equation of loop 1 E2 should be added to E1, not subtracted.

3. ### hogesjzz30 Thread Starter New Member

Sep 1, 2013
12
0
When I do that I end up with I2 = 10.36A....?

Aug 17, 2013
858
39
You got most signs in the first equation wrong (I guess in all equations).... you need to imagine a direction of the current (either clockwise or counterclockwise) on each loop and then go through each element looking on whether the sign of the voltage is in the direction of the current, or against it. If the the element has a negative to positive in the direction of the current, then it's a positive element on the equation; if not, then it's a negative; like this...

E2 + E1 + (I1*R4) - (I2*R2) + (I1*R1) = 0

Last edited: Sep 3, 2013
5. ### hogesjzz30 Thread Starter New Member

Sep 1, 2013
12
0
So from the assumed direction of the currents given the current would flow anticlockwise in both loops. So that gives E1 + E2 in Loop 1 as you say, and which I have changed it to, and E2 - E3 in the second loop, which is what I have. Since voltage always drops over a resistor shouldn't all of the values be (-Ix)Rx..?

Aug 17, 2013
858
39
Sorry, like this....

E2 + E1 + (I1*R4) - (I2*R2) + (I1*R1) = 0

Last edited: Sep 3, 2013

Aug 17, 2013
858
39
You go element by element in a direction and check if the current goes in one or another direction; you had the E1 and E2 right, it's I2*R2 you got wrong.

8. ### Raymond kunda New Member

Aug 26, 2013
7
0
have got the correct answer now?am trying to analyze the question
hey man if the the arrows are showing the assumed direction of current...............
taking it like that is not right because current cant flow from negative to positive

Last edited: Sep 3, 2013
9. ### hogesjzz30 Thread Starter New Member

Sep 1, 2013
12
0
wait. so now I'm adding E1 and E2 again? I think I am ready to throw out this whole question, This is the 3rd piece of advice I've had and every time I keep getting told different things, and none of it matches up with my lecture notes or textbook >_<

10. ### hogesjzz30 Thread Starter New Member

Sep 1, 2013
12
0
So then for loop 2 I should have E2 + E3 + (I3*R3) + (I2*R2) = 0 ?

Aug 17, 2013
858
39
E3 goes in the opposite direction, so it should be subtracted.

I need some sleep now... just check the signs according to the directions you were given the currents.

I'll be back later.

12. ### hogesjzz30 Thread Starter New Member

Sep 1, 2013
12
0
No, still getting the wrong answer, and I can't figure out why

Aug 17, 2013
858
39
14. ### Raymond kunda New Member

Aug 26, 2013
7
0
its the thing of coming up with good equations and solve them simultaneously

Aug 17, 2013
858
39
This is the part you got wrong....

Yes, voltage drops on all resistors; but this means that on each loop:
E1 + E2 + ..... n = (I1*R1) + (I1*R2).... n
or what is the same
Et - (It*Rt) = 0
But what we need to find out is the direction of the current to know whether or not we write a positive or a negative sign. For example: if the arrow of Ix goes in the same direction, then you write a positive sign, and if it goes in the opposite direction, then a negative sign. But you do it that way in the first equation I wrote; then change the signs when you pass them from side to side (the second equation up here).

Last edited: Sep 3, 2013
16. ### hogesjzz30 Thread Starter New Member

Sep 1, 2013
12
0
SO my equations should be:
For Loop 1: E1 + (I1*R4) - E2 - (I2*R2) + (I1*R4) = 0
For Loop 2: E2 - E3 + (I3*R3) + (I2*R2) = 0

17. ### hogesjzz30 Thread Starter New Member

Sep 1, 2013
12
0
Yeah I know that's how you do it, I just have absolutely no idea how to get the right equations! I have been at this one stupid problem for 3 days now and I am starting to lose it!

Aug 17, 2013
858
39
Last post.... (you only got 2 wrong)

E1 + (I1*R4) + E2 - (I2*R2) + (I1*R4) = 0
E3 - E2 + (I2*R2) + (I3*R3) = 0

I forgot to tell you something above: the arrows are for the I signs on each resistor, for the batteries it depends on the direction of the loop. So, if 2 batteries are on the same direction (the direction of the loop) they are added, and if they are in opposite directions they are subtracted; that's common sense.

Last edited: Sep 3, 2013