Finding Current with Kirchoffs Laws

Thread Starter

hogesjzz30

Joined Sep 1, 2013
12
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




 

adam555

Joined Aug 17, 2013
858
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:

Thread Starter

hogesjzz30

Joined Sep 1, 2013
12
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..?
 

adam555

Joined Aug 17, 2013
858
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..?
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.
 
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:

Thread Starter

hogesjzz30

Joined Sep 1, 2013
12
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 >_<
 

adam555

Joined Aug 17, 2013
858
So then for loop 2 I should have E2 + E3 + (I3*R3) + (I2*R2) = 0 ?
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.
 

adam555

Joined Aug 17, 2013
858
Since voltage always drops over a resistor shouldn't all of the values be (-Ix)Rx..?
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:

Thread Starter

hogesjzz30

Joined Sep 1, 2013
12
This is the part you got wrong....

Yes, voltage drops on all resistors; but this means that:
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.
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
 

adam555

Joined Aug 17, 2013
858
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:
Top