Understanding Current Direction and Voltage Drop in Kirchoff's Laws

[muffinccc]

Hi,

I've attached an image of the question I'm confused about. I don't know how to determine which way the current flows in each loop.

So far I've determined that I1 flows up and I2 goes down. So the loop on the right will be counter clockwise.

Because I2 goes down I3 should go up? I think??

So the loop in the middle goes clockwise.

Now I thought that since I3 goes up, I4 should go down making the left loop counter clockwise. But apparently its supposed to be clockwise.

The thing that's really confusing me is which way the current is flowing in I4.

Thanks

 

[turin]

I can't see the image yet, but if you are doing a mesh current analysis, then I don't need to see it. I will tell you that you simply choose the formal direction of the current in each mesh. When you solve for the mesh currents, you may find that some of them are negative, which simply means that your formal choice is the opposite of the physical choice, but that is no problem matter.

 

[muffinccc]

Sry, it's actually in pdf format so you might have to download it first.

I'm still kinda confused. Does the sign on the resistors depend on whether or not you're going along or against the current?

Cuz even if i chose a direction, don't i need to know the actual direction of flow in order to determine the sign? And the problem is i don't know how to find the actual direction of flow.

Thanks

 

[n!kofeyn]

http://math.fullerton.edu/mathews/n2003/kirchoff/KirchoffMod/Images/KirchoffMod_gr_1.gif

(Your attachment is still pending approval is why we can't see it.)

Here is an example. You just pick a direction for the loops, either clockwise or counter-clockwise. Then you sum up the voltage gains (- to + for voltage sources) and voltage drops (voltage of resistors and + to - for voltage sources) as you go around each loop.

For loop i_1:
e_1 - r_1i_1 - r_3(i_1-i_2) - r_5i_1 = 0

For loop i_2:
-r_2i_2 - r_4i_2 - r_6i_2 - r_3(i_2-i_1) = 0

Combining like terms gives the two equations:
\begin{cases}<br /> (r_1+r_3+r_5)i_1 - r_3i_2 = e_1 \\<br /> -r_3i_1 + (r_2+r_3+r_4+r_6)i_2 = 0<br /> \end{cases}

You now have two equations and two unknowns (if given the values for the voltage source and the resistors). You would then solve for i_1 and i_2[/tex]. Note that the resistor r_3 is a special case. Say for instance that we solved the currents to be i_1 = 0.7 ,\, i_2 = 0.1. Then the current over the resistor r_3[/tex] would be 0.6 and pointing downwards.&lt;br /&gt; &lt;br /&gt; (I found this figure on this http://math.fullerton.edu/mathews/n2003/KirchoffMod.html&amp;quot; .)

 

[muffinccc]

I can see how you did that. But like for example this one.

http://math.fullerton.edu/mathews/n2003/kirchoff/KirchoffMod/Images/KirchoffMod_gr_107.gif

I don't get how you can determine the sign for R6. Cuz i can't see how to find the actual flow of the current in that bottom loop.

Thanks

 

[n!kofeyn]

This is the figure you linked to:
http://math.fullerton.edu/mathews/n2003/kirchoff/KirchoffMod/Images/KirchoffMod_gr_107.gif

If we apply Kirchoff's Voltage Law to the loop for i_3, we would get:
-r_6i_3-r_5(i_3-i_2) - r_4(i_3+i_1) = 0
You travel around the loop subtracting off all voltage drops and adding all voltage gains (none in this particular loop).

Combining that with the other three loops, you will be able to solve for i_3. Then the resistor r_6 will have current i_3. From the way we set up the loop, if this number is positive, then the arrow will point to the right at r_6. If this number is negative, then the arrow will point to the left.

 

[turin]

Does the sign on the resistors depend on whether or not you're going along or against the current?

I don't know what you mean by "sign on the resistors". I will assume that you mean "determination of which side of the resistor is at a higher voltage". For that, again, it doesn't matter. If your formal choice matches the physical direction, then Ohm's law will be straightforward. If your formal choice is opposite to the physical direction, then the minus sign will multiply the negative voltage drop, and two negatives (when multiplied, of course) make a positive. Those are the only two possibilities.