Solving Circuits with Kirchhoff's Rules

[physicssux]

Homework Statement

Apply kirchhoffs rules to find the currents in the circuit shown

http://img297.imageshack.us/img297/5048/kirchvm8.jpg

Homework Equations

here's what the book says

i3(25 ohms) = 12 V

i2(15 ohms) = 6.0V + 12 V

i1 = i2 + i3

The Attempt at a Solution

i have tried over and over again on hoow they got the equations from but i don't understand.

where did i3 come from there are only 2 resistors. so only 2 current.

i just don't understand kirchhoff loop rule, is there any place where i can a ismple explanation because the book DOES NOT HELP

 

[denverdoc]

You might review the lecture notes found here.

https://www.physicsforums.com/showthread.php?t=95448

For this problem, a couple of observations:

the voltage change down the middle path must be 12V. It is in parallel with a a 12 Volt limb, so IR=12 where R=25.

Now along the outside loop, we have two batteries and a single resistance.
The drop must be zero so i2*15 must equal the sum of the two voltages. (Take everything else away except an 18 volt battery and the resistance.)

 

[physicssux]

hey thanks for the notes

but i still can't get understand it.

i think it's already built in in my head that i won't understand.

i just don't know how to get the equations.

how do you know if the current is positive or not, or if the voltage is positive or not, i just don't get t hose.

 

[denverdoc]

well your attitude re physics probably isn't helping.

But don't give up. I certainly didn't get those rules at first, and for many of us here at PF, physics generally is a hard fought battle for small chunks of understandings that over time can grow into an actual appreciation and love of the subject. Yea, sure you say. I just want to get thru this friggin class and have nothing to do with it again. Hey that's OK as well.

The signs aren't that hard but can be at first and maybe why you are getting confused.

lets try the small loop from lowerleft corner of circuit, since the wide line is on top as we move upwards, this is positive voltage

so 12V-i3*25=0

When current flows across a resistor we call these voltage drops, hence the negative sign. The direction of i3 is chosen as clockwise but this is arbitrary. We could just as easily gone the other way round the loop,
then -i3*25-12V=0. Now since we are going the other way across the battery, its -12. So the answer is different but that only tells us that its going in the opposite direction we assumed, so no penalty for guessing.

The other big loop is treated the same, again starting from lower left

12V-i2*15+6V=0 Had the voltage signs of the 6V battery been different, would have had to count that as -6V.

For the current let's consider the area in common between the loops. The currrent in that last stretch before we get to the lower left hand corner consists of contributions from the inside and outside loop. Technically we consider nodes, but intuitively one can see the total current there consists of contributions from both loops which were both treated as clockwise positive. So i1=i2+i3.

Look here for some more examples

:http://www.ualr.edu/dcwold/phys2122/p24man/p24man.html