# Kirchoff's Rule problem stuck

1. Mar 4, 2009

### zeldajae

Kirchoff's Rule...problem stuck!!!

1. The problem statement, all variables and given/known data

http://img5.imageshack.us/img5/5307/physishelp.jpg [Broken]

The Picture Explain above explains it all, I have attached a same file if you might need it! It is the same picture.

2. Relevant equations

V=IR
Thats all I know!

3. The attempt at a solution

V=IR=(3)(4+8)=36V

36V=I(6[ohms])-----I=6A????

I hope that's how you find amps!

Then I don't know how to find the voltage.

Do you find the voltage using the same equation???

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2. Mar 4, 2009

### yosefrajwulf

Re: Kirchoff's Rule...problem stuck!!!

Since charge is conserved then the electric current entering a node (where three or more lines join in your diagram) has to be the same as the current leaving the node.

Thus for the leftmost node I + Current through the 6 Ohm resistor = 3 A

Remember that energy is conserved, therefore the the sum of voltages (with their sign) through a closed loop is equal to zero.

Therefore you have

24 - 6 x Current throught the 6 ohm resistor - 12 -24 =0

Which means that the current through the 6 ohm resistor is -2 A (that is it runs in the direction opposite to what we supposed was for setting up the solution).

Which means that I= 5 A.
Applying again the voltage rule for the lower loop will give you V.

V=6(-2)+24+2(5)= 22 Volt

Last edited: Mar 4, 2009
3. Mar 4, 2009

### lanedance

Re: Kirchoff's Rule...problem stuck!!!

Hi Zeldajae

Do you know how to use kichoffs law in terms of current nodes & voltage loops?

not the easiest page to start, but have a look at
http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

basically once you have set up you conventions:
sum of all currents into each node is zero
sum of all voltage drops around a loop is zero

in terms of this problem - once you have found the voltage difference between the 2 nodes, you know this must be the voltage across each path between nodes

you have done this in you first step 36V.

use this to find the voltage across the 6ohm resistor, then think about the current through the resistor

from there you should be close to the finding the current through the last path & then the voltage of the source

4. Mar 4, 2009

### zeldajae

Re: Kirchoff's Rule...problem stuck!!!

Thank for replying. To find the amps through the resistor can't you just you this:

V=IR
36v=I(6)
I=6 Amps through the resistor.

Shouldn't that be the amps going through 6 ohms resistor? After that would you find the amps going through the 8 ohms resistor? Then the answer to that would be the current we are looking for...right?

And I still don't get what to do about finding the voltage????

5. Mar 4, 2009

### lanedance

Re: Kirchoff's Rule...problem stuck!!!

No, there is not 36V across the resistor

there is 36V across the resistor and the voltage source

you need to solve for the voltage drop across the resistor

6. Mar 4, 2009

### zeldajae

Re: Kirchoff's Rule...problem stuck!!!

ok, I got it. Thanks for the help guys!

7. Mar 4, 2009

### yosefrajwulf

Re: Kirchoff's Rule...problem stuck!!!

I was sloppy and originally solved it incorrectly. Please reread the post.