# Kirchoff's law

Discussion in 'Homework Help' started by GE03STY, Oct 1, 2010.

1. ### GE03STY Thread Starter New Member

Oct 1, 2010
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I have been given a ring circuit to work out for my homework and i cant seem to do it. i have been able to do a radial circuit which is adding up all the voltages and what goes in comes out. But i cant seem to get it to work on this and i am after a bit of help and advice on what i am missing
here is a link to the circuit in question!!
http://s573.photobucket.com/albums/ss175/jupaluma/?action=view&current=ring_main.jpg

thanks matt

2. ### Georacer Moderator

Nov 25, 2009
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You gave us the circuit, alright! But what is the question?

3. ### GE03STY Thread Starter New Member

Oct 1, 2010
13
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sorry the question is, use the reults on the ring circuit diagram to confirm the above laws ohms law and kirchoffs law.

4. ### Georacer Moderator

Nov 25, 2009
5,175
1,284
Concerning the Ohm's Law, all you have to do is, for each resistance whose voltage is measured with a voltmeter, revise if the current, the voltage and the resistance revise the equation $V=I\cdot R$. There is the case of R6 where the current is uknown. In this case all I suggest finding the current through it, as this will enable you to find easily the current flowing through R5 and R7.

From then on, it's a matter of how thoroughly you want to revise the laws of Kirchoff. You can take any number of meshes or all of them and add the voltages around each one. The sum should be zero. Also take any number of nodes or all of them and add the currents flowing in and out of each one. The sum should be zero. Don't forget to take polarities into account!

I don't see anything more to this exercise.

5. ### GE03STY Thread Starter New Member

Oct 1, 2010
13
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when we got tought how to work Kirchoffs law out it was to add all the values together and they should add up to the supply voltage this is how i worked my other circuit out but the same method is not working on this. is this something to do with the inside ring circuit are am i doing something rong. the idea of the excersice is to show what goes into the circuit come out and i can get this to work.

6. ### Georacer Moderator

Nov 25, 2009
5,175
1,284
I have noted on the image with red the meshes where you should apply the Voltage Law and with green the nodes where you should apply the Current Law. Note that nodes that have nothing but sole wire between them can be viewed as a single node.

Post your efforts so we can see what you do wrong.

7. ### GE03STY Thread Starter New Member

Oct 1, 2010
13
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i dont understand what you mean about meshes and nodes? the bit in green is to apply kerchoffs current law and the (squares) with red arrows in around the edges of the circuit it to apply kerchoffs voltage law? sorry to be a pain just cant get my head around this

8. ### Georacer Moderator

Nov 25, 2009
5,175
1,284
Yes, as you said it. A mesh is a loop around a part of the circuit where you should apply the voltage law. A node is a point of the circuit where you should apply the current law.

Remember the words mesh and node. They are used very often and are essential part of the electrician's jargon. If you are not english and don't understand some words use google translation to convert them to your native language.

9. ### GE03STY Thread Starter New Member

Oct 1, 2010
13
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no i am english just not clued up on the electricans termonolagy. so how would i work out the voltage/current to get the values to do kerchoffs current law as seen as i only have resistor values in the nodes? what else needs to be worked out in the mesh a voltage accross each resistor? when this is completed should all my voltages add up to 12v?

10. ### GE03STY Thread Starter New Member

Oct 1, 2010
13
0
tryed to work the voltages out over the resistors but dont no if i have done it wright

* R4 = 0.1058mV
* R8 = 0.0453mV
* R10 = 0.045267mV
* R13 = 0.127v
* R2 = 2.1mV

Then to work the voltage accorss R5 do i do 10.58mA x 1k ohm
Then to work out the current in R7 do i do 275.10mV / 1K ohm
Then to work out the current in R11 do i do 452.67mV / 1k ohm
Then to work out the voltage in R12 do i do 20.52mV x 1K ohm

Last edited: Oct 1, 2010
11. ### JoeJester AAC Fanatic!

Apr 26, 2005
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The schematic is full of clues.

Do you know both Kirchoff's laws?

12. ### GE03STY Thread Starter New Member

Oct 1, 2010
13
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no we have just done the voltage law but gather that current law is simalier, please could you shed some light on because i am confused at this circuit

Apr 26, 2005
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14. ### GE03STY Thread Starter New Member

Oct 1, 2010
13
0
when you say re draw the digram could i get rid of the inside box and but the resistors were it is connected EG; put R5 between R3 and R4 are will it not work out the inside box is causing me lots of confusion

Oct 1, 2010
13
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16. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,912
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No. I meant start at the positive terminal of the battery (or the negative) and make the drawing go from left (battery) to right, following each connection. In the end you'd have clearly defined series and series-parallel circuits.

The battery's negative terminal connects to the inner ring and there are four resistors connected to that "node".

Here's the begining of my redraw ....

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17. ### Georacer Moderator

Nov 25, 2009
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Ooookay. It seems we do have a problem. I suggest you pay more attention to your operations. For example, for R8 we have $V_{R8}=I_{R8} \cdot R8=4.53mA \cdot 100\Omega=0.00453 \cdot 100 V=0.453V$. You have several such mistakes. Please revise all your operations.

Now, in order to solve your problem. First, I want you to find all the currents flowing through all the branches. You will find the current flowing thourgh R6 and R4 by $I_{R6}=\frac{V_{R6}}{R6}$. You will find the current on the branches that go towards the center by finding how much current escapes from the outer loop. That means that you should apply Kirchoffs Current Law for every node I have marked with green on the outer loop.

When you have found all the currents, I want you to find the voltage on each resistor. Use Ohm's Law: V=I*R.

Do the above and post your results.

18. ### GE03STY Thread Starter New Member

Oct 1, 2010
13
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were did you get the 100v from for that eqation $image=http://forum.allaboutcircuits.com/mimetex.cgi?V_{R8}=I_{R8}+\cdot+R8=4.53mA+\cdot+100\Omega=0.00453+\cdot+100+V=0.453V&hash=b81a3956fe8d1476f05c8ea80334557d$.

19. ### GE03STY Thread Starter New Member

Oct 1, 2010
13
0
$image=http://forum.allaboutcircuits.com/mimetex.cgi?I_{R6}=\frac{V_{R6}}{R6}&hash=a7846834d6a3efd7a80a9b16d4de3d94$ 0.2751v / 100ohm = 0.002751A = 2.751mA

0.002751A / 100ohms = 0.000027 x 100v = 0.00275v so the voltage over R4 is 2.7mV??? then work the current out
V / R = I then this is the current drop over R4???

Last edited: Oct 1, 2010
20. ### Georacer Moderator

Nov 25, 2009
5,175
1,284
First of all, sorry for the misconception with Volts and Ohms. V and Ω are on the same key, so I messed up a bit the language switches. I meant Ω as the previous expression.

Other than that, I suggest you read a bit on the Site's E-Book that JoeJester proposed. It seems you don't get along yet well with basic concepts of electronics.

You got the correct formula to calculate the voltage drop over R4 but you must keep the precision given by the problem. The correct answer is 0.2751V and not 0.27V.
Also, it is a waste of time to try to calculate the current through R4. Since we found it for R6 and this resistance is connected in series with R4, the answer is the same by default.
Additionaly, note that the phrase "current drop" is wrong. Current is not reduced while going through a resistor. Only voltage is.

If this exercise seems too hard for you, I suggest you leave it for now and concentrate on reading. If you want to try harder however, just find all the voltages and the currents and note them on the circuit's schematic. Pay attention to the polarities!