# Kirchoff's rules and RC circuits [closed]

You can view the image bigger: http://i.stack.imgur.com/4moUP.png

I'm studying RC circuits and I can't understand the part when we use Kirchoff's rule. When working with batteries, when we're going through a branch, when we hit on the negative plate first we take it's voltage to be positive. How does that work with capacitors ? How do we know which plate will be positive? Also, the person who solved this wrote:

$Vc + I1 * 1 Ω- I10 * 10 Ω = 0$

Wouldn't it be :

$Vc - I1 * 1 Ω + I10 * 10 Ω = 0$

Because, if we are going clockwise, when we're going through the $1Ω$ resistor we're going in the opposite direction of the current and for the $10 Ω$ resistor we're going in the same direction as the current. Also what if I went counterclockwise would the $Vc$ be negative? And finally is this solved correctly?

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## closed as off-topic by jinawee, Brandon Enright, Valter Moretti, John Rennie, Kyle KanosMar 30 '14 at 16:43

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Consider a circuit (a single loop) with only a battery and a capacitor in it. That will tell you what the convention for the capacitor has to be, and you will always be able to recover it when needed. – dmckee Mar 29 '14 at 18:23
Sorry, I don't understand. Can you please explain it in more detail? – user1949350 Mar 29 '14 at 18:30
Kirchoff's Loop rule comes from the notion that each part of the circuit must have exactly one value of potential, so around any loop the changes in potential must add up to zero. From that understanding you can figure out the convention for yourself by setting up a very simple circuit where you know all the other contributions. The simplest one involves only a battery and the capacitor. Once you have learned to do this you will never have to wonder about the conventions for Kirchoff's rules again. – dmckee Mar 29 '14 at 19:37
Took me a while, but I understand it now. Thank you a lot! – user1949350 Mar 29 '14 at 21:06

You need to assume how the current gets divided across the whole circuit. For example, take the current coming out of the main battery as $I$, then when it reaches the loop that you have selected, it breaks into $I_1$ and $I-I_1$. Do the same for all the loops, then apply Kirchoff's Rule. For capacitors, apply it exactly the same way you do it for batteries.

Remember, you just need to see when the current is going from a higher potential to a lower potential and write positive voltage for it. Exactly the opposite when it is lower to higher.

The loop where you have made the red mark will have the equation as $V_c+(1)(I_1)-(10)(I-I_1)=0$

Do the same for the others.

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How am I going in the direction of the current in both cases? If you look at the picture and the assumed directions of the current which are correct according to kirchoff's rule. I've added a picture for simplicity. I'm sorry if I'm misunderstanding something. i.stack.imgur.com/TXbNs.png – user1949350 Mar 29 '14 at 19:08
Wait if you are considering that loop, why are there two terms with 10 omh resistors? You need to use multiple loops and then solve for I. – uncertainty_principle Mar 29 '14 at 19:34
Please look at the bigger picture, the current is already solved. – user1949350 Mar 29 '14 at 19:36
Read my edited answer and the comment by dmckee on the quesion. Figure out the conventions for using negative and positive signs yourself. – uncertainty_principle Mar 29 '14 at 19:39