# Homework Help: Kirchhoffs 2nd for Capacitor

1. Jan 13, 2010

### Apteronotus

1. The problem statement, all variables and given/known data
Suppose we have the circuit as in the diagram where the potential on the capacitor is larger than that on the battery.
ie. $$V_c>V_b$$

What is Kirchhoff's 2nd law for the circuit?

2. The attempt at a solution

$$V_b+IR-\frac{V_c}{C}=0$$

My reasoning is as follows:
--going in a clockwise direction

$$+V_b$$ -- since taking positive charge from cathode(-) to diode(+) would add to the potential

$$+IR$$ -- since we are moving against the direction of current (ie. clockwise)

$$-V_c /C$$ -- since taking a positve charge from the top plate to the bottom plate would reduce the potential on the capacitor

2. Jan 13, 2010

### RoyalCat

It is $$-IR$$ and not plus. You are experiencing a voltage DROP as you go across the resistor.

I'm not sure what you've been taught, but what I've always found is best, is to look at potential in terms of energy. And to look at moving charges as either requiring energy, or as gaining energy.

In moving a charge from high potential to low potential, work is performed by the electric field on the charge, and you gain energy at the expense of the field. And in moving from low potential to high potential, you must invest energy so as to bring the charge across the voltage difference.

Now let's go over the circuit elements one by one, starting with the battery and going clockwise.
There is a voltage increase across the battery. In going across the battery from the low potential end (-) to the high potential end (+), the charge is accelerated and we gain energy.

On the other hand, due to resistance inside the resistor, you must constantly invest energy in moving through the resistor. Since you have lost energy by going through the resistor, $$-IR$$

Going further on, you encounter a potential drop, in moving from the high potential of the positively charged capacitor plate, to the low potential of the negatively charged capacitor plate, you gain energy, so $$+V_c$$

Also note that you wrote that the voltage drop across a capacitor is $$\frac{V_c}{C}$$, you got the formula a bit mixed up: $$V_c=\frac{q}{C}$$

It's been a while since I've done this, but I'm pretty sure I'm right..

Last edited: Jan 13, 2010
3. Jan 13, 2010

### Apteronotus

RoyalCat, first I'd like to thank you for your thorough response. I really appreciate it.
I have a couple of questions regarding your explanation.

1. Battery: since moving in a clockwise direction, and the (+) end of the battery is at the top, do we not have to invest energy to move a charge from the (-) end to the (+)?

2. Resistor: since every time with come across a resistor we have to invest energy to go through it do we always have $$-IR$$ when every we come across a resistor?

3. Lastly, if $$V_b>V_c$$ would this make any difference in our calculations?

Thank you again.

4. Jan 13, 2010

### Staff: Mentor

Does a sign matter? Once you solve, you may get the negative value for current - that simply means it flows in the opposite direction. You don't have to correctly guess the result before calculations.

5. Jan 13, 2010

### RoyalCat

Like Borek said, the actual sign doesn't actually matter. In choosing a direction to go in, we're assuming that that is the direction of current! (That's why a resistor is always considered a voltage drop in going around, even though for our chosen direction the opposite may be true)
If we end up with a positive sign for the current once all is said and done, then our assumption was right. If however we get a negative sign, that means that we were wrong about the direction of current and that a minus sign is in order.

What matters is not the actual signs, but the relationships between the signs you assign to the different circuit elements in going around. As long as you use a consistent approach, imagining what a test particle in your pocket would experience as you go around, you'll get a correct result.

We do need to invest energy to move across its terminals. But a battery is the source of the voltage and energy that's required! If you go across its terminals from (-) to (+), you are forcibly raised from low potential to high potential, you gain energy from the battery by going across!

As an exercise, let's now go in the other direction, again starting from the battery, and going CCW.

In going across the battery terminals we go across a voltage drop, but here's the tricky part: We are going against the direction of the field in the battery, so in fact, we have to invest energy to go across! Remember that a free charge would be accelerated by the battery from the (-) terminal to the (+) terminal, so in going the opposite way, we must overcome the force with which the battery tries to do this.
So our sign for the battery is $$-V_b$$

We now approach the negative capacitor plate and try to move up to the positive capacitor plate. To do that, we have to "climb up" a voltage, which means, to invest energy in doing so. So our sign for the capacitor is $$-V_c$$

And since we're overcoming the resistance of the resistor as we go across it, the sign is, as always, $$-IR$$

First let's go back to the CW result. For the CW result we found: $$V_b-IR+V_c=0$$
Rearranging:
CW: $$V_b+V_c=IR$$

And now we find: $$-V_b-V_c-IR=0$$
Rearranging:
CCW : $$V_b+V_c=-IR$$

Now you would come to me, and say, "But that's inconsistent!" since we got different signs for the current in each case.

But that would be overlooking the origin of each of those equations. A negative current CCW, is the same as a positive current CW!

Notice that we are dealing with positive quantities $$(V_b, V_c)$$ and so we find that for our CW equation, $$IR$$ is also a positive quantity. So the current $$I$$ is also positive (This vindicates our choice of direction for the current).
For our CCW expression, however, we find that the sum of two positive quantities gives us a negative! How can that be? The only way out of the conundrum is to understand that we were wrong about the direction of the current and that in fact, it proceeds opposite the direction we marched in.
And so, we've shown that the two solutions are perfectly consistent as long as we're peachy with being wrong about the direction of current once.

A good way to avoid confusion over this issue is to assume a direction for the current and to draw it as a CCW/CW arrow inside the circuit loop and then use that as your meter stick in going around. So when you come to a resistor, you aren't always marching in the direction of current, you could be marching against your chosen direction, and actually experience a voltage increase in going across its contacts $$(+IR)$$!

Though I find it best to stick to one method so as not to become too confused. Find whatever works for you, either assuming an explicit direction for current, or going through the energy considerations for every component. With each of these you should be able to solve circuits quickly after some practice.