# Homework Help: Approaching capacitor circuit question

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1. Nov 4, 2016

### RoboNerd

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

The problem is above, I am having a hard time figuring out why the right answer for part F is "greater than."

2. Relevant equations

Ucapacitor = 0.5 * C * V^2

3. The attempt at a solution

My problem is figuring out the logic in how the capacitor's energy would increase. Now, I know that because I have a resistor with a smaller resistance that the total current in the circuit will increase. That is not a problem.

The constant of the capacitor remains the same, and because the R1 and C are in parallel, they have the same voltage drop across it. The battery's voltage remains the same so by Kirchoff, I would have:

EMF - (Itot * R2New) - (votlage drop across R1 or capacitor). The EMF remains the same. Itot increases while R2NEW decreases, so I do not know if the voltage drop across the capacitor will increase or decrease.

2. Nov 4, 2016

### Staff: Mentor

What determines the final voltage across the capacitor?

3. Nov 4, 2016

### RoboNerd

Voltage on capacitor = The charge on it divided by its capacitance.

4. Nov 4, 2016

### Staff: Mentor

Allow me to clarify my question. What circuit elements determine the final voltage on the capacitor?

5. Nov 4, 2016

### RoboNerd

the emf of the battery and the voltage drop across Resistor R2 by Kirchoff's voltage law

6. Nov 4, 2016

### Staff: Mentor

Okay, how do you determine that voltage drop across R2 at steady state?

7. Nov 4, 2016

R2 * Itotal

8. Nov 4, 2016

### Staff: Mentor

And what is Itotal?

9. Nov 4, 2016

### RoboNerd

EMF / R1?

IDK. ITOTAL varies with time

10. Nov 4, 2016

### Staff: Mentor

Ah. No. The total current varies with time only until the circuit reaches steady state (which is why the problem statement uses words like "after a long time", or "the switch remains closed for a long time"). When the circuit reaches steady state all transient phenomena have finished and only steady currents and voltages remain. No more changes to currents or voltages anywhere in the circuit.

I'm getting the feeling that perhaps you're missing a key concept that would make this problem much clearer, and easier for you to figure out.

The thing about capacitors in DC circuits is that, when they reach steady state they also reach their final charge and potential difference; their current goes to zero as their charge stops changing. With no current flowing into or out of the capacitor it is effectively an open circuit. At that point you can in fact remove the capacitor from a circuit without disturbing the steady state of the circuit. So you can analyze the circuit by pretending that the capacitor isn't there! Just determine what the potential would be between nodes where the capacitor connects if the capacitor were not there.

11. Nov 6, 2016

### RoboNerd

Yes, meant that, but I forgot to verbalize it specifically in words. Sorry about that.

When the total charge builds up on the capacitor, there is no current going across it, so we are going to have the "effective current" going through the battery and through R1 and R2.

I total is thus: EMF / (R1 + R2), right?

But then I do not understand how this would lead me to conclude that there is greater energy in the capacitor

12. Nov 6, 2016

### Staff: Mentor

How does the energy stored in a capacitor relate to the potential difference across the capacitor? (What's the formula?)

Sure. So how does this relate to the potential across the capacitor? What potential difference in that string of components (EMF, R2, R1) is the same as that of the capacitor? How does it vary if R2 is varied?

13. Nov 7, 2016

### CWatters

Take a look at the graph. It shows the capacitor is charging towards 12V rather than the 20V of the battery. Ask yourself what controls the final voltage on the capacitor.

14. Nov 8, 2016

### RoboNerd

Energy = 1/2 * C * V^2
I have no idea.

I do know that current through R1 multiplied by R1 equals voltage dropa cross capacitor.

If R2 varies, the potential across R1 and thus the capacitor varies. If the R2 decreases, the potential difference across it decreases, but potential difference across the capacitor increases?

This discrepancy looks fishy. I noticed that but I do not know what to do about it
I have no idea. The charge on it controls it. How do I move on with figuring it out?

15. Nov 9, 2016

### CWatters

Correct, and that allows you to state how the energy stored on the capacitor changes and answer the question.

It's not fishy at all. It tells you that the voltage at t=∞ on the junction of the two resistors is 12V. When a capacitor is fully charged no current flows into it so it can be removed from the circuit without effecting the voltage at the junction of the two resistors. If you are you familiar with the "potential divider" circuit (eg your circuit but without the capacitor) you can calculate R2.

16. Nov 11, 2016

### RoboNerd

Yes, I can calculate R2. Doing that led me to figure out that R2 = 1000 ohms because 12 volts = 15000 ohms * Itotal [considering that the total current does not flow through the cap. anymore because it has reached a steady state]

OK.

Now for my question in response to:
I assume that my capacitor has reached steady state. we thus have a closed circuit composed of a battery and R2 and R1, where we have a current running through it and the voltage drop across R1 is a constant number.

I thus need to have a voltage drop of 8 volts across R2. Decreasing the resistance of R2 will increase the total current, so the voltage drop over R2 will be:

V = Itot * R2. But then with one variable decreasing in value and the other variable increasing in value, how can we know if the voltage drop across it will increase or decrease?

Thanks for putting up with me and I apologize if it takes me some time in trying to figure out the answer and why the other answers are wrong

17. Nov 11, 2016

### TomHart

Can't you just use the voltage divider of R1 and R2?
To get the voltage drop across R2, you multiply the source voltage (20 V, in your case) times the divider ratio of (R2)/(R1 + R2).
When R2 decreases, both the numerator and the denominator of that ratio decreases. But which one decreases faster - the numerator or the denominator? That will tell you whether the voltage is increasing or decreasing.

18. Nov 11, 2016

### CWatters

+1

Just use the potential divider rule. If R2 reduces the voltage at the junction of R1 and R2 increases. So the energy in the capacitor increases. It's that simple. No need to worry about the current....

Vc=20*R1/(R1+R2)

So if R2 gets smaller Vc gets bigger.

Last edited: Nov 11, 2016
19. Nov 13, 2016

### RoboNerd

I can not say that I am familiar with the concept. Would you be so kind as to recommend a good resource that you know is the best for understanding this concept?
I think if I get this voltage divider concept, I can then progress and solve this problem. Please correct me if I am wrong.

Thanks for putting up with me and my slow progress in trying to reach the goal of cracking this problem.
Thanks for the help and patience!

20. Nov 13, 2016

### TomHart

It really is a simple concept so I will try to explain it. If the voltage across two series resistors is known, the voltage drop across each individual resistor can be calculated by this "voltage divider" calculation.

Let's say I have a circuit that consists of a DC voltage source at a voltage of Vdc Volts, connected to a resistor R2, that is connected to another resistor R1, which connects to the low potential side of the DC voltage source, thus completing the circuit. (It would be the exact circuit you have in your original problem statement if the switch and capacitor were removed.)

For that circuit, the voltage gets divided across those 2 resistors as follows:
The voltage across R2 will be VR2 = (Vdc)(R2)/(R1 + R2).
And the voltage across R1 will be VR1 = (Vdc)(R1)/(R1 + R2).

The voltage gets divided between the two resistors according to the size of the resistors - the larger resistor having the larger voltage drop.

21. Nov 13, 2016

### RoboNerd

So it is sort of like the fraction of the resistor in question divided by the total resistance. Right?
Instead of having voltage drop across resistor 2 = Itotal * R2 we just solve for Itotal using the same expression for resistor 1 and plug it in.

I understand the concept. Thanks for explaining it!!

voltage drop across r2 is thus = total voltage * R2 / (R1 + R2)
voltage drop across r1 is thus = total voltage * R1 / (R1 + R2)

Thus the voltage drop across r2 will decrease and voltage drop across r1 will increase as R2 decreases, so the voltage drop across the capacitor increases, increasing its energy.

Right?

22. Nov 13, 2016

### TomHart

When the switch is closed and the circuit has reached steady state (in other words, when the capacitor is fully charged), there will be no current flowing to the capacitor and the voltage across the cap will be equal to the voltage across R1 - whatever that voltage happens to be.

NOTE: The voltage divider only applies when the circuit has reached steady state.

23. Nov 14, 2016

### CWatters

24. Nov 18, 2016

### RoboNerd

Thanks for the help everyone.

So changing resistor R2 will make the voltage divider increase and the energy in the cap. thus increases!

Thanks a lot for the help everyone!

25. Nov 18, 2016

### Cutter Ketch

as stated the voltage divider rule looks like it came out of thin air. The point is the resistors are in series. They must carry the same current. The current across both together is

I = V / (R1+R2)

So that is also the current through each individually

V1 = I R1 = V R1 / (R1 + R2)

And there is your voltage divider rule.