# C,R,V--argh

Discussion in 'MCAT Study Question Q&A' started by inaccensa, Aug 13, 2011.

1. ### inaccensa 7+ Year Member

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Ok so want to completely understand the relationship between C,Q,R,V. Based on other threads, I get the following idea.

-Capacitance and Resistance are both independent variables
C=kA/d and R=rhol/A, so it is sufficient to say in simple words that any changes in circuits does not affect C and R?

Voltage is set by the battery,
-->so based V=IR, if I increase this voltage, the current will increase and resistance will remain constant.
--> based on C=Q/V=Q/IR, R is not affected by changing voltage, but V increase the current and the charge of the capacitance.
In short, V doesnt change the C and R of a circuit??

-->Additionally, to decrease the time needed to charge a capacitor, can we increase the voltage?

Resistors in parallel allow for more work for the same amount of current?

What exactly is meant by the statement that as the voltage between the parallel plates increase, the capacitance increases?

many many thanks

2. ### liveoak

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yes, capacitance (C) and resistance (R) are based off of physical properties, those values are set in stone.

you're right on everything except increasing the voltage will not decrease the charge time. it will simply stack more charge onto the capacitor plates. we need to increase current to decrease charge time. (i think...someone help out here)

3. ### MD Odyssey 2+ Year Member

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The discharge rate of a capacitor depends both upon the capacitance as well as the resistive load that it is being discharged through. This is usually referred to as a time constant. It also depends upon the voltage that it was initially charge up to. Assuming that we haven't broken down the dielectric material in the capacitor, you can show that the charge on the capacitor at any given time t is

$image=http://latex.codecogs.com/png.latex?%5Clarge%20Q%20=%20C%20V_0%20%281%20-%20e%5E%7B-%5Cfrac%7Bt%7D%7Brc%7D%7D%29&hash=cd7ac82fb5a7e98cfae9615d581f6153$

If you want to discharge the capacitor through the load R, then the rate is just the current from the capacitor through the resistor, which is

$image=http://latex.codecogs.com/png.latex?%5Clarge%20I%20=%20%5Cfrac%7BV_0%7D%7BR%7De%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D&hash=53bfdfee70f87f1ee062c5da8cb5e8e2$

So, if you want to increase the initial rate of discharge off a fixed capacitance, increasing the initial voltage you charge it up to will do that. Note that we're talking discharge rates here, not overall discharge time. Clearly, from the equation, we can see that the time to discharge any capacitive load is infinite.

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4. ### MD Odyssey 2+ Year Member

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One can show that for two sheets separated by a distance that is much smaller than their length and width, the capacitance of those sheets is given by

$image=http://latex.codecogs.com/png.latex?%5Clarge%20C%20=%20%5Cepsilon_0%5Cfrac%7BA%7D%7Bd%7D&hash=5ccd99697340909d4e9e2296885e5c06$

where A is the area of the overlapping plates and d is the distance of separation. Similar results can be derived for other systems and in general, the capacitance is always defined by the geometry of the system. Applying a voltage to a capacitor doesn't change the physical arrangement of the system, so the capacitance is a fixed constant.

So, the statement that "as voltage increases, capacitance increases" is incorrect. Capacitors are typically made with a fixed value which you can read off the side, usually in the pF range. I've seen large capacitors that were the size of a D battery that may have been in the 500 mF range or maybe even larger. It doesn't matter what voltage you put on them or what resistance you discharge them through - their capacity to hold charge is fixed by the geometry of the system and their dielectric constant.

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5. ### plzNOCarribbean

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yeah, I don't think that increasing the voltage of the battery will decrease the time it takes to fully charge the capacitor.

Q=CV.... I don't think it holds true that "as the voltage between the plates increases, the capacitance increases". We see that if we re-arrange this equation, that Q/C=V; Therefore, V is inversely proportional to capacitance. Again, though, capacitance is an inherent, intrinsic property of that plate that depends on cross sectional area of each plate and the distance that separates them, so I don't believe that it would hold true that increasing the voltage would increase the capacitance (the ability to store charge)

***I think what implies is that as the voltage, or the potential difference between the two plates increases, there is a greater tendency for current (charge) to flow. This *may* mean that charge wants to the plate faster, which *may* mean that the time it takes to charge the capacitor is decreased, but again I'm not sure.

Given a constant capacitance, higher voltage would mean that more charge would store on the plate; ie C=constant, so based on Q=CV, increasing V would mean increasing Q, or the charge on the capacitor.

lastly, work= Power x time; P= IV = I^2 R = V^2/R; given that the current is constant (you said the same amount of current), then resistors in parallel decrease the overall resistance; So, were looking for an equation for power than includes both current and resistance; this would be equation #2:

P= I^2 x R; we see that R decreases in the case where resistors are in parrallel, so the power must also decrease; if power decreases, then the work done would also decrease if the time interval has not changed.

hope that helps; would be curious to see what others say about the first situation where if increasing V results in shorter time period to full charge the capacitor (although I believe that has to do with the time constant)

6. ### plzNOCarribbean

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could you elaborate? I am a little confused; I think I understand what your saying. you said were talking about "discharge rate, not overall discharge time". So you mean that by increasing the voltage of a fixed capacitance, that initially the rate of discharge would be faster but the overall discharge time is not? The overall discharge time doesn't change?

I understand that a capacitor cannot ever be fully discharged because the time constant graphs and the discharge graphs level out but never, theoretically ever reach a certain maximum. but if we increase the voltage, and thus that translated into increasing the rate with which charge leaves the capacitor (initial discharge rate), then wouldn't the time of discharge decrease, since rate is increasing? (rate = 1/time) I *think* what you mean is that the time at that particular interval with which the rate increases would decrease for discharge, but the OVERALL time of discharge for the capacitor is infinite???? < is that correct MD-ODYSSEY?

7. ### MD Odyssey 2+ Year Member

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Notice that I said the initial rate. If you look at the equation showing the discharge rate (i.e., the current) it isn't a constant. The differential equation which describes a capacitor discharging has the same form as that of a bacterial colony that is multiplying. Since it's not a constant, it makes more sense to talk about altering the initial rate. The initial rate increases as you increase the voltage it begins at or decrease the resistance you're discharging the capacitor through. Notice that the time constant changes if you mess with the resistance, so the charge left after a particular time has elapsed will be different.

I realize it's a subtle point, but try using your intuition about how a bacterial colony works. It's exponential growth and decay, just different contexts.

The same equation governs radioactive decay too. The initial rate of decay (number of particles that decay per time) is clearly dependent upon the number you start with. Let me try to explain.

If I have a compound with a 1 minute half life and I start with 2 molecules, after one minute, I'll have on average one molecule left. This gives me an average initial rate of 1 molecule / minute. On the other hand, if I start with 200 molecules, after one minute, I'll have 100 molecules left on average, which gives an average initial rate of 100 molecules / minute. The initial rates are different, even though the half life is the same.

Make sense? It's a different context, but the same physics apply because the ODE that we use to model their behavior is the same.

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Last edited: Aug 13, 2011
8. ### MD Odyssey 2+ Year Member

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Let me try to describe a bit more clearly what I'm talking about. I take a capacitor and hook it up to a voltage source V and then wait a long time so that I know the voltage on the capacitor is V. Then, I unhook it from the battery - the voltage is still V - and attach it to a switch and a resistor in parallel. The voltage on the capacitor is still V, because I kept the switch open. Now, at t = 0, I close the switch. The initial rate of discharge, which is just the rate of charge being pulled off the capacitor, is just V/R - initially, it looks just like Ohm's law. But, after that initial event, things get more complicated and you have to use some calculus to actually figure out what's going on. That's what the current equation I gave earlier does - it tells you the flow rate of charge off the capacitor as a function of time - it isn't a constant, so to speak of "discharging faster" or "discharging slower" would only make sense if you told me the time you were interested in. But, since the function for the discharge rate goes all the way out to infinite, the discharge rate of any two systems will eventually be arbitrarily close to one another (it might just take a while to get there).

Hope this helps.

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9. ### MD Odyssey 2+ Year Member

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There is no such a thing as a "full charge" on an ideal capacitor. It can take as much charge as you are willing to put on it. In fact, the charge you are able to put on it is going to be proportional to the voltage you charge it at. That's what capacitance is - it's the constant of proportionality for charging!

Of course, in practice, there is a finite limit to how large a voltage you can apply to a capacitor, since eventually the electric field between the plates becomes large enough to cause dielectric breakdown. At this point, you've burnt out your capacitor, usually causing the magical blue smoke to escape through the side of the cap. To prevent this, capacitors are usually rated to a particular voltage which is dependent upon the nature of the dielectric. Often times capacitors are intentionally used at a lower voltage than rated because turn-on transients are often large and can lead to dielectric breakdown over time.

Wow...I just realized that I must sound like a huge nerd.

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10. ### liveoak

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regardless, anything regarding charging/discharging TIME will not be tested on the mcat.

get that EE **** outta here!

11. ### MD Odyssey 2+ Year Member

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I hate to sound like a know-it-all, but after re-reading some of the earlier posts, I think I know where some of the confusion might be coming from.

This equation:

$image=http://latex.codecogs.com/png.latex?%5Clarge%20Q%20=%20CV&hash=83a6de4c85383c965878c012aae80e37$

describes the amount of charge that is deposited upon a capacitor at a particular voltage. It's true that you can rearrange it as

$image=http://latex.codecogs.com/png.latex?%5Clarge%20C%20=%20%5Cfrac%7BQ%7D%7BV%7D&hash=b0b3cb673d8a4c4029d1dc29729b4a20$

The misunderstanding some of you may have is in thinking that Q and V are independent variables. They are not. The rearranged equation simply lets you know what the capacitance of the two-plate system is if you know the charge on the plates and their voltage difference. It is incorrect to think of this as a system where I can change the voltage and get a different capacitance. It doesn't work like that - the expression is relating charge on the plates to the voltage of the plates. That proportionality is described explicitly by the first equation and, as I've already pointed out once before, the capacitance of the system is the constant proportionality between the charge and the voltage.

Make sense now?

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12. ### MD Odyssey 2+ Year Member

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I am not an EE. You cut me deep dude.

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13. ### MD Odyssey 2+ Year Member

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Here's what a capacitor looks like when you put too large of a voltage across it:

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14. ### plzNOCarribbean

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I think this is what is confusing people. So, if the capacitance is constant (which it is because the plates doesn't all of a sudden get bigger or have the distance between them change), then by increasing the voltage, we can increase the amount of charge that can be stored but that there is a limit to how much charge can be stored, which is dictated by capacitance of the capacitors in the circuit. is this correct?

15. ### MD Odyssey 2+ Year Member

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No. For an ideal (MCAT) capacitor, there is NO limit to the amount of charge you can put on it. The capacitance tells you what the voltage will be for a given amount of charge. Other than your underlined statement, you're correct. Capacitance is a constant, altering the voltage just alters the charge you are storing at that voltage.

For realistic devices in the real world, there is a limit to how much you can put on the device because if the charge is too large, the electric field between the plates becomes huge enough to allow charges to tear through the material separating the two plates, which destroys the device.

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16. ### plzNOCarribbean

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perfect, thanks. Yeah I remember reading that you can have dielectric breakdown when the potential difference between the two plates become too large due to the amount of charge that is stored.

17. ### MD Odyssey 2+ Year Member

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Good. I'm glad you get it now.

Yeah, that's the rub with capacitors - to get really big capacitance, you need the plates to be really close together, but then the risk of dielectric breakdown becomes huge. There is a lot of science behind the actual fabrication of circuit components. I built a 300 pF capacitor using a couple of 1 sq. m squares of aluminum plates separated by a 3" piece of foam. If you want a REALLY big capacitor, find an old tube TV set and dig around in the back of it. I've got a 1 F capacitor on my desk that I got from a TV in a junk yard. It's about the size of a D battery and it can hold 100,000 V for hours before the charge bleeds off.

There is a guy that works down the hall from me that does nothing but work on this kind of thing, which I would find terribly boring.

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