Capacitor Charge/Discharge question - Electronic Circuits Projects Diagrams Free

ThermalRunaway · 30th May 2007, 12:27 PM

Hi everyone,

Please see the attached file for reference to a question I've recently been given as part of a test.

The question involves a capacitor charging from a supply at the changeover of a switch and I would normally find this kind of question very easy - you just use the formula VC = VS . 1- e-t/rc and transpose if you need any of the other variables (taking natural logs of both sides if necessary)

However, this particular question has confused me because you've got a capacitor charding via a fixed potential divider circuit. With the values given, this means that the capacitor can only charge to a maximum of 9V, because when fully charged the capacitor will be an open circuit and at that time the output of the potential divider will be 9V... right?

The first part of the question asks what the initial current will be. Well, if we're assuming that the capacitor starts off completely discharged, then initially it will be a short circuit so the current would be the supply voltage (18V) divided by the resistance of R1. Negligable current would flow through R2 due to the short circuit of the capacitor.

And this is where I start becoming confused. As the capacitor charges up, it will become more and more resistive. But then it's in parallel with R2, and I was unsure how this would effect the formula I originally quoted. The rest of the question goes on to ask how long it'd take to charge to a particular voltage, what voltage would be across the cap after a certain period of time etc but I'm guessing you can't use that formula anymore? If you can, what do you put for R in the equation?!?!?!

If only he had left out R2, I'd have flown that question.

Brian

lord loh. · 30th May 2007, 12:35 PM

It is a question of transient analysis...

I do not recollect the formulae right now... But once you switch on the cap,

There is a short transient respose and later there is a sustained steady state.

So do not look at the circuit from DC point of view.

Added later...
You can consider the cap short to being with.
charging with the time constant R1 C

later a resonance condition is to be considered of R2||C

and teh steadt state is when C is charged to a potential equal to Es x R2/(R1+R2) ... I think...

ThermalRunaway · 30th May 2007, 01:08 PM

Thanks for the response. I would agree with your comments, and infact I'm not confused on those matters. I understand that the initial condition would be for the capacitor to be short circuit, and I understand that the steady-state condition would be for the capacitor to be an open circuit with it charged to the output of the potential divider (or close to it)

My confusion surrounds what happens in between these two extremes. It's no longer as simple as using VC = VS . 1- e-t/rc because the charging resistor is no longer a constant, it changes as the resistance of the capacitor changes because the cap is in parallel with R2.

Brian

lord loh. · 30th May 2007, 01:27 PM

The in-between condition is the transient state... I do not recollect the formulae.

But the graph of current versus time shall take the form of a damped oscillation. and finally be a DC line.

The in between state is that of the damped oscillation or the transient state. and the DC state is the steady state.

Sorry for not being able to help you further. I am not good at remembering complicated (even the simple ones) differential equations such as the one which represent the inbetween transient state.

eng1 · 30th May 2007, 01:30 PM

Quote:

Originally Posted by ThermalRunaway

My confusion surrounds what happens in between these two extremes. It's no longer as simple as using VC = VS . 1- e-t/rc because the charging resistor is no longer a constant, it changes as the resistance of the capacitor changes because the cap is in parallel with R2.

Do you think that Thevenin's theorem might help finding the solution? Vs, R1, R2 can be replaced with an equivalent voltage source Veq with an equivalent series resistor Req. You might see if the equivalent circuit leads to sensible results.

Also, the similator might be a great help.

ThermalRunaway · 30th May 2007, 01:37 PM

Quote:

Originally Posted by lord loh.

But the graph of current versus time shall take the form of a damped oscillation. and finally be a DC line.

I disagree with this comment. A graph of current versus time would show an initial peak current decaying until finally no current was flowing at all. Unless of course you're talking about the small current which would flow through the potential divider circuit?

Brian

ericgibbs · 30th May 2007, 01:55 PM

hi brian,

Consider the resistance R1 as the source impedance of the voltage source.

Use Thevenin's theorem to solve the problem, the voltage rise across the Cap will be exponential. [which I suspect you already know].

eng1 · 30th May 2007, 02:00 PM

See the attachment.

ThermalRunaway · 30th May 2007, 02:02 PM

Eric, Eng1.

Hmmm, I think you're right. R1 could be considered the internal resistance of the battery, although it's a very high value. In any case, I guess you could determine the equivalent circuit and answer the question that way.

Thevenin's theorem it is, then. Pity I couldn't have thought of that during the test!!!

Thanks all,

Brian

ericgibbs · 30th May 2007, 02:09 PM

hi brian,
A final thought, why dont you construct the circuit, you have the values, use a CRO??... It will prove the result.
Eric

lord loh. · 31st May 2007, 04:03 AM

Quote:

Originally Posted by ThermalRunaway

I disagree with this comment. A graph of current versus time would show an initial peak current decaying until finally no current was flowing at all. Unless of course you're talking about the small current which would flow through the potential divider circuit?

Brian

Sorry... I guess I worded my thoughts wrong. That is what I meant.

And I think you need a real good CRO to to transient analysis - a storage one.

Roff · 31st May 2007, 12:18 PM

Quote:

Originally Posted by lord loh.

The in-between condition is the transient state... I do not recollect the formulae.

But the graph of current versus time shall take the form of a damped oscillation. and finally be a DC line.

The in between state is that of the damped oscillation or the transient state. and the DC state is the steady state.

Sorry for not being able to help you further. I am not good at remembering complicated (even the simple ones) differential equations such as the one which represent the inbetween transient state.

In the absence of inductance, there is no resonance.

lord loh. · 31st May 2007, 12:22 PM

Sorry Ron H, I do not follow. How do RC Oscillators like the wein's bridge, Phase Shift and Colpitt's work?

3iMaJ · 31st May 2007, 02:56 PM

Simply take the Laplace transform, solve the circuit with appropriate IC and take the inverse. Its the blind mathematical way to do it.

ThermalRunaway · 31st May 2007, 06:06 PM

LaPlace Transform?? Please don't sware at me!!!

hehe thanks for the suggestion.

Brian

30th May 2007, 12:35 PM	(permalink)
lord loh. Experienced Member	It is a question of transient analysis... I do not recollect the formulae right now... But once you switch on the cap, There is a short transient respose and later there is a sustained steady state. So do not look at the circuit from DC point of view. Added later... You can consider the cap short to being with. charging with the time constant R1 C later a resonance condition is to be considered of R2\|\|C and teh steadt state is when C is charged to a potential equal to Es x R2/(R1+R2) ... I think... __________________ Bharath Bhushan Lohray. M.Sc. Electronics. Last edited by lord loh.; 30th May 2007 at 12:39 PM.

30th May 2007, 01:08 PM	(permalink)
ThermalRunaway Experienced Member	Thanks for the response. I would agree with your comments, and infact I'm not confused on those matters. I understand that the initial condition would be for the capacitor to be short circuit, and I understand that the steady-state condition would be for the capacitor to be an open circuit with it charged to the output of the potential divider (or close to it) My confusion surrounds what happens in between these two extremes. It's no longer as simple as using VC = VS . 1- e-t/rc because the charging resistor is no longer a constant, it changes as the resistance of the capacitor changes because the cap is in parallel with R2. Brian __________________ --------------------------------- Electronics Test Development Engineer ---------------------------------

30th May 2007, 01:27 PM	(permalink)
lord loh. Experienced Member	The in-between condition is the transient state... I do not recollect the formulae. But the graph of current versus time shall take the form of a damped oscillation. and finally be a DC line. The in between state is that of the damped oscillation or the transient state. and the DC state is the steady state. Sorry for not being able to help you further. I am not good at remembering complicated (even the simple ones) differential equations such as the one which represent the inbetween transient state. __________________ Bharath Bhushan Lohray. M.Sc. Electronics.

30th May 2007, 01:55 PM	(permalink)
ericgibbs Experienced Member	hi brian, Consider the resistance R1 as the source impedance of the voltage source. Use Thevenin's theorem to solve the problem, the voltage rise across the Cap will be exponential. [which I suspect you already know]. __________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/

30th May 2007, 02:02 PM	(permalink)
ThermalRunaway Experienced Member	Eric, Eng1. Hmmm, I think you're right. R1 could be considered the internal resistance of the battery, although it's a very high value. In any case, I guess you could determine the equivalent circuit and answer the question that way. Thevenin's theorem it is, then. Pity I couldn't have thought of that during the test!!! Thanks all, Brian __________________ --------------------------------- Electronics Test Development Engineer ---------------------------------

30th May 2007, 02:09 PM	(permalink)
ericgibbs Experienced Member	hi brian, A final thought, why dont you construct the circuit, you have the values, use a CRO??... It will prove the result. Eric __________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/

31st May 2007, 12:22 PM	(permalink)
lord loh. Experienced Member	Sorry Ron H, I do not follow. How do RC Oscillators like the wein's bridge, Phase Shift and Colpitt's work? __________________ Bharath Bhushan Lohray. M.Sc. Electronics.

31st May 2007, 02:56 PM	(permalink)
3iMaJ Experienced Member	Simply take the Laplace transform, solve the circuit with appropriate IC and take the inverse. Its the blind mathematical way to do it.

31st May 2007, 06:06 PM	(permalink)
ThermalRunaway Experienced Member	LaPlace Transform?? Please don't sware at me!!! hehe thanks for the suggestion. Brian __________________ --------------------------------- Electronics Test Development Engineer ---------------------------------

Thread Tools
Show Printable Version Email this Page
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

Similar Threads
Thread	Thread Starter	Forum	Replies	Latest
Open Book Test Question	raywl	General Electronics Chat	3	5th November 2006 06:41 PM
Basic Transmission question	CamNuge	General Electronics Chat	2	8th March 2006 10:08 PM
High speed data logging question	cubdh23	Micro Controllers	3	19th August 2004 11:54 AM
Data Logging question with high speed!!!!	cubdh23	General Electronics Chat	4	16th August 2004 06:33 PM
PBX question	waqar	Electronic Projects Design/Ideas/Reviews	0	30th March 2003 06:21 PM