# The unusual problem with two capacitors

Discussion in 'Homework Help' started by MrRockchip, May 29, 2010.

1. ### MrRockchip Thread Starter New Member

May 25, 2010
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This problem is very difficult, because there're TWO capacitors!
And I can't find any examples for this kind of problem.

http://myimgs.net/images/pbah.jpg

How to solve this problem: using differential equations, or Laplas transform?
Please, show me the beginning of solution, and I will do the rest by myself.

Apr 20, 2004
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3. ### MrRockchip Thread Starter New Member

May 25, 2010
18
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No, I must not substitute two Capacitors with one.
It doesn't help me.

4. ### Markd77 Senior Member

Sep 7, 2009
2,803
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Am I reading the switch right? E charges C1 through R1 while C2 discharges through R2. When the switch is flipped C1 is in parallel with C2 and R2 and E and R1 are disconnected?

5. ### MrRockchip Thread Starter New Member

May 25, 2010
18
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Yes, after switch is changed, E and R1 are disconnected.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Then you are going to have some difficulty solving it. Seems to me you don't have a choice.

7. ### Ghar Active Member

Mar 8, 2010
655
73
This question is unsolvable. You cannot directly connect two capacitors, it's akin to paralleling two different voltage sources. You can't do it without accepting an infinite pulse of current.

Not only that but C1 has an undefined voltage without specifying a time. Current source into a capacitor gives infinite voltage.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Good point Ghar. It is unsolvable. Maybe the switch was drawn incorrectly. Which it probably wasn't.

Mind you - the problem could be solved assuming conservation of charge - at the instant t=0+. While the circuit violates the conservation of energy one could come up with an answer.

This issue of instantaneous charge redistribution on an ideal parallel capacitor network is a very old problem. My understanding of the energy disparity is that one postulates the emission of electromagnetic wave from the network - which is obviously beyond the scope of the problem at hand.

Last edited: May 29, 2010
9. ### MrRockchip Thread Starter New Member

May 25, 2010
18
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This problem can't be unsolvable: it's not real circuit,
so infinite jumps of voltage/current are possible.
Just show me the start of solution, and I will do the rest.

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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It may not be a real circuit but it is highly improbable. In my opinion it's a flawed question that should never have been set by the examiner. As I said earlier, at its heart the problem requires one to dispense with the fundamental principle of conservation of energy.

The only "solution" possible then is one based on the conservation of circuit charge at the completion of the switch transition.

The charge on C1 prior to switch transition will be Q1=C1E.

This charge will redistribute instantaneously at t=0 such that the combined parallel capacitance voltage at t=0+ satisfies the condition

V=V1=V2=Q1/(C1+C2)=E*C1/(C1+C2)

The solution is then a trivial exercise in solving for the discharge of the effective parallel capacitance from V to zero via R2.

The proviso here is that the switch orientation as drawn is correct.

It would make far more sense if the switch was merely transferring the centre source branch from one side branch of the circuit to the other side branch.

11. ### Ghar Active Member

Mar 8, 2010
655
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Problem - C1 is at infinite voltage unless there is a time specified for the switching event and you know the initial voltage.

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12. ### MrRockchip Thread Starter New Member

May 25, 2010
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At the lesson we have solved a very similar problem :
the only difference is that we hadn't the branch with a second resistor.

Is there a way to "transform" this solution to the solution of my problem ?

13. ### Ghar Active Member

Mar 8, 2010
655
73
Ok, so the possibly infinite initial condition is just ignored entirely...

What is R and U supposed to be?
Once the switch changes state there is no voltage source nor a resistance. Something is missing here.

The way you modify this is to add the resistor current to the first equation:
$C_1\frac{du_1}{dt} = -C_2\frac{du_2}{dt} - \frac{u_2}{R_2}$

And then because u2 isn't an impulse you just get:
$\int^{+0}_{-0}\frac{u_2}{R_2} = 0$

Making the term disappear and doing nothing for you...

The difference is that you actually do have a time constant with this resistor which will just discharge the the +0 voltage.

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14. ### MrRockchip Thread Starter New Member

May 25, 2010
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So, the answer will be the same ?

15. ### Ghar Active Member

Mar 8, 2010
655
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Sorry, I was pretty vague.

No, they won't be the same. I just meant that the +0 voltage calculation will be the same.
The differential equation for current has the R2 term that wasn't there before.

The solution you posted I don't agree with unless that schematic is incomplete. The charge will be redistribute and so the voltage will change, however, the solution is using a resistor which doesn't exist and a voltage source which doesn't exist. I have no idea what they're doing.
The schematic implies the voltage will stay the same forever after you close the switch.

In your problem with R2 there actually is a time constant. The charge will redistribute and then the capacitors will discharge through R2.

I made up values for the components and here is a comparison with R2 (yellow) and without R2 (blue). Red is the initially charged capacitor (C1) while yellow/blue the uncharged (C2). Of course after the switch C1 and C2 have the same voltage so the colors overlap.

Last edited: May 30, 2010
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16. ### MrRockchip Thread Starter New Member

May 25, 2010
18
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Sorry for the typo; in the solution - there's only one resistor - R1.

The voltage source is used only to power up the first capacitor - that's why E value is not used.

17. ### Ghar Active Member

Mar 8, 2010
655
73
It still doesn't make sense.
The solution is implying that a voltage source U is charging the combination of C1 and C2 with a resistance of R.

E and R1 are disconnected after the switch changes states so it can't be R1 in that equation. If U is what initially charges C1 then E and R1 never do anything.

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18. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Hi Ghar,

I'm not sure here whether you have assumed the source is a current source - I have assumed it's a voltage source, which is a matter also at issue with respect to the drawing notation. That arrow with a voltage symbol E next to it is a distraction.

If it is a voltage source then .....

The usual assumption in this type of problem involving switching operations at t=0 is that the circuit is at steady state conditions just prior to the switch transition.

In this problem it would be a reasonable expectation that C1 had been "allowed" to charge to voltage E via the switch position shown in the schematic. The switch operation would then be made.

If it's a current source then it means this whole thing just a plain stupid question.

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19. ### Ghar Active Member

Mar 8, 2010
655
73
I know, I agree. The questions are always like that however they do often enjoy putting less obvious things such a current source for a finite time (which is perfectly reasonable). They labeled voltages as U so I figured E might as well be a current. I've never seen an arrow as a voltage source before either.

This issue goes further though. The solution posted continues charging the capacitors after the switch changes meaning there's some missing information.
Maybe the switch flips for 0 time and then goes back (momentary switch)? That would also explain why the capacitors continue charging.
This is a strange and incomplete problem no matter how look at it.

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20. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Hi MrRockchip,

I'd be seriously concerned about what you are being taught by way of the solution outlined in this other example.

Let's come back to a realistic case of this other example in which we allow for the existence of a series resistor R somewhere in the circuit after the switch condition changes.

Suppose C1 has an initial voltage E volts and C2 has zero volts.

A series loop current will flow according to the relationship

$i(t)=\frac{E}{R}e^{-\frac{(C_{1}+C_{2})t}{RC_{1}C_{2}}}$

The voltage on C2 would be given by

$V_{2}(t)=E \frac{C_{1}}{(C_{1}+C_{2})}[1-e^{-\frac{(C_{1}+C_{2})t}{RC_{1}C_{2}}}]$

as t tends to ∞

$V_{2}(inf)=E \frac{C_{1}}{(C_{1}+C_{2})}$

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