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Suppose there are two capacitors ( of capacitance suppose C1 and C2 ) .There is a switch S1 between them which is open for lone time .Initially the capacitor C1 is charged to a voltage V1 volt .Now at time t=0 , the switch is closed instantly .What will be ultimate resultant voltage of the two capacitors after closing the switch ?

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  • \$\begingroup\$ What do you think will happen? Do you know what happens to the total charge in the system? Do you know how capacitors combine in series? \$\endgroup\$ – Roger Rowland Jun 24 '15 at 10:51
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    \$\begingroup\$ What is the other end of each capacitor (the end not connected to the switch) connected to? \$\endgroup\$ – Dave Tweed Jun 24 '15 at 12:01
  • \$\begingroup\$ What voltage is V2 at? \$\endgroup\$ – rsaxvc Jun 25 '15 at 3:37
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If the capacitors with different voltage and capacitance are connected in parallel,
we have Q= CV
Qcombined = Q1 + Q2 = (C1*V1 + C2*V2)
Ccombined = C1 + C2
Vresultant = Qcombined/Ccombined
What happens when they are connected in series?

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  • \$\begingroup\$ The question doesn't say they were connected in parallell. \$\endgroup\$ – avl_sweden Jun 24 '15 at 19:29
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This has been asked before (I can't find the link) BUT the correct answer is to analyse the charge before and after (it's the same by the way). The incorrect way to deal with this is to assume that the total energy stored in the two capacitors eventually becomes what was the original energy stored in the first capacitor.

It's wrong to analyse the energy because infinities are involved - you connect a charged cap to a discharged cap and infinite current flows. Then you might argue that there is a small resistor between the two and, what you will find is that the final energy in both caps is less than the energy in the single charged capacitor and, fully predicted by analyzing charge!

You might then assume that the interconnect is inductive but then the system oscillates and any attempt to dampen this oscillation results in removing energy - the bottom line is you might as well have just analyzed charge.

You can however, connect via inductors and diodes and get a final steady solution where the total energy remaining is almost the original energy less the small energy lost when the diode is conducting. Using a diode avoids indefinite oscillations taking place AND, what you have replicated is what a buck or boost convertor does!

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  • \$\begingroup\$ If you add a diode+inductor, you get a solution to a completely different problem. Besides, it sounds like you're talking about capacitors in parallel, not in series. \$\endgroup\$ – Dave Tweed Jun 24 '15 at 14:46
  • \$\begingroup\$ @dave I think, for the question to make sence, both caps should share a common 0 volts. \$\endgroup\$ – Andy aka Jun 24 '15 at 20:58
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If you connect two capacitors, each with voltage V1, V2 and charge C1*V1 and C2*V2, in series, then the total voltage across the two capacitors will be V1 + V2.

The total capacitance of the two capacitors is

Ct = \$\frac{C1 C2}{C1+C2}\$, so the charge must be

Qt = \$\frac{C1 C2(V1+V2)}{C1+C2}\$

I don't think it is particularly useful to think of conservation of charge here.

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In an ideal world, no current will flow, since there is no closed circuit.

So the charge of the two capacitors will be unchanged during the experiment.

In the real world there will be a capacitance between the air and the capacitors, and between the two capacitors, but I think the effect of that is negligible.

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