Say we have two capacitors C1 and C2, initially isolated from each other and each having an initial voltage Vc1 and Vc2 respectively, where Vc1>Vc2. When suddenly connected to each other in parallel, obviously C1 will discharge to C2 until their voltages meet at a steady state value. Assuming zero resistance between the two capacitors, what is the expression for this steady state voltage Vc in terms of Vc1 and Vc2? Is it just the average of Vc1 and Vc2?
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1\$\begingroup\$ I think if you assume zero resistance, things get complicated. It's better to assume a very small resistance for the math to make sense \$\endgroup\$– SRRCommented Mar 28, 2019 at 6:24
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1\$\begingroup\$ before you connect them there will small air gap between them that will ionize air and start conducting some charge. after you connect them you will have infiniti amp for 0s. how? i dont know. anyway all caps have resistance within them you cant have 0 resistance. bottom line its intresting question but doesnt apply to real life and i dont know the answer. but just wanted to share my thoughts \$\endgroup\$– Hasan alattarCommented Mar 28, 2019 at 6:37
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3\$\begingroup\$ The charge is conserved. There is excess energy that gets radiated. \$\endgroup\$– analogsystemsrfCommented Mar 28, 2019 at 6:46
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\$\begingroup\$ @analogsystemsrf Thanks! had you come a little earlier I would have spared my friend the trouble (and myself the embarrassment)! \$\endgroup\$– FartVaderCommented Mar 28, 2019 at 6:49
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2 Answers
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Just got a likely answer to my own question by asking a friend, who pointed out that charge is always conserved.
The total charge in our circuit (two capacitors in parallel) is Q = Vc1*C1 + Vc2*C2. We can then equate this to the total charge when voltage has settled, Vc*(C1+C2), giving a final expression of
Vc = (Vc1*C1 + Vc2*C2)/(C1+C2)
Sure the circuit in question doesn't have real life relevance, but I think this equation is very handy when you're just concerned with the steady state values.
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2\$\begingroup\$ The electrons cannot escape. So Q remains the same. But there is an energy balance problem, unless we think about sparks during the connection, or L+C ringing that radiates. \$\endgroup\$ Commented Mar 28, 2019 at 6:52
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2\$\begingroup\$ I think we just bumped into the "Two-Capacitor Paradox" \$\endgroup\$ Commented Mar 28, 2019 at 6:54
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\$\begingroup\$ If initial voltages are different, KVL is violated in \$t=0\$. There an impulsive (\$\delta(t)\$) current in \$t = 0\$. Charge is conserved, energy not. The energy before closing the switch is \$>\$ than the energy after closing the switch (except for equal initial voltages). \$\endgroup\$ Commented Mar 29, 2019 at 1:44
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This question involves the capacitor paradoxon, where some part of the energy stored in both capacitors seems to disappear.
In theory, i.e. in the case of ideal capacitors, without resistances or inductivities or radiation of HF or other losses involved, the capacitors are strictly parallel after connection, which is a contradiction, since the initial value/voltage of the differential equation would be defined to be Vc1 and Vc2 at the same time, and Vc1 > Vc2.
The paralleled capacitors can be seen to be 1 capacitor after connection, having Vc1 and Vc2 at the same time. So this case is simply impossible, it is a mathematical contradiction concerning the initial values. The 2 conditions are Vc1 = Vc2 and Vc1 > Vc2.
In reality, there are always resistors, inductivities, spark gaps, other capacities, HF radiation etc. involved, i.e. these 2 capacitors are never directly connected in parallel as it is the case in theory.
