I wanted to add but my morning poop time was over so never had a chance to write this.
If you have two capacitors, say C1 and C2, and say their capacitance is both C.
Say C1 is charged up to 2V. The energy stored in that cap is 1/2 CV^2 = 2C.
Say C2 is not charged up.
The charge on C1 is Q = CV = 2C. The charge on C2 is 0.
Say suddenly you connect C1 to C2 via a lossless wire.
The charge on C1 and C2 must be equal before and after the connection since charge cannot be destroyed.
After the connection, the voltage on both caps will be equal, and the charge on them will therefore be equal.
Charge on each cap is 1C; voltage is therefore Q/C = V, V = 1V on both caps..
Now let's look at the energy on either cap. 1/2 CV^2 = 0.5C.
The combined energy on both caps are 1C. Where'd the energy go? We just said we connected the capacitors with a lossless wire, so it can't be dissipated there, and capacitors by definition cannot dissipate power.
The answer is that lossless wires cannot exist, and if you do the math more carefully, this time with real resistance and take the limit as R->0, you will see that the power-time integral of the wire (dissipated energy) will approach 1C.
The same argument can be made for integrated circuits; as your resistance drops, more and more of the portion of loss will be dissipated in the "other" sources of loss.
edit: superconductors are lossy at AC (but not as much as regular conductors get lossy at AC), and the capacitor connection is an AC phenomena so even with superconductors there will be a teeny loss even with superconducting wires. The rest of the loss will happen in other ways (EM radiation, dielectric loss, loss from capacitor resistance, etc)
If you have two capacitors, say C1 and C2, and say their capacitance is both C.
Say C1 is charged up to 2V. The energy stored in that cap is 1/2 CV^2 = 2C.
Say C2 is not charged up.
The charge on C1 is Q = CV = 2C. The charge on C2 is 0.
Say suddenly you connect C1 to C2 via a lossless wire.
The charge on C1 and C2 must be equal before and after the connection since charge cannot be destroyed.
After the connection, the voltage on both caps will be equal, and the charge on them will therefore be equal.
Charge on each cap is 1C; voltage is therefore Q/C = V, V = 1V on both caps..
Now let's look at the energy on either cap. 1/2 CV^2 = 0.5C.
The combined energy on both caps are 1C. Where'd the energy go? We just said we connected the capacitors with a lossless wire, so it can't be dissipated there, and capacitors by definition cannot dissipate power.
The answer is that lossless wires cannot exist, and if you do the math more carefully, this time with real resistance and take the limit as R->0, you will see that the power-time integral of the wire (dissipated energy) will approach 1C.
The same argument can be made for integrated circuits; as your resistance drops, more and more of the portion of loss will be dissipated in the "other" sources of loss.
edit: superconductors are lossy at AC (but not as much as regular conductors get lossy at AC), and the capacitor connection is an AC phenomena so even with superconductors there will be a teeny loss even with superconducting wires. The rest of the loss will happen in other ways (EM radiation, dielectric loss, loss from capacitor resistance, etc)