The rest of the energy was lost in the battery's internal resistance. At least in theory. Then you'll have to add wire resistance, cap's internal resistance etc..But the formula for power or work done for capacitor is: 1/2 VQ. It's half the work done by battery. Where did the rest of the energy, which is half of the supplied energy, go?
The rest of the energy was lost in the battery's internal resistance. At least in theory. Then you'll have to add wire resistance, cap's internal resistance etc..
Just think of a resistor and a cap in serie. When charging that cap, the current will be the same both for the resistor and the cap. Therefore, the resistor will steal half of the energy from the voltage source.
Energy = VI x t. Therefore, we can write, Energy = VQ.
To the Ineffable All,
OK, let's do a constant voltage (E) energy analysis across a capacitor (C) and resistor (R) in series. R represents the sum total of all the resistance in the circuit.
The instantaneous current is the well known i = (E/R)*e^(-t/RC)
First lets calculate the energy stored in C . The voltage across C is the constant voltage minus the voltage across R. So that is E-Ri. The incremental energy stored in the capacitor is dW = (E-Ri)*i*dt . Integrating dW across a time interval T, we get W = ½E²C-½E²C*e^(-T/RC)+½E²C*e^(-2T/RC) as the energy stored in the capacitor. Letting T run to infinity gives ½E²C .
Now let's calculate the energy dissipated in R . dW = i²*R*dt , so integrating dW across T gives us W = ½E²C-½E²C*e^(-2T/RC) as the energy dissipated by the resistor, or ½E²C when T goes to infinity.
The same results are found when the RC circuit is driven by a constant current source.
In conclusion, half the energy from E will be lost across the resistor, no matter what its value is.
Ratch
Empty capacitor and a constant voltage source gives you a potential energy of ½E²C. ½E²C is the amount of potential energy you have at your disposal. After you used it all, you've lost it all. You don't "waste the half". You just use it all.
This energy will be dissipated as heat in the inivitable resistance of the circuit no matter what its value is.
Ratch
The math is correct whatever the resistance is. And zero resistance does not dissipate any energy.
misterT,
Now you are discussing infinities (current) and infinitesimals (resistance). Energizing a capacitor without any resistance will take a infinite amount of current. That is not what practical problems are about.
Ratch
Still.. that is what the math says. I know real circuits always have resistance. The problem here is in the constant voltage supply. Every action has a reaction.. where is the reaction in this example?
Imagine if I am floating in space with a steel ball (1 kg). I push the ball away from me at 1 m/s. The ball now has kinetic energy ½mv². But when I pushed the ball forward, I also pushed myself backwards and I also have kinetic energy ½mv².
Now.. imagine a situation that I pushed the ball forward, but I managed to remain still. Where did one of the ½mv² go? That is exactly the same situation when you use an ideal constant voltage source in your calculation. The constant voltage supply brings energy to the system and therefore you do not have a closed system.
What does the math say? What is the problem with a constant voltage supply?
How do you figure that? You won't have the same velocity and mass as the ball.
Who cares if the system is open or closed?
The constant voltage supply brings energy to the system.
Different mass, different velocity, but the same kinetic energy. See my signature for a proof.
Every scientist I know. It is important to know if the system is open or closed. See my signature for a proof.
Ok. Can you show me a calculation for this: I have a charged capacitor C1 (100uF, 10V in it). I know that it has energy in it (½E²C) and I want to use that energy to heat up something. I decide to connect a resistor R1 and a capacitor C2 (100uF, 0V in it) in series with the charged capacitor. How much of the energy will get dissipated to heat? My intuition says "half".
"Why only half"? If I connect the resistor directly to C1, I will get all of the energy to heat up my stuff. So why does the capacitor C2 "waste" half of the energy. Well it does not waste it.. it only stores it.
Can you calculate the energy I will get if, after the first step, I connect the resistor R1 across both of the capacitors C1, C2? How much energy will I get out of them?
[After the two calculations, could you add up how much total heat I got? My intuition says 3/4 E²C. How did I get more heat out of the system than was initially in the capacitor C1?
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