Physics----capacitance

Question

a capacitor is connected with a battery and stores energy U. after removing the battery, it is connected to a similar capacitor in parallel.

what is the new energy stored in each capacitor

Answer 1

Energy is not conserved in this operation, what is conserved is charge. This is a very difficult question, similar to one I used to use when interviewing new grad engineers, and they mostly got it wrong. Some teachers get it wrong, and I wouldn't be surprised if a text book got it wrong.

Remember that charge is always conserved unless you physically remove it. energy can be lost via lots of different mechanisms.

So the charge is divided between the two caps, assuming they have equal value. The voltage is cut in half, and the energy is:

E = ½CV²

Which works out to 1/4 the energy on each cap, because of the squared term.

In more detail:

Q = CV

assuming Q stays constant, and it has to, then

V = Q/C

Double the C, and V is cut in half.

With half the V:

E = ½C(V/2)² = (1/2)(1/4)C(V)² = 1/4 (½CV²)

So the E is 1/4 it's original value.

Why is energy not conserved? Because it is lost as heat or radiation when you connect the two caps together.  There is a very high instantaneous current when you first connect them together, because the only thing limiting their current is the resistance of the wires and the internal resistance of the caps. That high current will resonate with the wiring inductance and cause a resonant oscillation which will cause EM radiation.

The current will surge back and forth, with energy being lost due to heat and radiation, until it finally settles out with half the voltage on each cap.

Bottom line: either charge is conserved or energy is conserved, they can't both be as you would get different numbers. There is no way for charge to be lost, so it is energy that is lost, as described above.

.

Answer 2

Energy stored in a capacitor is given by equestion, U=C(V^2)/2

Now if the two capacitors are conected in parallel they have the same common voltage across them (V).

The new energy stored in each will be U=C(V^2)/2

But the total energy will be U=C(V^2) because the two capacitors are similar, hence multiplied the above equation by 2.

                                                                                

Answer 3

if the capacitors are the same, each will have 1/2 U

The first capacitor will have voltage across it equal to the battery voltage, and a charge q given by q=c*v

when that is connected in parallel with and identical capacitor, which is uncharged, half of the charge will distribute itself to the other capacitor.  But,  since capacitors don't dissipate energy, the total energy wont change, so half of it must end up on each capacitor.

Answer 4

Considering the second capacitor is fully discharged, after connecting both together in parallel, each capacitor will have the initial energy divided by 2, less the energy necessary to move electrons from the charge to the empty one.  This lost energy during transfer is mostly caused by the ESR (Equivalent Series Resistance) of both capacitors, and the not so good electric contact when you connect both.  The lost energy is calculated based on the square of the summ of the distribution current (that may go back and forth for a while until stabilization) times the added ESR from both the capacitors.

This technique is used to double the voltage of a battery.  A special electronic switching circuit is used to do the trick.  First one capacitor is connected to the battery or voltage source.  Then this capacitor is moved to stack over the voltage source, together they add almost double the original voltage, then a second capacitor is connected to the stack, it charges with the double voltage. The cycle repeats and the second capacitor keep charging with (almost) double voltage and feed its electric load.

Answer 5

We make the ideal condition that you have no resistance.So the energystored in the first capacitor will be,E=1/2CV^2,and the voltage accross it V=Q/C,so the amount of energy accumulated is;        E=1/2C(Q/C)^2 =1/2(Q^2)/C.After connection to the second cacitor,you have voltage(v)=q1/C=q2/C=Q/2C,(q1=q2,capacit... in parallel.)So the electrical energy in them is E1+E2=1/2C(Q^2/4C^2) +

1/2C(Q^2/4C^2)=1/4Q^2/C.Before connection E=1/2(Q^2)/C.

The amount of (1/2)Q^2/C - (1/4)Q^2/C=(1/4)Q^2/C is the WORK FUNCTION which causes to force the electrons to the first capacitor=

1/2 Mv^2 M is the mass of the electrons moved and v is their velocity.