Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Question about energy in capacitors

Status
Not open for further replies.

luckypearl25

Member level 2
Joined
Jan 6, 2006
Messages
46
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Activity points
1,618
energy in the capacitor

this is quite a common question asked in many interviews.There are two capacitors C.Intially one capacitor is charged to V volts while the other is at zero volts.At t=0,a switch connects both of them in parallel.Due to charge sharing the voltage at the node connecting both of the transistors become V/2 volts.Now the question is the energy before the switch was closed was 1/2 CV^2 and the total energy after the switch was closed is 1/4 CV^2.WHere is has the other half of the energy gone ?????
 

energy in the capacitor

how do you say the voltage is gonna settle at V/2.... there will be surge current surely and so it will oscillate with V amplitude and hence the rms value is V/root 2....
so you get 1/2 CV^2....
 

Re: energy in the capacitor

The energy has been dissipated in the connecting wires, as the cap with the lower voltage was charging up.

To understand this, try solving the same problem with a resistor connecting the caps, instead of and ideal wire.
Calculate the energy lost in that resistor. You will see that it does not depend on the value of the resistor, only on that of the caps and the voltages.
That tells you that even with low-resistance wires connecting the caps, the same energy is lost and it's simply lost in the wiring resistance.
 

Re: energy in the capacitor

luckypearl25 said:
this is quite a common question asked in many interviews.There are two capacitors C.Intially one capacitor is charged to V volts while the other is at zero volts.At t=0,a switch connects both of them in parallel.Due to charge sharing the voltage at the node connecting both of the transistors become V/2 volts.Now the question is the energy before the switch was closed was 1/2 CV^2 and the total energy after the switch was closed is 1/4 CV^2.WHere is has the other half of the energy gone ?????

Very Nice !!

but what is the solution ??? :D :D :D
 

Re: energy in the capacitor

Anand,

write the charge balance equation in the node we will get it to be V/2.where does surge current come here ?

vvv,
does this mean ,that irrespective of the wiring resistance exaclty half of the energy will get lost is it ?It seems a little abstract.can u explain it better
 

energy in the capacitor

how do you say it will settle at V/2... the capacitor is a reactive component... it will surely oscillate since the capacitor will not allow the C(dv/dt) term to zero quickly....
 

Re: energy in the capacitor

A.Anand Srinivasan said:
how do you say it will settle at V/2... the capacitor is a reactive component... it will surely oscillate since the capacitor will not allow the C(dv/dt) term to zero quickly....

reactive component does not dissipate power , it has an imaginery power , Only resistores have real power
 

Re: energy in the capacitor

There has been no energy lost.

There are 2 capacitors,

We'll call them C1 and C2 and we know that C1=C2 since when charge sharing takes place the voltage drops to V/2

So initially, before charge sharing the energy in the system is
1/2 C1V^2+1/2 C2V2^2 and V2 = 0V thus
1/2 C1V^2+0 = 1/2 CV^2

after charge sharing

1/2 C1(V/2)^2+1/2 C2(V/2)^2
1/4C1V^2+1/4C1V^2 and C1=C2
thus
energy in the system
1/2 CV^2
 

Re: energy in the capacitor

Old Nick said:
There has been no energy lost.

There are 2 capacitors,

We'll call them C1 and C2 and we know that C1=C2 since when changew sharing takes place the voltage drops to V/2

So initially, before charge sharing the energy in the system is
1/2 C1V^2+1/2 C2V2^2 and V2 = 0V thus
1/2 C1V^2+0 = 1/2 CV^2

after charge sharing

1/2 C1(V/2)^2+1/2 C2(V/2)^2
1/4C1V^2+1/4C1V^2 and C1=C2
thus
energy in the system
1/2 CV^2


something wrong in calc.

you said :
after charge sharing

1/2 C1(V/2)^2+1/2 C2(V/2)^2


it is (v/2)^2 => (V^2)/4
so (1/2)*C1*(V^2) /4 ==> (1/8 ) *C1*(V^2)

since it is two same cap so above * 2 ===>> (1/4)*C*(V^2)
 

Re: energy in the capacitor

ahmed osama said:
Old Nick said:
There has been no energy lost.

There are 2 capacitors,

We'll call them C1 and C2 and we know that C1=C2 since when changew sharing takes place the voltage drops to V/2

So initially, before charge sharing the energy in the system is
1/2 C1V^2+1/2 C2V2^2 and V2 = 0V thus
1/2 C1V^2+0 = 1/2 CV^2

after charge sharing

1/2 C1(V/2)^2+1/2 C2(V/2)^2
1/4C1V^2+1/4C1V^2 and C1=C2
thus
energy in the system
1/2 CV^2


something wrong in calc.

you said :
after charge sharing

1/2 C1(V/2)^2+1/2 C2(V/2)^2


it is (v/2)^2 => (V^2)/4
so (1/2)*C1*(V^2) /4 ==> (1/8 ) *C1*(V^2)

since it is two same cap so above * 2 ===>> (1/4)*C*(V^2)


Ah I see my mistake.

Strange one though, as the number of electrons in the system has not changed and the energy of an electron is a definite value. (of course there will be some losses in the real world due to wire resistances etc., but these are not taken into account here)
It guess it's to do with the electrons spreading themselves over the plates of a the capacitor, thus the field strength changing relative to the area (which is of course a square term) thus the same amount of electrons spread over double the capacitance will create an electric field less than 1/2 of the original.

Although I'm dubious about that too, however it is as you've demonstrated to do with the square of the voltage. Which leads me to beleive it is related to the spread of electrons/Plate area.
 

energy in the capacitor

Suppose we use a resistor in series with switch.
So when switch closed resistor consume other half of energy as VVV said.
In this example, energy consumption in Resistor is not related to resistor value,
so we can approach resistor value to zero.

regards.
davood.
 

Re: energy in the capacitor

Davood Amerion said:
Suppose we use a resistor in series with switch.
So when switch closed resistor consume other half of energy as VVV said.
In this example, energy consumption in Resistor is not related to resistor value,
so we can approach resistor value to zero.

regards.
davood.

how is R is zero and there is a power cons. on it ??

P=I^2 *R

so if R=0 ==> P is also zero
 

Re: energy in the capacitor

ahmed osama said:
Davood Amerion said:
Suppose we use a resistor in series with switch.
So when switch closed resistor consume other half of energy as VVV said.
In this example, energy consumption in Resistor is not related to resistor value,
so we can approach resistor value to zero.

regards.
davood.

how is R is zero and there is a power cons. on it ??

P=I^2 *R

so if R=0 ==> P is also zero

indeed,

It has nothing to do with resistances etc, since it is described in a 1st order equation which contains no reference to resistances or the like.
energy with voltage changing, it'll not be linear. Which makes sense, if you look at the charging curve of a capacitor under a constant current. Doubling the number of electrons stored in the capacitor doesn't double the voltage across it, which leads to afore mentioned effect of the 'lost energy', however I can't actually come up with a rational as to exactly what is happening, as energy certainly has not dissapeared.
Change the capacitance and the energy changes linearly with it, so my previous guess was wrong.
 

Re: energy in the capacitor

luckypearl25 said:
this is quite a common question asked in many interviews.There are two capacitors C.Intially one capacitor is charged to V volts while the other is at zero volts.At t=0,a switch connects both of them in parallel.Due to charge sharing the voltage at the node connecting both of the transistors become V/2 volts.Now the question is the energy before the switch was closed was 1/2 CV^2 and the total energy after the switch was closed is 1/4 CV^2.WHere is has the other half of the energy gone ?????

someone ask me a very nice ques. are they two C or two trans. ??

"..........connecting both of the transistors become V/2 volts..................."
 

energy in the capacitor

work is needed to move the charges from one cap to another cap, using a simple equation u can verify that, use the work=integration(deltaV*dq) where dq changes from 0 to final value of charges on second cap, and put the change in voltage as a function of charges.
i remember that i saw that question and the full answer here on the board before, try to search for it!
 

Re: energy in the capacitor

safwatonline said:
work is needed to move the charges from one cap to another cap, using a simple equation u can verify that, use the work=integration(deltaV*dq) where dq changes from 0 to final value of charges on second cap, and put the change in voltage as a function of charges.
i remember that i saw that question and the full answer here on the board before, try to search for it!

two links inded :D :D
 
Last edited by a moderator:

Re: energy in the capacitor

See the attachment for my two cents.
 

Re: energy in the capacitor

ahmed osama said:
Davood Amerion said:
Suppose we use a resistor in series with switch.
So when switch closed resistor consume other half of energy as VVV said.
In this example, energy consumption in Resistor is not related to resistor value,
so we can approach resistor value to zero.

regards.
davood.

how is R is zero and there is a power cons. on it ??

P=I^2 *R

so if R=0 ==> P is also zero

you must consider r=ε (i mean R gets close to zero)
This is similar to a circuit which you connect two ideal battery (with different voltages) insted of capacitors.
what is the final voltage !!!!!!! can you explain it?
Actually this question is not a case.
 

Re: energy in the capacitor

If you look at the equations, you will see that as the resistor decreases the peak current increases. For a truly zero resistance you get infinite current. Except that truly zero resistances do not exist, as do not infinite currents.
But for really low resistances, such as those encountered when you are shorting caps through heavy wires, the peak current will EASILY get in hundreds of amperes, even with ordinary capacitors, charged at only a few volts.

The power dissipation is not zero in any case, though, because the energy dissipated does not depend on the resistance.
So the result is always valid.
 

Re: energy in the capacitor

The electric charge in the capacitor is conserved. However to move the electric charge from one capacitor to another work has to be done. The missing 1/4*C*V² is spent for moving the charge from one capacitor to the other capacitor.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top