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 ?????
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....
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
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)
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.
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
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 ?????
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!
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
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