Initial Energy/Final energy on 2 capacitors

AI Thread Summary
Two capacitors, C1 (4μF) and C2 (9μF), are connected in series to a 30 V battery, and after disconnecting the battery, their plates are connected to equalize charge. The equivalent capacitance (Ceq) is calculated as 2.77μF, with an initial charge of approximately 93.077μC on each capacitor. Upon reconnection, the total charge becomes 166.154μC, leading to final voltages of about 12.78V for both capacitors. The final energies calculated for C1 and C2 are approximately 325.125 J and 735 J, respectively, with a suggestion to use an alternative energy formula for verification.
MarcL
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Homework Statement



Two capacitors, C1 = 4μF and C2 = 9μF, are connected in series with a 30 V battery. The battery is removed and the plates of like sign are connected. Find the initial and final energies for each capacitor.

Homework Equations



Q=C*V
U=q^2/2C
Ceq = ∑Ci ( don't know how to note it on the computer) --> in parallel
1/Ceq= ∑1/Ci
I think that's about it..?


The Attempt at a Solution


1) find ceq ( in series) =>2.77uF
2)Find totaly charge ( same on both cause in series) =>93.077uC => Get initial energy from this (?)
3)Voltage on C1 => q/c => 20.78 V
4)Voltage on C2 => 1/c => 9.23 V
5)After they are reconnected, it needs to reach equilibrium ( I think?) So my total charge is
2*83.077 = 166.154 uC ??
Sooooo then my voltage on capacitor 1 would be 166.154/13 => 12.78V ( approx.)
6)Q1 final => c*v => 4uF*12.78 =>51uC
7)Q2 final => c*v => 9uF*12.78 =>115.02uC
9) Energy Final on 1 => q^2/2C => 325.125 J
10) Energy Final on 1=> q^2/2C=> 735 J
 
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MarcL said:

Homework Statement



Two capacitors, C1 = 4μF and C2 = 9μF, are connected in series with a 30 V battery. The battery is removed and the plates of like sign are connected. Find the initial and final energies for each capacitor.

Homework Equations



Q=C*V
U=q^2/2C
Ceq = ∑Ci ( don't know how to note it on the computer) --> in parallel
1/Ceq= ∑1/Ci
I think that's about it..?

The Attempt at a Solution


1) find ceq ( in series) =>2.77uF
2)Find totaly charge ( same on both cause in series) =>93.077uC => Get initial energy from this (?)
3)Voltage on C1 => q/c => 20.78 V
4)Voltage on C2 => 1/c => 9.23 V
5)After they are reconnected, it needs to reach equilibrium ( I think?) So my total charge is
2*83.077 = 166.154 uC ??
Sooooo then my voltage on capacitor 1 would be 166.154/13 => 12.78V ( approx.)
6)Q1 final => c*v => 4uF*12.78 =>51uC
7)Q2 final => c*v => 9uF*12.78 =>115.02uC
9) Energy Final on 1 => q^2/2C => 325.125 J
10) Energy Final on 1=> q^2/2C=> 735 J

It is correct, but what are the initial energies?
You could have used also the formula Energy=U^2*C/2 for the final energies.

ehild
 
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