Initial Energy/Final energy on 2 capacitors

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SUMMARY

The discussion focuses on calculating the initial and final energies of two capacitors, C1 (4μF) and C2 (9μF), connected in series with a 30 V battery. The equivalent capacitance (Ceq) is determined to be 2.77μF, leading to an initial charge of 93.077μC on both capacitors. After reconnecting the plates of like sign, the final charges are calculated as 51μC for C1 and 115.02μC for C2, with final energies of 325.125 J and 735 J, respectively. The alternative energy formula, U^2*C/2, is also suggested for calculating final energies.

PREREQUISITES
  • Understanding of capacitor theory and formulas, specifically Q=C*V and U=q^2/2C.
  • Knowledge of series and parallel capacitor configurations, including equivalent capacitance calculations.
  • Familiarity with energy calculations in electrical circuits.
  • Basic algebra for manipulating equations and solving for unknowns.
NEXT STEPS
  • Learn about energy storage in capacitors and the implications of series vs. parallel configurations.
  • Study the derivation and application of the energy formula U=q^2/2C in various circuit scenarios.
  • Explore the concept of charge redistribution in capacitors when connected in different configurations.
  • Investigate practical applications of capacitors in electronic circuits and energy storage systems.
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone involved in circuit design or analysis will benefit from this discussion.

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 totally 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 totally 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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