Initial Energy/Final energy on 2 capacitors

In summary, two capacitors of values 4μF and 9μF are connected in series with a 30 V battery. When the battery is removed and the plates of like sign are connected, the initial and final energies for each capacitor can be found by using the equations Q=C*V and U=q^2/2C, as well as Ceq = ∑Ci and 1/Ceq= ∑1/Ci. The initial energy is calculated to be 325.125 J for the first capacitor and 735 J for the second capacitor. The final energies are found to be 51 μC and 115.02 μC for the first and second capacitors, respectively, using the formula Energy=
  • #1
MarcL
170
2

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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  • #2
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
 

What is initial energy and final energy on 2 capacitors?

The initial energy and final energy on 2 capacitors refers to the amount of energy stored in two capacitors at the beginning and end of a process, such as charging or discharging.

How do you calculate the initial and final energy on 2 capacitors?

The initial energy on a capacitor can be calculated using the formula 1/2CV^2, where C is the capacitance in Farads and V is the voltage in volts. The final energy can be calculated using the same formula after the capacitor has been charged or discharged.

What is the relationship between initial and final energy on 2 capacitors?

The relationship between initial and final energy on 2 capacitors is that the initial energy is equal to the final energy. This is known as the conservation of energy principle, which states that energy cannot be created or destroyed, only transferred or transformed.

How does the capacitance affect the initial and final energy on 2 capacitors?

The capacitance affects the initial and final energy on 2 capacitors by determining the amount of charge that can be stored on the capacitor at a given voltage. A higher capacitance will result in a higher initial and final energy.

What are some real-life applications of understanding initial and final energy on 2 capacitors?

Understanding initial and final energy on 2 capacitors is important in various electronic devices, such as cameras, computers, and smartphones, which use capacitors for energy storage. It is also crucial in designing and optimizing circuits for efficient energy usage.

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