Stored Energy Difference Between Capacitors

In summary: yes, the stored energy in the parallel configuration is 176.42 microjoules more than the series configuration.
  • #1
8008jsmith
15
0

Homework Statement


Three capacitors of C1 = 6.7 micro-farads, C2 = 19.4 microfarads, and C3 = 9.3 microfarads are all connected in series to a 9 volt battery. The stored energy in C1 is found (U1s). The same three capacitors are connected in parallel to the same battery. The stored energy in C1 is found again in this configuration (U1p). What is the difference in stored energy in the parallel configuration compared to the series configuration in microjoules? That is determine U1p - U1s=

Homework Equations


U1p = 1/2C1V^2
U1s = 1/2 Q^2/2C1
VCeq = Q

The Attempt at a Solution


[/B]
Using the formulas I got U1p = 271.35 and U1s = 64.24. Thus U1p - U1s should equal 207.15 but the correct answer is 176.42. Can anyone see what I'm doing wrong?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
8008jsmith said:

Homework Statement


Three capacitors of C1 = 6.7 micro-farads, C2 = 19.4 microfarads, and C3 = 9.3 microfarads are all connected in series to a 9 volt battery. The stored energy in C1 is found (U1s). The same three capacitors are connected in parallel to the same battery. The stored energy in C1 is found again in this configuration (U1p). What is the difference in stored energy in the parallel configuration compared to the series configuration in microjoules? That is determine U1p - U1s=

Homework Equations


U1p = 1/2C1V^2
U1s = 1/2 Q^2/2C1
VCeq = Q

The Attempt at a Solution


[/B]
Using the formulas I got U1p = 271.35 and U1s = 64.24. Thus U1p - U1s should equal 207.15 but the correct answer is 176.42. Can anyone see what I'm doing wrong?

I'm not understanding how you are handling the series case. What are the voltages across the 3 caps in the series case?
 
  • #3
berkeman said:
I'm not understanding how you are handling the series case. What are the voltages across the 3 caps in the series case?

I wasn't given the voltage across each, just the voltage of the battery. For a series I found the equivalent capacitance using (1/C1 + 1/Cn)^-1. And then I plugged that into the Q = CeqV equation to find Q. Then I used Q in the stored energy equation with C1. I hope that made sense.
 
  • #4
8008jsmith said:
I wasn't given the voltage across each, just the voltage of the battery. For a series I found the equivalent capacitance using (1/C1 + 1/Cn)^-1. And then I plugged that into the Q = CeqV equation to find Q. Then I used Q in the stored energy equation with C1. I hope that made sense.

I don't think that will work. You need to find the voltage division among the 3 series capacitors, so you can calculate how much energy is stored on C1 in the series configuration. The total series capacitance doesn't help, I don't believe.

What equation(s) would you use to figure out how the 9V source divides among the 3 series capacitors? Since they are in series, what can you say about charge flow as they are charged up to 9V total across all 3 caps?
 
  • #5
berkeman said:
I don't think that will work. You need to find the voltage division among the 3 series capacitors, so you can calculate how much energy is stored on C1 in the series configuration. The total series capacitance doesn't help, I don't believe.

What equation(s) would you use to figure out how the 9V source divides among the 3 series capacitors? Since they are in series, what can you say about charge flow as they are charged up to 9V total across all 3 caps?

I'm not really sure what equation I could use. But I know in series the charge in each would be the same which is why I used the equivalent capacitance to find the charge using Q=CV.
 
  • #6
8008jsmith said:
I'm not really sure what equation I could use. But I know in series the charge in each would be the same which is why I used the equivalent capacitance to find the charge using Q=CV.

Right, since the series current is the same for all 3 caps, and current I = dQ/dt, then each cap will get the same amount of charge on it during the charging process. And since Q=CV, you know what the voltage is across each cap. If you use that voltage for C1, do you get the right answer for the ratio of the stored energies?
 
  • #7
berkeman said:
Right, since the series current is the same for all 3 caps, and current I = dQ/dt, then each cap will get the same amount of charge on it during the charging process. And since Q=CV, you know what the voltage is across each cap. If you use that voltage for C1, do you get the right answer for the ratio of the stored energies?

Well when I do the math I get Q = 29.34. Then I plug that into 1/2 (Q^2/C1) and I get 64.24 for the stored energy of the series capacitor. Then the difference between the two is 207.15 but the correct answer is 176.42.
 
  • #8
You can use the voltage or the charge to find the energy stored on a capacitor:

$$E_c = \frac{1}{2} C V_c^2 = \frac{1}{2} \frac{Q_c^2}{C}$$

I'm seeing a result that's very close to yours (rounding of intermediate results could explain the difference). It could be that values in the problem have been "updated" without changing the answer key.
 

Related to Stored Energy Difference Between Capacitors

1. What is stored energy difference between capacitors?

The stored energy difference between capacitors is the difference in the amount of energy stored in two capacitors with the same capacitance but different voltages. This difference in energy is due to the fact that the energy stored in a capacitor is directly proportional to the square of the voltage.

2. How do you calculate the stored energy difference between capacitors?

The stored energy difference between capacitors can be calculated using the formula:
E = 1/2 * C * (V12 - V22)
where E is the energy difference, C is the capacitance, and V1 and V2 are the voltages of the two capacitors.

3. Why is the stored energy difference between capacitors important?

The stored energy difference between capacitors is important because it determines the amount of energy that can be stored and released by a capacitor. This can affect the performance and efficiency of electronic devices that use capacitors, such as power supplies and audio equipment.

4. How does the stored energy difference between capacitors affect their charging and discharging times?

The stored energy difference between capacitors affects their charging and discharging times because the higher the voltage, the faster the capacitor can charge and discharge. This is because the rate of change of voltage, or voltage gradient, is directly proportional to the amount of energy stored in the capacitor.

5. Can the stored energy difference between capacitors be negative?

Yes, the stored energy difference between capacitors can be negative. This occurs when the voltage on one capacitor is lower than the voltage on the other capacitor. In this case, one capacitor will have a positive stored energy, while the other will have a negative stored energy, resulting in a negative energy difference.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
924
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
12K
  • Introductory Physics Homework Help
Replies
5
Views
3K
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
4K
Back
Top