Capacitor network finding potential

In summary: PART B:i then used my answer from the top two and the lower capacitator as a parallel network. so in the last calculation i did 4E-6. AH! there is my mistake. before i had my old answer in there. NOW it should be 4E-6 + 3.5E-6 = 7.6E-6. for the final answer i get 0.001
  • #1
kirby2
43
0
For the capacitor network shown in the figure, the potential difference across ab is 12V. Find (a) the total energy stored in this network and (b) the energy stored in the 4.8 uF capacitor.

figure: http://i.imgur.com/Wxbuu.png

ATTEMPT:

PART A:
i first did the leftmost 2 capacitors and got C=3.1E-6.
then i did the top 2 capacitors and got C=1.4E-6.
then i used those top 2 as a parallel capacitor with the bottom one getting: 4.9E-6.
i added them all together to get C=9.39E-6.
using the forumla u=.5CV^2 i got U = 6.76E-4 J. did i do this right?
 
Physics news on Phys.org
  • #2
Your value for the top two capacitors looks a bit low. Check your calculation.

Otherwise, your method looks okay for finding the energy.
 
  • #3
i re-did it and got 4E-6. is this right?
 
  • #4
i then got a total capatance of 1.21E-5 F and a total energy of 0.0027 J
 
  • #5
kirby2 said:
i re-did it and got 4E-6. is this right?

That's better. 4.06E-6 would be a bit better... try to keep enough significant figures in intermediate values so that roundoff and truncation errors don't sneak into your results.

kirby2 said:
i then got a total capatance of 1.21E-5 F and a total energy of 0.0027 J

The total capacitance looks a bit large. Can you show your steps?
 
  • #6
my bad. i don't know what i did wrong, but i redid it and got u=8.64E-4.
 
  • #7
kirby2 said:
my bad. i don't know what i did wrong, but i redid it and got u=8.64E-4.

That doesn't match what I'm seeing. Again, why not show your steps and the intermediate values?
 
  • #8
ok. adding all the Cs gives me 1.2E-5. i plugged this into the formula U=.5CV^2. for v i used 12 and solved for u
 
  • #9
kirby2 said:
ok. adding all the Cs gives me 1.2E-5. i plugged this into the formula U=.5CV^2. for v i used 12 and solved for u

That's not the total for C that I get. Show the steps that arrive at the total.
 
  • #10
3.1e-6 + 4e-6 + 4.9e-6
 
  • #11
kirby2 said:
3.1e-6 + 4e-6 + 4.9e-6

You haven't specified what capacitor combinations arrive at those numbers, or why you think they should be added in that way. I recognize the first two from your earlier posts (again, you should keep more significant figures). But I don't see any justification for the last one or why they are being added together.
 
  • #12
i did this in three steps. i found C from the first two as a series. then i found c from the top two as a series. then i used my answer from the top two and the lower capacitator as a parallel network. so in the last calculation i did 4E-6. AH! there is my mistake. before i had my old answer in there. NOW it should be 4E-6 + 3.5E-6 = 7.6E-6. for the final answer i get 0.001 J
 
  • #13
kirby2 said:
i did this in three steps. i found C from the first two as a series. then i found c from the top two as a series. then i used my answer from the top two and the lower capacitator as a parallel network. so in the last calculation i did 4E-6. AH! there is my mistake. before i had my old answer in there. NOW it should be 4E-6 + 3.5E-6 = 7.6E-6. for the final answer i get 0.001 J

Things are starting to get cleared up. What value did you get for the final equivalent capacitance, and what values did you combine (and how) to get it?
 
  • #14
final capacitance = 1.48E-5 F. i combined 3.1E-6, 4.1E-6 and 7.6E-6
 
  • #15
kirby2 said:
final capacitance = 1.48E-5 F. i combined 3.1E-6, 4.1E-6 and 7.6E-6

Okay, HOW were they combined? What is the configuration of the capacitors represented by those values? Didn't the 4.1μF value already get combined with another capacitance to arrive at the 7.6μF value?
 
  • #16
3.1E-6 is from series. so you add them by 1/C and the answer is 1/c. so for the first one i did 1/8.6E-6 + 1/4.8E-6 = 1/C where C is 3.1E-6

4.1E-6 is also from series. i did 1/6.2E-6 + 1/11.8E-6 = 1/C where C is 4.1E-6.

7.6E-6 is parallel so you just add them. i added 4.1E-6 to 3.5E-6
 
  • #17
kirby2 said:
3.1E-6 is from series. so you add them by 1/C and the answer is 1/c. so for the first one i did 1/8.6E-6 + 1/4.8E-6 = 1/C where C is 3.1E-6

4.1E-6 is also from series. i did 1/6.2E-6 + 1/11.8E-6 = 1/C where C is 4.1E-6.

7.6E-6 is parallel so you just add them. i added 4.1E-6 to 3.5E-6

Okay, and then what? What's left to be combined?
 

1. What is a capacitor network?

A capacitor network is a group of capacitors connected in a circuit. This can be a series network, where the capacitors are connected end-to-end, or a parallel network, where the capacitors are connected side-by-side.

2. How do you find the potential of a capacitor network?

The potential or voltage across a capacitor network can be found by using Kirchhoff's voltage law, which states that the sum of the potential differences around a closed loop must equal zero. This can also be calculated using the formula V = Q/C, where V is the potential, Q is the charge stored in the capacitor, and C is the capacitance.

3. What factors can affect the potential of a capacitor network?

The potential of a capacitor network can be affected by the capacitance of the individual capacitors, the arrangement of the capacitors (series or parallel), and the voltage of the power source connected to the network.

4. How can you increase the potential of a capacitor network?

The potential of a capacitor network can be increased by adding additional capacitors in series or parallel, increasing the capacitance of the individual capacitors, or increasing the voltage of the power source. However, it is important to note that there may be limitations based on the maximum voltage rating of the capacitors.

5. Can a capacitor network have a negative potential?

No, a capacitor network cannot have a negative potential. Capacitors are passive components that can only store and release energy, they cannot generate energy. Therefore, the potential across a capacitor network will always be positive, and the negative sign in equations is used to indicate the direction of the current flow.

Similar threads

Finding charge on a capacitor given potential difference across two points
    • Introductory Physics Homework Help
Replies
4
Views
925
Calculations involving different Dielectrics and Capacitors
    • Introductory Physics Homework Help
Replies
7
Views
708
Energy stored in Capacitor Network
    • Introductory Physics Homework Help
Replies
2
Views
12K
Determining Inductance L in an LC Circuit
    • Introductory Physics Homework Help
Replies
5
Views
204
Potential difference across capacitor in a network
    • Introductory Physics Homework Help
Replies
8
Views
6K
Uncharged capacitors connected in series
    • Introductory Physics Homework Help
Replies
18
Views
1K
How Do You Calculate Energy Stored in Specific Capacitors Within a Network?
    • Introductory Physics Homework Help
Replies
1
Views
3K
Capacitors in series and Parallel
    • Introductory Physics Homework Help
Replies
9
Views
1K
Effects of capacitors on potential difference
    • Introductory Physics Homework Help
Replies
7
Views
2K
Time dependant Potential Energy in ion trap
    • Introductory Physics Homework Help
Replies
1
Views
720

Hot Threads

Recent Insights

Back
Top