Capacitance (Potential Difference and Energy)

In summary, the maximum potential difference that can exist between points A and B in the given three-capacitor arrangement is 130 volts. The maximum energy that can be stored in the arrangement can be calculated by adding the individual potential energies of each capacitor, resulting in a value higher than if only the potential energy from the first capacitor was used.
  • #1
kilnvzol
33
0

Homework Statement


In the figure C1 = 10.0 μF,C2 = 20.0 μF and C3 = 27.0 μF. If no capacitor can withstand a potential difference of more than 130 V without failure, what is (a) the magnitude of the maximum potential difference that can exist between points A and B and (b) the maximum energy that can be stored in the three-capacitor arrangement?
http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c25/q37.jpg
Image link - http://www.webassign.net/hrw8/25-45.gif

Homework Equations


1/Cseries= 1/C1 + 1/C2 ...
U = q2/(2C) = (0.5)CV2

The Attempt at a Solution


For part A, I know that 130V goes into C1 and I think half that goes into C2 so 65V but I'm not sure how much goes into C3.
For part B, I need the solution to part A to figure that out.

Thank you!
 
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  • #2
kilnvzol said:
For part A, I know that 130V goes into C1 and I think half that goes into C2 so 65V ...
Can you justify that claim?
... but I'm not sure how much goes into C3.
For part B, I need the solution to part A to figure that out.
Consider how charge is distributed amongst series-connected capacitors (when they are charged simultaneously by the same current).

What relationship is there between voltage, charge, and capacitance?

Edit: For part (b), consider a clever way to put as much energy as possible onto the arrangement.
 
  • #3
Actually I think I got part A. C = Q/V => V=Q/C the voltage for the first one is 130 so Q = (130)(10E-6) = 1.3E-3
Then you use the Q value to find the rest? So C3 is 48.15?

For part B, I just use the equation U = q2/(2C) = (0.5)CV2?
 
  • #4
kilnvzol said:
Actually I think I got part A. C = Q/V => V=Q/C the voltage for the first one is 130 so Q = (130)(10E-6) = 1.3E-3
Then you use the Q value to find the rest? So C3 is 48.15?
That looks good.
For part B, I just use the equation U = q2/(2C) = (0.5)CV2?
Yes,... but the question as posed does not restrict you to using the charges or voltages from part A...
 
  • #5
gneill said:
Yes,... but the question as posed does not restrict you to using the charges or voltages from part A...
Do you just plug in each voltage with its respective capacitance to the equation and see which is higher or what is the other way?
 
  • #6
kilnvzol said:
Do you just plug in each voltage with its respective capacitance to the equation and see which is higher or what is the other way?
You want to maximize the total energy stored (while respecting the voltage tolerance of the individual components). So what would be the ideal situation?
 
  • #7
gneill said:
You want to maximize the total energy stored (while respecting the voltage tolerance of the individual components). So what would be the ideal situation?
Only using the potential energy from the first capacitor?
 
  • #8
kilnvzol said:
Only using the potential energy from the first capacitor?
Wouldn't that be a waste of the energy storage potential of the other two?
 
  • #9
gneill said:
Wouldn't that be a waste of the energy storage potential of the other two?
Add all of the potential energies together?
 
  • #10
kilnvzol said:
Add all of the potential energies together?
Sure. After all, you have three "containers" and each can store some. Otherwise it's like having three boxes to put things in and only using the smallest one.
 
  • #11
gneill said:
Sure. After all, you have three "containers" and each can store some. Otherwise it's like having three boxes to put things in and only using the smallest one.
Thank you i got it!
 

Related to Capacitance (Potential Difference and Energy)

1. What is capacitance?

Capacitance is the ability of a system to store electrical charge. It is represented by the letter C and is measured in units of farads (F).

2. How is capacitance calculated?

Capacitance is calculated by dividing the amount of charge stored on a conductor by the potential difference (voltage) between the two conductors. It is also affected by the size and distance between the conductors.

3. What is potential difference?

Potential difference, also known as voltage, is the difference in electric potential between two points in an electric field. It is measured in volts (V) and represents the work needed to move a unit of electric charge from one point to another.

4. How does capacitance affect energy storage?

The higher the capacitance, the more energy can be stored in a system. This is because capacitance is directly proportional to the amount of charge stored, and the amount of charge is directly proportional to the energy stored.

5. What are some real-world applications of capacitance?

Capacitance has many practical applications, such as in electronic devices (e.g. capacitors), power grids, and energy storage systems. It is also used in sensors, filters, and tuning circuits in radios and televisions.

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