How to find the energy of capacitance?

In summary: Both approaches are valid; just be aware of the limitations of each.Both approaches are valid; just be aware of the limitations of each.
  • #1
Coderhk
59
2

Homework Statement


Find the energy in the 2uf capacitor in the diagram attached in uploads.

Homework Equations


E=(1/2)CV^2
E=QV
C=Q/V
Capacitor in series : 1/(Equivalent Capacitor) =Σ1/(Ci)*
Capacitor in parallel: Equivalent Capacitor =∑Ci*
*i is the index of summation

The Attempt at a Solution


Attached in uploads, but here's a summary of what I did.
First we find the equivalent capacitance of 2.55micro farads
using Q=CV I get net charge is 1.533E-5 Coulombs
Knowing voltage drop of 1uf, find charge
Then the charge on 1.55uf is total charge minus charge on 1uf.
Afterwards I revert back to second diagram and all capacitor in series have same charge.
Find V at 2uf
Plug in E= QV
Done.[/B]

Textbook answer is 2.2E-7 J
 

Attachments

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Last edited:
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  • #2
I don't understand the number you've used for the voltage in your final calculation. Where did it come from?
upload_2018-1-30_0-18-3.png


Also, the answer you say your textbook gives looks to me to be incorrect (the power of ten used).
 

Attachments

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  • #3
gneill said:
I don't understand the number you've used for the voltage in your final calculation. Where did it come from?
View attachment 219370

Also, the answer you say your textbook gives looks to me to be incorrect (the power of ten used).
Oops that number was suppose to be 4.665V. The 4.665 is from the line right before the last line.
 
  • #4
Coderhk said:
Oops that number was suppose to be 4.665V. The 4.665 is from the line right before the last line.
Okay, you might want to run the numbers (for that final calculation) through your calculator again though.
 
  • #5
gneill said:
Okay, you might want to run the numbers (for that final calculation) through your calculator again though.
I got 2.17E-5. After significant figures I get 2.2E-5. I guess the book just has the wrong exponent. Thanks for you help.
 
  • #6
You're welcome.

I should mention that you did a lot of extra work to get to the solution, finding the total equivalent capacitance, the total charge, then removing the charge for the 1 μF capacitor, then finding the voltage on the 2 μF capacitor, then finally calculating its energy.

Since the series connection of C2 and the pair C3||C4 is a branch connected across the 6.0 V battery, you could have ignored C1 entirely for this problem. Nothing C1 does can affect that potential across the branch.

As an alternative approach, you might have used the potential divider formula for series capacitors to go directly to the voltage across C2, then applied your energy formula.
 
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Related to How to find the energy of capacitance?

1. What is capacitance?

Capacitance is the measure of an object's ability to store electrical charge. It is represented by the symbol C and is measured in the unit farad (F).

2. How do you calculate the energy of capacitance?

The energy stored in a capacitor is calculated using the formula E = 1/2CV^2, where C is the capacitance in farads and V is the voltage across the capacitor in volts.

3. What factors affect the energy of capacitance?

The energy of capacitance is affected by the capacitance value and the voltage applied across the capacitor. A higher capacitance or voltage will result in a higher energy stored in the capacitor.

4. How is capacitance measured?

Capacitance can be measured using a device called a capacitance meter or by using the formula C = Q/V, where Q is the charge stored in the capacitor and V is the voltage across the capacitor.

5. How is the energy of capacitance used in practical applications?

The energy of capacitance is used in various practical applications such as in electronic circuits, power grids, and energy storage devices. Capacitors are also used in energy conversion and filtering applications.

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