Capacitors in series and Parallel

  • Thread starter KieranRC
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  • #1
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There are a few problems here but it would be helpful if The below solutions could be checked and some insight provided for the last one/any other mistakes

The problem & The Solution Attempts

A circuit with a 12V battery then on the row below in series with the battery is a 120nF capacitor and a
3.2(mu)F capacitor then in parallel with those resistors is another resistor of 50nF. On the immediate left of the battery is a 270Ohm resistor.

First I am asked to calculate the equivalent capacitance of the two capacitors in series for which I did:
(1.2x10^-7) + (3.2x10^-6) = 3.32x10^-6 F

Next I was asked to work out the equivalent capacitance of this capacitor network
For which I did:

((1.2x10^-7)x(3.2x10^-6)) / ((1.2x10^-7)+(3.2x10^-6))= 1.16x10^-7

(1.16x10^-7) + (5x10^-8) = 1.66x10-7F

After this I was then asked to calculate the maximum charge that can be stored in this capacitor network for which i did:

(1.66x10-7)(12)=1.992x10^-6C

Then last but not least I was asked to calculate the maximum energy that could be stored in the smallest capacitor and I have no idea where to start.

Thank You
 

Answers and Replies

  • #2
stevendaryl
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There are a few problems here but it would be helpful if The below solutions could be checked and some insight provided for the last one/any other mistakes

The problem & The Solution Attempts

A circuit with a 12V battery then on the row below in series with the battery is a 120nF capacitor and a
3.2(mu)F capacitor then in parallel with those resistors is another resistor of 50nF. On the immediate left of the battery is a 270Ohm resistor.

First I am asked to calculate the equivalent capacitance of the two capacitors in series for which I did:
(1.2x10^-7) + (3.2x10^-6) = 3.32x10^-6 F

Next I was asked to work out the equivalent capacitance of this capacitor network
For which I did:

((1.2x10^-7)x(3.2x10^-6)) / ((1.2x10^-7)+(3.2x10^-6))= 1.16x10^-7

(1.16x10^-7) + (5x10^-8) = 1.66x10-7F

After this I was then asked to calculate the maximum charge that can be stored in this capacitor network for which i did:

(1.66x10-7)(12)=1.992x10^-6C

Then last but not least I was asked to calculate the maximum energy that could be stored in the smallest capacitor and I have no idea where to start.

Thank You
 
  • #3
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
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I think the only additional fact you need is the formula for the energy in a capacitor: [itex]E = \frac{1}{2} C V^2[/itex]
 
  • #4
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I think the only additional fact you need is the formula for the energy in a capacitor: [itex]E = \frac{1}{2} C V^2[/itex]
Okay great, thank you.
Does everything else look right to you?
 
  • #5
gneill
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There are a few problems here but it would be helpful if The below solutions could be checked and some insight provided for the last one/any other mistakes

The problem & The Solution Attempts

A circuit with a 12V battery then on the row below in series with the battery is a 120nF capacitor and a
3.2(mu)F capacitor then in parallel with those resistors is another resistor of 50nF. On the immediate left of the battery is a 270Ohm resistor.
This is unclear. You are mentioning resistors that you haven't described yet. How does what you describe form a closed path ("circuit")? What does the resistor in parallel with the battery do? (Is it a resistor?).

Please upload a drawing of your circuit.
First I am asked to calculate the equivalent capacitance of the two capacitors in series for which I did:
(1.2x10^-7) + (3.2x10^-6) = 3.32x10^-6 F
That is not correct. Series capacitors do not simply sum. Capacitors in parallel sum. Where is your Relevant Equations section of the required formatting template? You should have listed the equations required to combine capacitors in series and in parallel.

Please use the formatting template when posting questions in the homework area.
 
  • #6
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This is unclear. You are mentioning resistors that you haven't described yet. How does what you describe form a closed path ("circuit")? What does the resistor in parallel with the battery do? (Is it a resistor?).

Please upload a drawing of your circuit.

That is not correct. Series capacitors do not simply sum. Capacitors in parallel sum. Where is your Relevant Equations section of the required formatting template? You should have listed the equations required to combine capacitors in series and in parallel.

Please use the formatting template when posting questions in the homework area.
How do you go about adding a photo?
 
  • #7
gneill
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How do you go about adding a photo?
Use the UPLOAD button at the bottom right of the edit panel.
 
  • #8
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image.jpg
This is unclear. You are mentioning resistors that you haven't described yet. How does what you describe form a closed path ("circuit")? What does the resistor in parallel with the battery do? (Is it a resistor?).

Please upload a drawing of your circuit.

That is not correct. Series capacitors do not simply sum. Capacitors in parallel sum. Where is your Relevant Equations section of the required formatting template? You should have listed the equations required to combine capacitors in series and in parallel.

Please use the formatting template when posting questions in the homework area.

Homework Statement


Shown in

Homework Equations




The Attempt at a Solution


There are a few problems here but it would be helpful if The below solutions could be checked and some insight provided for the last one/any other mistakes

The problem & The Solution Attempts

A circuit with a 12V battery then on the row below in series with the battery is a 120nF capacitor and a
3.2(mu)F capacitor then in parallel with those resistors is another resistor of 50nF. On the immediate left of the battery is a 270Ohm resistor.

First I am asked to calculate the equivalent capacitance of the two capacitors in series for which I did:
(1.2x10^-7) + (3.2x10^-6) = 3.32x10^-6 F

Next I was asked to work out the equivalent capacitance of this capacitor network
For which I did:

((1.2x10^-7)x(3.2x10^-6)) / ((1.2x10^-7)+(3.2x10^-6))= 1.16x10^-7

(1.16x10^-7) + (5x10^-8) = 1.66x10-7F

After this I was then asked to calculate the maximum charge that can be stored in this capacitor network for which i did:

(1.66x10-7)(12)=1.992x10^-6C

Then last but not least I was asked to calculate the maximum energy that could be stored in the smallest capacitor and I have no idea where to start.

Thank You
[/B]
 
  • #9
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0
Use the UPLOAD button at the bottom right of the edit panel.
Is this okay? I have uploaded a photo of the problem
 
  • #10
gneill
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20,913
2,862
Is this okay? I have uploaded a photo of the problem
Yes, it's fine.

Now we can go through your answers:
First I am asked to calculate the equivalent capacitance of the two capacitors in series for which I did:
(1.2x10^-7) + (3.2x10^-6) = 3.32x10^-6 F
That's not right. Note that capacitors in series do not combine by addition. How do they combine?
Next I was asked to work out the equivalent capacitance of this capacitor network
For which I did:

((1.2x10^-7)x(3.2x10^-6)) / ((1.2x10^-7)+(3.2x10^-6))= 1.16x10^-7

(1.16x10^-7) + (5x10^-8) = 1.66x10-7F
That looks fine. This time you combined the series capacitors correctly.
After this I was then asked to calculate the maximum charge that can be stored in this capacitor network for which i did:

(1.66x10-7)(12)=1.992x10^-6C
That looks fine.
Then last but not least I was asked to calculate the maximum energy that could be stored in the smallest capacitor and I have no idea where to start.
Start by finding the steady-state charge or voltage on the smallest capacitor (that is, after some long time has passed after the circuit is connected and current has stopped flowing the capacitors reach their final charges or voltages).

What formulas do you know for determining the energy stored in a capacitor?
 

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