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For each homework question and lab report we will make a post, this will probably contain a few tips on what the problems are about and how to solve them. If you are stuck on something then instead of emailing us directly you should post a comment in reply to the relevant post. We will try to guide you through tough points and help you understand the problems and the concepts behind them.
Sunday, September 5, 2010
Chapter 16, questions 10, 11
These two questions refer to material in SV8 section 16.9.
4 comments:
Vito
said...
Professor,
This is a question based on example 16.8 in the textbook, page 553.
In the process of doing part (a) "calculate the equivalent capacitance of the series," I do exactly as the book instructs: 1/Ceq= 1/3.0mF+1/6mF+1/12mF+1/24mF The book calculates an answer of 1.6mF but I calculate .0625 as my answer. I am not sure what I'm doing wrong.
Also, for part (b)the book states the Ceq to be (1.6e-6F) but when you multiply that by 18V you get 2.8e-5, while the book says the answer is 29mC. Wouldn't that equate to 2.88e-6? not 2.88e-5? Thanks!
Check your numbers again. 1/3 + 1/6 + 1/12 + 1/24 = 0.625. But that's 1/Ceq, so Ceq = 1/0.625 = 1.6. By the way, since we can't type greek on this blog, I think it's better to write mu for micro than m. m is the usual abbreviation for milli, 1/1000.
As for part b, remember that each capacitor will have the same charge, and that is the same as the charge that a 1.6muF capacitor would have connected to an 18V battery, Q=V*C=28.8 muC. So each capacitor has a charge of 28.8 muC, and we can use that to find the voltage across the 12muF capacitor: V=Q/C=28.8muC / 12muF = 2.4V.
Going one step further, you can find the voltage drop across each capacitor: 28.8muC / 3muF = 9.6V 28.8muC / 6muF = 4.8V 28.8muC / 12muF = 2.4V 28.8muC / 24muF = 1.2V Note that those four voltages add up to 18V. How cool is that!
I have yet another question: WA #11. Part (c) states "The battery is subsequently reattached to the capacitor, but the plate separation remains as in part (b). How much energy is stored?"
I've figured out the answer, but my question is, what happens if a NEW battery were to be reconnected to this circuit that has a much higher voltage, say 100Volts. Would the energy stored still decrease the same amount?
In the first part of the question, the charge is fixed, and so changing the separation of the plates changes the voltage according to V=C/Q. If you connect it to a battery - the original or a new one, that will fix V at the battery voltage, so charge may have to flow one way or the other to get the charge on each plate to be Q=C/Vbatt. So whether energy is stored or released depends on which way the charge flows.
4 comments:
Professor,
This is a question based on example 16.8 in the textbook, page 553.
In the process of doing part (a) "calculate the equivalent capacitance of the series," I do exactly as the book instructs:
1/Ceq= 1/3.0mF+1/6mF+1/12mF+1/24mF
The book calculates an answer of 1.6mF but I calculate .0625 as my answer. I am not sure what I'm doing wrong.
Also, for part (b)the book states the Ceq to be (1.6e-6F) but when you multiply that by 18V you get 2.8e-5, while the book says the answer is 29mC. Wouldn't that equate to 2.88e-6? not 2.88e-5?
Thanks!
Check your numbers again.
1/3 + 1/6 + 1/12 + 1/24 = 0.625.
But that's 1/Ceq, so Ceq = 1/0.625 = 1.6.
By the way, since we can't type greek on this blog, I think it's better to write mu for micro than m. m is the usual abbreviation for milli, 1/1000.
As for part b, remember that each capacitor will have the same charge, and that is the same as the charge that a 1.6muF capacitor would have connected to an 18V battery, Q=V*C=28.8 muC. So each capacitor has a charge of 28.8 muC, and we can use that to find the voltage across the 12muF capacitor:
V=Q/C=28.8muC / 12muF = 2.4V.
Going one step further, you can find the voltage drop across each capacitor:
28.8muC / 3muF = 9.6V
28.8muC / 6muF = 4.8V
28.8muC / 12muF = 2.4V
28.8muC / 24muF = 1.2V
Note that those four voltages add up to 18V. How cool is that!
Thank you professor!
I think I understand it now.
I have yet another question:
WA #11. Part (c) states "The battery is subsequently reattached to the capacitor, but the plate separation remains as in part (b). How much energy is stored?"
I've figured out the answer, but my question is, what happens if a NEW battery were to be reconnected to this circuit that has a much higher voltage, say 100Volts. Would the energy stored still decrease the same amount?
In the first part of the question, the charge is fixed, and so changing the separation of the plates changes the voltage according to V=C/Q.
If you connect it to a battery - the original or a new one, that will fix V at the battery voltage, so charge may have to flow one way or the other to get the charge on each plate to be Q=C/Vbatt. So whether energy is stored or released depends on which way the charge flows.
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