Circuit Lab B/C

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Creationist127
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Re: Circuit Lab B/C

Postby Creationist127 » February 12th, 2019, 1:31 pm

Nydauron wrote:Oh no! Dr. Doom has just finished building his capacitor! When designing the build he decided to make the plates have an area of 10 m^2 separated by 1 mm. As the dielectric, he used strontium titanate (k = 310). Dr. Doom plans on using his battery with an EMF = 20kV connected to a set of 20 parallel 100 Mohm resistors in series to charge the capacitor.
1) How long will it take to "fully" charge the capacitor?
2) How much charge will the capacitor hold when it reaches maximum voltage?
3) How much energy is stored in the charged capacitor? How much energy is lost during the process of charging the capacitor?

Is this correct?
Part One
C=K*e0*A/d
C=310*8.85e-12*10/.001
C=2.7435e-5 F
R eq=1/(20/1e8)=5e6 ohm
t charge=4*R*C
t charge=4*5e6*2.7435e-5
t charge=548.7 seconds, or about 9 minutes. (charges to 98%)

Part Two
Q=CV
Q=2.7435e-5*20000
Q=.5487 coulombs.

Part Three
PEcapacitor=1/2*Q*V
PE capacitor=1/2*.5487*20000
PE capacitor=5487 joules.
PE dissipated=1/2*Q*V
PE dissipated=5487 joules.
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Re: Circuit Lab B/C

Postby Nydauron » February 12th, 2019, 7:40 pm

Creationist127 wrote:
Nydauron wrote:Oh no! Dr. Doom has just finished building his capacitor! When designing the build he decided to make the plates have an area of 10 m^2 separated by 1 mm. As the dielectric, he used strontium titanate (k = 310). Dr. Doom plans on using his battery with an EMF = 20kV connected to a set of 20 parallel 100 Mohm resistors in series to charge the capacitor.
1) How long will it take to "fully" charge the capacitor?
2) How much charge will the capacitor hold when it reaches maximum voltage?
3) How much energy is stored in the charged capacitor? How much energy is lost during the process of charging the capacitor?

Is this correct?
Part One
C=K*e0*A/d
C=310*8.85e-12*10/.001
C=2.7435e-5 F
R eq=1/(20/1e8)=5e6 ohm
t charge=4*R*C
t charge=4*5e6*2.7435e-5
t charge=548.7 seconds, or about 9 minutes. (charges to 98%)

Part Two
Q=CV
Q=2.7435e-5*20000
Q=.5487 coulombs.

Part Three
PEcapacitor=1/2*Q*V
PE capacitor=1/2*.5487*20000
PE capacitor=5487 joules.
PE dissipated=1/2*Q*V
PE dissipated=5487 joules.

1 is close...
In general, 5 time constants usually yield a fully charged capacitor. You just simply did 4 time constants which will charge the capacitor to 98% as you stated. It is practically the same, but I don't think people will count 98% as "full". ;)
With the change, you get:


So, 685.9 seconds or 11 minutes and 25.9 seconds.

Either way, your turn!
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Re: Circuit Lab B/C

Postby PM2017 » February 12th, 2019, 10:06 pm

Nydauron wrote:
Creationist127 wrote:
Nydauron wrote:Oh no! Dr. Doom has just finished building his capacitor! When designing the build he decided to make the plates have an area of 10 m^2 separated by 1 mm. As the dielectric, he used strontium titanate (k = 310). Dr. Doom plans on using his battery with an EMF = 20kV connected to a set of 20 parallel 100 Mohm resistors in series to charge the capacitor.
1) How long will it take to "fully" charge the capacitor?
2) How much charge will the capacitor hold when it reaches maximum voltage?
3) How much energy is stored in the charged capacitor? How much energy is lost during the process of charging the capacitor?

Is this correct?
Part One
C=K*e0*A/d
C=310*8.85e-12*10/.001
C=2.7435e-5 F
R eq=1/(20/1e8)=5e6 ohm
t charge=4*R*C
t charge=4*5e6*2.7435e-5
t charge=548.7 seconds, or about 9 minutes. (charges to 98%)

Part Two
Q=CV
Q=2.7435e-5*20000
Q=.5487 coulombs.

Part Three
PEcapacitor=1/2*Q*V
PE capacitor=1/2*.5487*20000
PE capacitor=5487 joules.
PE dissipated=1/2*Q*V
PE dissipated=5487 joules.

1 is close...
In general, 5 time constants usually yield a fully charged capacitor. You just simply did 4 time constants which will charge the capacitor to 98% as you stated. It is practically the same, but I don't think people will count 98% as "full". ;)
With the change, you get:


So, 685.9 seconds or 11 minutes and 25.9 seconds.

Either way, your turn!

Actually, I think it was exactly right
My resources said 4-time constants is considered full...
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Re: Circuit Lab B/C

Postby Nydauron » February 12th, 2019, 10:44 pm

PM2017 wrote:
Nydauron wrote:
Creationist127 wrote:

Actually, I think it was exactly right
My resources said 4-time constants is considered full...

Huh, that's so weird...
Literally every search I've done has said that it takes 5 time constants to "fully" charge just like it is shown in the top graph, but there are some images like this one that show the steady state period beginning at 4T even though they marked "fully" charged at 5T. What is this madness! :x

Image
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Re: Circuit Lab B/C

Postby PM2017 » February 13th, 2019, 6:48 am

Nydauron wrote:
PM2017 wrote:
Nydauron wrote:

Actually, I think it was exactly right
My resources said 4-time constants is considered full...

Huh, that's so weird...
Literally every search I've done has said that it takes 5 time constants to "fully" charge just like it is shown in the top graph, but there are some images like this one that show the steady state period beginning at 4T even though they marked "fully" charged at 5T. What is this madness! :x

Image

So do we just both answers and specify?
https://www.electronics-tutorials.ws/rc/rc_1.html
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Re: Circuit Lab B/C

Postby Creationist127 » February 13th, 2019, 7:08 am

Nydauron wrote:
PM2017 wrote:
Nydauron wrote:

Actually, I think it was exactly right
My resources said 4-time constants is considered full...

Huh, that's so weird...
Literally every search I've done has said that it takes 5 time constants to "fully" charge just like it is shown in the top graph, but there are some images like this one that show the steady state period beginning at 4T even though they marked "fully" charged at 5T. What is this madness! :x

Image

Math indicates...
that a capacitor will never truly reach full charge, without infinite time. Quick research finds that different websites use different values for accepted "full charge". I used four time constants, because that's what I was taught, but it may be different depending on the teacher/competition, I suppose. So both answers can be correct, based on context? Is this accurate?
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Re: Circuit Lab B/C

Postby Nydauron » February 13th, 2019, 7:32 am

Creationist127 wrote:
Nydauron wrote:
PM2017 wrote:
Actually, I think it was exactly right
My resources said 4-time constants is considered full...

Huh, that's so weird...
Literally every search I've done has said that it takes 5 time constants to "fully" charge just like it is shown in the top graph, but there are some images like this one that show the steady state period beginning at 4T even though they marked "fully" charged at 5T. What is this madness! :x

Image

Math indicates...
that a capacitor will never truly reach full charge, without infinite time. Quick research finds that different websites use different values for accepted "full charge". I used four time constants, because that's what I was taught, but it may be different depending on the teacher/competition, I suppose. So both answers can be correct, based on context? Is this accurate?

Yes, I would assume...
both answers are correct then. Eh, we all live and learn. :)

Let's continue with the next question.
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Re: Circuit Lab B/C

Postby Creationist127 » February 13th, 2019, 9:15 am

When connected to a 9.0 volt battery and a copper wire (resistivity 1.68e-8 ohm-meters, diameter 2.0 mm), a capacitor charges to a potential drop of 6.0 volts in 81.89 ps. The capacitor's plates are square, 5.0 meters to a side, and are separated by 1.0 millimeter of air (k=1.0). How long is the wire?
EDIT: Fixed bad sig figs.
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Re: Circuit Lab B/C

Postby Nydauron » February 13th, 2019, 3:26 pm

Creationist127 wrote:When connected to a 9.0 volt battery and a copper wire (resistivity 1.68e-8 ohm-meters, diameter 2.0 mm), a capacitor charges to a potential drop of 6.0 volts in 81.89 ps. The capacitor's plates are square, 5.0 meters to a side, and are separated by 1.0 millimeter of air (k=1.0). How long is the wire?
EDIT: Fixed bad sig figs.

Answer







Substituting for , , , , and ...


Now solving for ...




(2 sig figs)
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Re: Circuit Lab B/C

Postby Creationist127 » February 14th, 2019, 6:54 am

Nydauron wrote:
Creationist127 wrote:When connected to a 9.0 volt battery and a copper wire (resistivity 1.68e-8 ohm-meters, diameter 2.0 mm), a capacitor charges to a potential drop of 6.0 volts in 81.89 ps. The capacitor's plates are square, 5.0 meters to a side, and are separated by 1.0 millimeter of air (k=1.0). How long is the wire?
EDIT: Fixed bad sig figs.

Answer







Substituting for , , , , and ...


Now solving for ...




(2 sig figs)

Correct. Your turn.
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Re: Circuit Lab B/C

Postby mjcox2000 » February 21st, 2019, 7:53 pm

To revive this thread:

Diagram: https://drive.google.com/file/d/1-IqHS7w46-5SeewuyUZokkuVdnzl_l8d/view?usp=sharing

Resistors R1, R2, and R3, voltage source V1, and current source I1 are arranged as in the diagram. These circuit elements have the following values:

R1: 5Ω
R2: 3Ω
R3: 10Ω
V1: 8V
I1: 6A

What is the voltage across and current through each of the three resistors?
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Re: Circuit Lab B/C

Postby Cathy-TJ » February 21st, 2019, 8:39 pm

mjcox2000 wrote:To revive this thread:

Diagram: https://drive.google.com/file/d/1-IqHS7w46-5SeewuyUZokkuVdnzl_l8d/view?usp=sharing

Resistors R1, R2, and R3, voltage source V1, and current source I1 are arranged as in the diagram. These circuit elements have the following values:

R1: 5Ω
R2: 3Ω
R3: 10Ω
V1: 8V
I1: 6A

What is the voltage across and current through each of the three resistors?

Answer
Voltages:
R1: 65/9 V (Right is positive)
R2: 41/3 V (Bottom is positive)
R3: 130/9 V (Right is positive)

Currents:
R1: 13/9 A (Right to Left)
R2: 41/9 A (Bottom to Top)
R3: 13/9A (Bottom to Top)

Edited because oops
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Re: Circuit Lab B/C

Postby mjcox2000 » February 22nd, 2019, 12:38 pm

Cathy-TJ wrote:
mjcox2000 wrote:To revive this thread:

Diagram: https://drive.google.com/file/d/1-IqHS7w46-5SeewuyUZokkuVdnzl_l8d/view?usp=sharing

Resistors R1, R2, and R3, voltage source V1, and current source I1 are arranged as in the diagram. These circuit elements have the following values:

R1: 5Ω
R2: 3Ω
R3: 10Ω
V1: 8V
I1: 6A

What is the voltage across and current through each of the three resistors?

Answer
Voltages:
R1: 65/9 V (Right is positive)
R2: 41/3 V (Bottom is positive)
R3: 130/9 V (Right is positive)

Currents:
R1: 13/9 A (Right to Left)
R2: 41/9 A (Bottom to Top)
R3: 13/9A (Bottom to Top)

Edited because oops


That looks right! (Although, if this were a test, you might want to evaluate those fractions and truncate them to 1 sig fig.) Your turn!
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Re: Circuit Lab B/C

Postby mjcox2000 » February 22nd, 2019, 6:19 pm

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