## Circuit Lab B/C

Creationist127
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### Re: Circuit Lab B/C

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?
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 [b]about 9 minutes[/b]. (charges to 98%)
Q=CV
Q=2.7435e-5*20000
Q=[b].5487 coulombs.[/b]
PE[sub]capacitor[/sub]=1/2*Q*V
PE capacitor=1/2*.5487*20000
PE capacitor=[b]5487 joules[/b].
PE dissipated=1/2*Q*V
PE dissipated=[b]5487 joules[/b].
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Nydauron
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### Re: Circuit Lab B/C

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?
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 [b]about 9 minutes[/b]. (charges to 98%)
Q=CV
Q=2.7435e-5*20000
Q=[b].5487 coulombs.[/b]
PE[sub]capacitor[/sub]=1/2*Q*V
PE capacitor=1/2*.5487*20000
PE capacitor=[b]5487 joules[/b].
PE dissipated=1/2*Q*V
PE dissipated=[b]5487 joules[/b].
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:
$5 * \tau = 5(5* 10^6)(2.7435*10^{-5}) = 685.9$

So, 685.9 seconds or 11 minutes and 25.9 seconds.
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PM2017
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### Re: Circuit Lab B/C

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?
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 [b]about 9 minutes[/b]. (charges to 98%)
Q=CV
Q=2.7435e-5*20000
Q=[b].5487 coulombs.[/b]
PE[sub]capacitor[/sub]=1/2*Q*V
PE capacitor=1/2*.5487*20000
PE capacitor=[b]5487 joules[/b].
PE dissipated=1/2*Q*V
PE dissipated=[b]5487 joules[/b].
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:
$5 * \tau = 5(5* 10^6)(2.7435*10^{-5}) = 685.9$

So, 685.9 seconds or 11 minutes and 25.9 seconds.
My resources said 4-time constants is considered full...
Astronomy, Mousetrap Vehicle, Mission Possible, Fermi Questions
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Nydauron
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Joined: March 20th, 2018, 8:10 pm
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Location: Cornfields...

### Re: Circuit Lab B/C

My resources said 4-time constants is considered full...
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

[img]https://www.electronics-tutorials.ws/rc/rc2.gif[/img]
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PM2017
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Posts: 491
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State: CA

### Re: Circuit Lab B/C

My resources said 4-time constants is considered full...
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

[img]https://www.electronics-tutorials.ws/rc/rc2.gif[/img]
[url]https://www.electronics-tutorials.ws/rc/rc_1.html[/url]
Astronomy, Mousetrap Vehicle, Mission Possible, Fermi Questions
Astronomy, Mousetrap Vehicle, Mission Possible, [strikethrough]Fermi Questions[/strikethrough] :cry: , Circuit Lab
--
West High '19
UC Berkeley '23

Creationist127
Member
Posts: 32
Joined: August 14th, 2018, 3:21 pm
Division: C
State: IN
Location: On a train, traveling from Kalos to Galar

### Re: Circuit Lab B/C

My resources said 4-time constants is considered full...
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

[img]https://www.electronics-tutorials.ws/rc/rc2.gif[/img]
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?
2018: Hovercraft, Thermo, Coaster, Solar System
2019: Thermo, Circuit Lab, Sounds, Wright Stuff
2020: Circuit Lab, Chem Lab, Wright Stuff, Machines

The grind begins now!

Nydauron
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Posts: 18
Joined: March 20th, 2018, 8:10 pm
State: IL
Location: Cornfields...

### Re: Circuit Lab B/C

My resources said 4-time constants is considered full...
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

[img]https://www.electronics-tutorials.ws/rc/rc2.gif[/img]
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?
both answers are correct then. Eh, we all live and learn. :)
Let's continue with the next question.
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Creationist127
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### Re: Circuit Lab B/C

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?
2018: Hovercraft, Thermo, Coaster, Solar System
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Nydauron
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Posts: 18
Joined: March 20th, 2018, 8:10 pm
State: IL
Location: Cornfields...

### Re: Circuit Lab B/C

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?
$C = (1.0)\epsilon _0\frac{5.0^2}{1.0*10^{-3}}$
$C = 2.2 * 10^{-7}F$

$R = \rho \frac{l}{A}$
$R = (1.68*10^{-8})\frac{l}{(1.0*10^{-3})^2\pi}$
$V_c = V_s (1-e^{\frac{-t}{RC}})$

Substituting for $R$, $C$, $V_s$, $t$, and $V_c$...
$6.0 = 9.0 (1-e^{\frac{-81.89*10^{-9}}{((1.68*10^{-8})\frac{l}{(1.0*10^{-3})^2\pi})(2.2*10^{-7})}})$

Now solving for $l$...
$\frac{1}{3} = e^{\frac{-69.6063}{l}}$
$-ln(3) = \frac{-69.6063}{l}$
$l = \frac{69.6063}{ln(3)}$

$l = 63 m$ (2 sig figs)
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Creationist127
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Location: On a train, traveling from Kalos to Galar

### Re: Circuit Lab B/C

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?
$C = (1.0)\epsilon _0\frac{5.0^2}{1.0*10^{-3}}$
$C = 2.2 * 10^{-7}F$

$R = \rho \frac{l}{A}$
$R = (1.68*10^{-8})\frac{l}{(1.0*10^{-3})^2\pi}$
$V_c = V_s (1-e^{\frac{-t}{RC}})$

Substituting for $R$, $C$, $V_s$, $t$, and $V_c$...
$6.0 = 9.0 (1-e^{\frac{-81.89*10^{-9}}{((1.68*10^{-8})\frac{l}{(1.0*10^{-3})^2\pi})(2.2*10^{-7})}})$

Now solving for $l$...
$\frac{1}{3} = e^{\frac{-69.6063}{l}}$
$-ln(3) = \frac{-69.6063}{l}$
$l = \frac{69.6063}{ln(3)}$

$l = 63 m$ (2 sig figs)