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Lordknukle
Posts: 12,788
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10/11/2013 10:03:52 AM
Posted: 3 years ago
Let's assume that you have two equivalent capacitors connected in parallel to a 9v battery supply.

Question 1:

If you let them charge, will each capacitor have 9v potential difference?

Question 2:

Does the current released from both of these two capacitors act as if they were two parallel batteries? For example, if the voltage on both is 9v and there are X amp hours, the amp hours would be doubled, instead of the voltage.

I know that capacitance increases in parallel, but I fail to see how that has anything to do with this.
"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
Enji
Posts: 1,022
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10/11/2013 10:27:45 AM
Posted: 3 years ago
1. Yes

2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.
Lordknukle
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10/11/2013 10:49:27 AM
Posted: 3 years ago
At 10/11/2013 10:27:45 AM, Enji wrote:
1. Yes

2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.

Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?
"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
Lordknukle
Posts: 12,788
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10/11/2013 10:50:38 AM
Posted: 3 years ago
Or can the capacitors only output at the input current?
"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
Enji
Posts: 1,022
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10/11/2013 11:00:38 AM
Posted: 3 years ago
At 10/11/2013 10:49:27 AM, Lordknukle wrote:
At 10/11/2013 10:27:45 AM, Enji wrote:
1. Yes

2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.

Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?

The current of the output wire would equal the sum of the currents of the two capacitors, so you'd get 9v and 2x amps where x is the current across one capacitor. This would be different if you connected the capacitors in series; the sum of the voltages across each capacitor would equal the voltage of the battery and the current would be the same across each component.
Lordknukle
Posts: 12,788
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10/11/2013 11:10:16 AM
Posted: 3 years ago
At 10/11/2013 11:00:38 AM, Enji wrote:
At 10/11/2013 10:49:27 AM, Lordknukle wrote:
At 10/11/2013 10:27:45 AM, Enji wrote:
1. Yes

2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.

Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?

The current of the output wire would equal the sum of the currents of the two capacitors, so you'd get 9v and 2x amps where x is the current across one capacitor. This would be different if you connected the capacitors in series; the sum of the voltages across each capacitor would equal the voltage of the battery and the current would be the same across each component.

So if I want my capacitors to produce 18v, I'd connect them in series; if I want them to produce 2x amp/hours I'd connect them in parallel?
"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
Lordknukle
Posts: 12,788
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10/11/2013 11:14:26 AM
Posted: 3 years ago
Also, just one more question, what exactly is the point of capacitance ratings? It seems as if you can get all the information you need by knowing the output voltage and the nature of the connection.
"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
drafterman
Posts: 18,870
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10/11/2013 11:15:31 AM
Posted: 3 years ago
Go here:

http://www.falstad.com...

When the app loads (new window) go to: File > Import

Paste:

$ 1 5.0E-6 10.20027730826997 50 5.0 43
v 464 256 464 320 0 0 40.0 9.0 0.0 0.0 0.5
w 464 208 512 208 0
w 464 160 512 160 0
c 512 160 576 160 0 9.999999999999999E-6 -8.999999999998936
c 512 208 576 208 0 1.0E-5 -8.999999999999291
w 576 160 624 160 0
w 576 208 624 208 0
w 624 160 624 208 0
w 624 208 624 368 0
w 624 368 464 368 0
w 464 368 464 320 0
r 464 256 464 208 0 100.0
r 464 208 464 160 0 100.0
o 0 64 0 35 10.0 9.765625E-5 0 -1
o 4 64 0 35 10.0 9.765625E-5 1 -1
o 3 64 0 35 10.0 9.765625E-5 2 -1

Click "Import"
Enji
Posts: 1,022
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10/11/2013 11:25:07 AM
Posted: 3 years ago
At 10/11/2013 11:10:16 AM, Lordknukle wrote:
At 10/11/2013 11:00:38 AM, Enji wrote:
At 10/11/2013 10:49:27 AM, Lordknukle wrote:
At 10/11/2013 10:27:45 AM, Enji wrote:
1. Yes

2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.

Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?

The current of the output wire would equal the sum of the currents of the two capacitors, so you'd get 9v and 2x amps where x is the current across one capacitor. This would be different if you connected the capacitors in series; the sum of the voltages across each capacitor would equal the voltage of the battery and the current would be the same across each component.

So if I want my capacitors to produce 18v, I'd connect them in series; if I want them to produce 2x amp/hours I'd connect them in parallel?

No, if you had a 9V battery and equivalent capacitors in series, then the total voltage across both capacitors would still be 9V; the sum of the voltages across the capacitors equals the total voltage in series [http://en.wikipedia.org...].
Enji
Posts: 1,022
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10/11/2013 11:27:57 AM
Posted: 3 years ago
At 10/11/2013 11:14:26 AM, Lordknukle wrote:
Also, just one more question, what exactly is the point of capacitance ratings? It seems as if you can get all the information you need by knowing the output voltage and the nature of the connection.

The discharge voltage varies with time, and the rate of decay depends upon the capacitance.
Lordknukle
Posts: 12,788
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10/11/2013 11:33:37 AM
Posted: 3 years ago
At 10/11/2013 11:25:07 AM, Enji wrote:
At 10/11/2013 11:10:16 AM, Lordknukle wrote:
At 10/11/2013 11:00:38 AM, Enji wrote:
At 10/11/2013 10:49:27 AM, Lordknukle wrote:
At 10/11/2013 10:27:45 AM, Enji wrote:
1. Yes

2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.

Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?

The current of the output wire would equal the sum of the currents of the two capacitors, so you'd get 9v and 2x amps where x is the current across one capacitor. This would be different if you connected the capacitors in series; the sum of the voltages across each capacitor would equal the voltage of the battery and the current would be the same across each component.

So if I want my capacitors to produce 18v, I'd connect them in series; if I want them to produce 2x amp/hours I'd connect them in parallel?

No, if you had a 9V battery and equivalent capacitors in series, then the total voltage across both capacitors would still be 9V; the sum of the voltages across the capacitors equals the total voltage in series [http://en.wikipedia.org...].

Ok. So I would get the 18v output with a parallel circuit?
"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
Lordknukle
Posts: 12,788
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10/11/2013 11:35:25 AM
Posted: 3 years ago
At 10/11/2013 11:15:31 AM, drafterman wrote:
Go here:

http://www.falstad.com...

When the app loads (new window) go to: File > Import

Paste:

$ 1 5.0E-6 10.20027730826997 50 5.0 43
v 464 256 464 320 0 0 40.0 9.0 0.0 0.0 0.5
w 464 208 512 208 0
w 464 160 512 160 0
c 512 160 576 160 0 9.999999999999999E-6 -8.999999999998936
c 512 208 576 208 0 1.0E-5 -8.999999999999291
w 576 160 624 160 0
w 576 208 624 208 0
w 624 160 624 208 0
w 624 208 624 368 0
w 624 368 464 368 0
w 464 368 464 320 0
r 464 256 464 208 0 100.0
r 464 208 464 160 0 100.0
o 0 64 0 35 10.0 9.765625E-5 0 -1
o 4 64 0 35 10.0 9.765625E-5 1 -1
o 3 64 0 35 10.0 9.765625E-5 2 -1

Click "Import"

Won't let me c/p.
"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
Enji
Posts: 1,022
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10/11/2013 11:36:14 AM
Posted: 3 years ago
At 10/11/2013 11:33:37 AM, Lordknukle wrote:
At 10/11/2013 11:25:07 AM, Enji wrote:
At 10/11/2013 11:10:16 AM, Lordknukle wrote:
At 10/11/2013 11:00:38 AM, Enji wrote:
At 10/11/2013 10:49:27 AM, Lordknukle wrote:
At 10/11/2013 10:27:45 AM, Enji wrote:
1. Yes

2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.

Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?

The current of the output wire would equal the sum of the currents of the two capacitors, so you'd get 9v and 2x amps where x is the current across one capacitor. This would be different if you connected the capacitors in series; the sum of the voltages across each capacitor would equal the voltage of the battery and the current would be the same across each component.

So if I want my capacitors to produce 18v, I'd connect them in series; if I want them to produce 2x amp/hours I'd connect them in parallel?

No, if you had a 9V battery and equivalent capacitors in series, then the total voltage across both capacitors would still be 9V; the sum of the voltages across the capacitors equals the total voltage in series [http://en.wikipedia.org...].

Ok. So I would get the 18v output with a parallel circuit?

In parallel the voltage across each capacitor would be the same as the voltage of the output wire.
drafterman
Posts: 18,870
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10/11/2013 11:38:30 AM
Posted: 3 years ago
At 10/11/2013 11:35:25 AM, Lordknukle wrote:
At 10/11/2013 11:15:31 AM, drafterman wrote:
Go here:

http://www.falstad.com...

When the app loads (new window) go to: File > Import

Paste:

$ 1 5.0E-6 10.20027730826997 50 5.0 43
v 464 256 464 320 0 0 40.0 9.0 0.0 0.0 0.5
w 464 208 512 208 0
w 464 160 512 160 0
c 512 160 576 160 0 9.999999999999999E-6 -8.999999999998936
c 512 208 576 208 0 1.0E-5 -8.999999999999291
w 576 160 624 160 0
w 576 208 624 208 0
w 624 160 624 208 0
w 624 208 624 368 0
w 624 368 464 368 0
w 464 368 464 320 0
r 464 256 464 208 0 100.0
r 464 208 464 160 0 100.0
o 0 64 0 35 10.0 9.765625E-5 0 -1
o 4 64 0 35 10.0 9.765625E-5 1 -1
o 3 64 0 35 10.0 9.765625E-5 2 -1

Click "Import"

Won't let me c/p.

You can build it manually. Right click on the workspace to add components. RIght click on a component to add it to the scope for monitoring.
Lordknukle
Posts: 12,788
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10/11/2013 11:45:01 AM
Posted: 3 years ago
At 10/11/2013 11:38:30 AM, drafterman wrote:
At 10/11/2013 11:35:25 AM, Lordknukle wrote:
At 10/11/2013 11:15:31 AM, drafterman wrote:
Go here:

http://www.falstad.com...

When the app loads (new window) go to: File > Import

Paste:

$ 1 5.0E-6 10.20027730826997 50 5.0 43
v 464 256 464 320 0 0 40.0 9.0 0.0 0.0 0.5
w 464 208 512 208 0
w 464 160 512 160 0
c 512 160 576 160 0 9.999999999999999E-6 -8.999999999998936
c 512 208 576 208 0 1.0E-5 -8.999999999999291
w 576 160 624 160 0
w 576 208 624 208 0
w 624 160 624 208 0
w 624 208 624 368 0
w 624 368 464 368 0
w 464 368 464 320 0
r 464 256 464 208 0 100.0
r 464 208 464 160 0 100.0
o 0 64 0 35 10.0 9.765625E-5 0 -1
o 4 64 0 35 10.0 9.765625E-5 1 -1
o 3 64 0 35 10.0 9.765625E-5 2 -1

Click "Import"

Won't let me c/p.

You can build it manually. Right click on the workspace to add components. RIght click on a component to add it to the scope for monitoring.

That website really helps, thanks. It has simulations of capacitors in series and parallel.

Basically, if in series, neither the voltage nor the current add up. If in parallel, only the current adds up.
"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
Enji
Posts: 1,022
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10/11/2013 11:51:06 AM
Posted: 3 years ago
At 10/11/2013 11:45:01 AM, Lordknukle wrote:

Basically, if in series, neither the voltage nor the current add up. If in parallel, only the current adds up.

In series, the total voltage across the circuit should be the sum of the voltages across each component and the current should be the same across the circuit.
RoyLatham
Posts: 4,488
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10/11/2013 12:16:54 PM
Posted: 3 years ago
Yes, each capacitor will have 9 volts across it. If either capacitors is disconnected, it will show 9 volts.

Yes, they will discharge like batteries in parallel. Real capacitors have an internal resistance in series with the capacitance. That prevents them from discharging instantly if shorted, and in a slow discharge one many provide slightly more current than the other. For just storing energy, the resistance is not likely to be any problem and can be ignored.
RoyLatham
Posts: 4,488
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10/11/2013 12:21:38 PM
Posted: 3 years ago
At 10/11/2013 10:49:27 AM, Lordknukle wrote:
Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?

You will always get 9 volts, at 2 amps rather than 1.

If the charged capacitors were disconnected then reconnected in series, then you could get 18 volts at 1 amp. The disconnecting and reconnecting can be done automatically with a voltage doubler circuit. http://en.wikipedia.org... Diodes perform the simple switching.
RoyLatham
Posts: 4,488
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10/11/2013 12:24:53 PM
Posted: 3 years ago
At 10/11/2013 10:50:38 AM, Lordknukle wrote:
Or can the capacitors only output at the input current?

The input current is irrelevant. If there were no internal resistance, the output current could be infinite. When I said 1 amp or 2 amps, I meant that as an example of the average current -- two capacitors store twice the energy at the same voltage as one capacitor of the same size.
RoyLatham
Posts: 4,488
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10/11/2013 12:30:18 PM
Posted: 3 years ago
At 10/11/2013 11:14:26 AM, Lordknukle wrote:
Also, just one more question, what exactly is the point of capacitance ratings? It seems as if you can get all the information you need by knowing the output voltage and the nature of the connection.

The capacitance tells you how much energy the capacitor can store. A small capacitor will only prove current for a short time, a larger capacitor will prove more. It's like the capacity of a battery.

"A capacitor's storage potential, or capacitance, is measured in units called farads. A 1-farad capacitor can store one coulomb (coo-lomb) of charge at 1 volt. A coulomb is 6.25e18 (6.25 * 10^18, or 6.25 billion billion) electrons. One amp represents a rate of electron flow of 1 coulomb of electrons per second, so a 1-farad capacitor can hold 1 amp-second of electrons at 1 volt.

A 1-farad capacitor would typically be pretty big. It might be as big as a can of tuna or a 1-liter soda bottle, depending on the voltage it can handle. For this reason, capacitors are typically measured in microfarads (millionths of a farad)." http://electronics.howstuffworks.com...

To store a lot of energy, it's more efficient to use a high voltage capacitor.