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Understanding capacitors

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Homework Statement
So there is a 30000 micro farad capacitor 16 volts connected in parallel in the 9v voltage source to charge it and series to a light bulb. Why did the light bulb lights up momentarily when the source is turned on and eventually turning off? The measured voltage on the capacitor also increases more than of the 9V supply why is that?
Homework Equations
none
I suspect that at first the electrons on the other side of the plate in the capacitors are pushed to the light bulb that is the plate becomes more positive to compensate for the increase of electron on the other plate. Was this right? And can anyone refer me to a study or law regarding it?
 
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kuruman

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I assume the capacitor is initially charged to 16 V when the 9 V battery is not connected. When the battery is connected, the capacitor will lose enough of its charge to bring down its voltage to 9 V. The law is ##Q=CV##, the more ##V## across the plates, the more ##Q## on the plates.
Questions for you to consider
1. Where does the charge that leaves the plates of the capacitor end up?
2. What happens to the light bulb while the capacitor is partially discharging from 16 V to 9 V?
3. What happens to the light bulb after the voltage across the capacitor has reached 9 V?
 
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I guess the circuit looks something like this:
Captura de pantalla (619).png


Red: switch symbol.
Green: battery symbol.
Yellow: Capacitor symbol.
Black: light bulb.
 

kuruman

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I would think more like this. The switch is initially at "A" to charge the capacitor through resistor R. It is then thrown to position "B" to partially discharge the capacitor through the light bulb resistance.

CabAndBulb.png
 

CWatters

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We should wait for the OP to clarify the problem. I suspect the capacitor is _rated_ for 16V but is only connected to a 9V battery.
 
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The connection when it was charging was like this: I do not know the rating of the load (light bulb), but the voltmeter measures 0V when it was completely discharges and as we turns on the power supply it climbs up to 12V and it climbs up higher and higher, but slower and slower over time. Does it have something to do with the power supply shoving off more charges? I assume that it would equalize with the supply though. There was also a compass between the light bulb and the capacitor putted above a wiggly wire(not coil).
 

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kuruman

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The connection when it was charging was like this: I do not know the rating of the load (light bulb), but the voltmeter measures 0V when it was completely discharges and as we turns on the power supply it climbs up to 12V and it climbs up higher and higher, but slower and slower over time. Does it have something to do with the power supply shoving off more charges? I assume that it would equalize with the supply though. There was also a compass between the light bulb and the capacitor putted above a wiggly wire(not coil).
Help me understand. You have the circuit that you show in post #6 with the switch initially open. Then you close the switch and the voltage across the capacitor starts from zero and reaches a final value of 12 V over some time. Is that correct?
If not a coil, what does this "wiggly wire" represent? Is it part of the circuit?
Does the resistor in the circuit you posted represent the light bulb? Is that what you call the "wiggly wire"?
Can you explain the role of the compass in this?
 
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kuruman

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We should wait for the OP to clarify the problem. I suspect the capacitor is _rated_ for 16V but is only connected to a 9V battery.
Considering post #6, good call but the mystery deepens. We now have an additional voltage value of 12 V to deal with.
 
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Help me understand. You have the circuit that you show in post #6 with the switch initially open. Then you close the switch and the voltage across the capacitor starts from zero and reaches a final value of 12 V over some time. Is that correct?
If not a coil, what does this "wiggly wire" represent? Is it part of the circuit?
Does the resistor in the circuit you posted represent the light bulb? Is that what you call the "wiggly wire"?
Can you explain the role of the compass in this?
The wiggly wire was in series the resistor acts as a light bulb. I believe we only used the compass to analyze the magnetic field produced in the circuit.
 

kuruman

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OK, the wiggly line may be a fuse. You didn't answer my previous question, though,
When you close the switch, does the voltage across the capacitor start from zero and eventually level off to a final value of 12 V, increasing at a decreasing rate over time?
 
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OK, the wiggly line may be a fuse. You didn't answer my previous question, though,
When you close the switch, does the voltage across the capacitor start from zero and eventually level off to a final value of 12 V, increasing at a decreasing rate over time?
Yes, the voltage across the capacitor starts from zero and starts rising after around 12V it decreases the rate at which it ascends. The wiggly line was just a conductor bent. I do not have a reason why though. My thinking goes that it increases the magnetic field so that the compass needle might be affected stronger for us to see it.
 

kuruman

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You should get a compass deflection as long as the capacitor is charging which means that there is a current in the connecting wire. Once the capacitor is fully charged, the compass should point in the direction of the Earth's magnetic field. However, I cannot explain how you can get 12 V or more from a 9 V battery on the basis of the circuit you have provided and what you have posted about it. Maybe someone else can figure it out.
 

CWatters

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What value is the capacitor? 30 microfarads? 30 milifarads? The circuit says mF which is slightly non standard notation.

I ask because of light bulbs are generally low resistance so the time constant should be quite fast? Perhaps too fast to see it charging using a meter?

Are you sure it wasn't a 12 v battery?
 

256bits

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OP states power supply of 9v. (?? ac, dc, filtered, ripple, ... )
root2 x 9v = 12.69v peak though.
 

gneill

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OP states power supply of 9v.
root2 x 9v = 12.69v peak.
The circuit diagram would indicate that this is a DC circuit, so RMS and Peak are values are not pertinent here.
 

256bits

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The circuit diagram would indicate that this is a DC circuit, so RMS and Peak are values are not pertinent here.
Explains a rise to and above 12 v.
 

gneill

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Explains a rise to and above 12 v.
I suspect that there's something else going on here, related to information about the problem that we haven't been explicitly given. The capacitor may indeed have been pre-charged to 16 V as others have surmised.
 

256bits

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The OP stated in #11
Quote
the voltage across the capacitor starts from zero and starts rising after around 12V it decreases the rate at which it ascends
Unquote

I do agree something is being unstated for this problem.
 

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