I am testing this simple circuit that when the switch is off the light slowly dims out instead of going out.
What I am wondering is when the capacitor is fully charged and switch is off does the current flow out of the negative or positive side of the capacitor?
The reason why I ask this is because I have the LED cathnode (short leg)
on the side of the - of the capacitor. It works but if I do it the other way around cathnode to postive Doesn't work.
I thought current in a LED flows from anode (long leg ) to cathnode (short leg). Because I would think since LED are special diodes they won't work if the current is flowing in the cathnode direction.
So the only thing I can think of is current must flow out of the positive side of the capacitor?
i.e electrons are flowing out of the positive side ?
In a battery, current (electrons flow from negative side to positive side )
Maybe I am getting something with conventional flow as oppose to actual flow.
Can anybody clear up this simple issue.
Also how can you figure out how long the light will stay on with a capacitor. Like if I had a 10uF capacitor how could I figure out how long it would take to fully discharge? i.e turn off the light.
Some of my capacitors also give a voltage number as well as the capacitance. For example I have 220uF 16v. Is this voltage to calculate the total charge the capacitor can hold C = Q/V => 220uF * 16v = Q
If this is true how can we calculate the time it would take to completely discharge. Won't this depend on the rate at which the current is flowing. How can we do it just from knowing 220uf , 16v.
Because I would like to beable to determine when the capacitor is fully discharged and how long it takes to fully charge it. So I know how long the led is going to stay on.
What I am wondering is when the capacitor is fully charged and switch is off does the current flow out of the negative or positive side of the capacitor?
The reason why I ask this is because I have the LED cathnode (short leg)
on the side of the - of the capacitor. It works but if I do it the other way around cathnode to postive Doesn't work.
I thought current in a LED flows from anode (long leg ) to cathnode (short leg). Because I would think since LED are special diodes they won't work if the current is flowing in the cathnode direction.
So the only thing I can think of is current must flow out of the positive side of the capacitor?
i.e electrons are flowing out of the positive side ?
In a battery, current (electrons flow from negative side to positive side )
Maybe I am getting something with conventional flow as oppose to actual flow.
Can anybody clear up this simple issue.
Also how can you figure out how long the light will stay on with a capacitor. Like if I had a 10uF capacitor how could I figure out how long it would take to fully discharge? i.e turn off the light.
Some of my capacitors also give a voltage number as well as the capacitance. For example I have 220uF 16v. Is this voltage to calculate the total charge the capacitor can hold C = Q/V => 220uF * 16v = Q
If this is true how can we calculate the time it would take to completely discharge. Won't this depend on the rate at which the current is flowing. How can we do it just from knowing 220uf , 16v.
Because I would like to beable to determine when the capacitor is fully discharged and how long it takes to fully charge it. So I know how long the led is going to stay on.
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