Time constance capacitors electricity

AI Thread Summary
The discussion centers on the concept of the time constant (RC) in capacitor circuits, specifically how it affects voltage charging over time. The time constant indicates how long it takes for a capacitor to charge to approximately 63% of its maximum voltage. Increasing the resistance (R) by a factor of two will also double the time constant, leading to a longer charging time, but the relationship is not linear due to the exponential nature of the charging equation. It is clarified that charging typically takes longer than discharging, depending on the circuit configuration. Capacitors are often used in applications like flash cameras, where they rapidly release stored energy.
seto6
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time constance "capacitors" electricity

was looking over my physics electricity
i got some question actually i just forgot them.
v(t)=V(1-e^(-t/RC))


where RC is the the time Constance,
1) i am not sure what the time Constance means.
2)If R is increased by a factor of 2 what in theory is the effect on time Constance

i tried reading my old notes but i guess i did not attend that lecture

any help is appreciated
 
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seto6 said:
was looking over my physics electricity
i got some question actually i just forgot them.
v(t)=V(1-e^(-t/RC))


where RC is the the time Constance,
1) i am not sure what the time Constance means.
2)If R is increased by a factor of 2 what in theory is the effect on time Constance

i tried reading my old notes but i guess i did not attend that lecture

any help is appreciated

First off you are dealing with an equation that describes the voltage across the plates of a capacitor with capacitance C with respect to time in relation to some initial voltage supplied most likely by a battery of voltage V. At long times t, after a capacitor is allowed to charge, the term to the right of V approaches 1. At smaller times, the capacitor will still be charging so the term to the right of V really describes what % of the max voltage the capacitor is at... (1- a number smaller than 1 but not less than zero). At t = 0 the capacitor is uncharged so the term to the right is (1-1) multiplied by V. So naturally the capacitor is uncharged.

If you have a resistor in series with the capacitor then the amount of current flowing will be smaller. So RC sort of describes how long it will take to get to approach a fully charged capacitor, or approaches max V. If RC is really big, then it will take a longer time, t, to become fully charged to a voltage, V.

Since e to the t/RC is exponential, if you make the bottom part of this exponent, RC larger it should take longer to charge. If for example, RC is equal to 1, then at t=1 the term to the right would equal about .63 or 63% of full charge... (1-0.37) e to the first power is about 2.718 and this divided into 1 is about 0.37...
Now if you kept t=1 and made RC = 2 then you would get e to the 1/2 power or the square root of e divided into 1 and this would give you a term to the right of V and multiplied by V of about 0.40... So increasing the RC by 2 would lead to only 40% of total charge instead of 63%... it takes longer to charge the capacitor to a full V.
 


Sorry that was long winded but I tried to use simple examples. You can also describe current flow through a capacitor with these same basic ideas. The capacitor is basically just a wire with very little resistance at t=0 but it soon builds with charge thus V gets big and I, the current flow, slows to a trickle and approaches O. Capacitors can be used effectively in certain circuits to stop current flow down a certain part of a circuit. Or to give off a lot of charge and energy quickly, like to a flash bulb.
 


thanks but i don't understand about the 63% and the 40%, if Resistance increase by factor of two, then time constant would increase by a factor of 2, that means it would take 50% more to charge.

and does it usually take more time to charge than discharge?
 


seto6 said:
thanks but i don't understand about the 63% and the 40%, if Resistance increase by factor of two, then time constant would increase by a factor of 2, that means it would take 50% more to charge.

and does it usually take more time to charge than discharge?

Its e raised to the power of t/RC all under 1. Thats what e^-t/RC is. So increasing RC by a factor of 2 means you are changing the exponent, its not a linear relationship.

On the last point it depends on how you have the capacitor set up in a circuit. You usually charge a capacitor with a battery and discharge it through a piece of the circuit without the battery as a part of it.

For practical purposes you can charge a capacitor with a battery and then "release" the charge rapidly, say, through a flashbulb. Thats how many flash cameras use them.
 
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