# Time constance capacitors electricity

time constance "capacitors" electricity

was looking over my physic 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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was looking over my physic 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?

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.