# capacitor time constant

Discussion in 'General Electronics Chat' started by alexmath, May 10, 2014.

1. ### alexmath Thread Starter New Member

May 2, 2014
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0
Hello everyone! If i try to charge up a capacitor for example 2F capacitor through a resistor of 1k, it will charge approximately 100% after 5T right? therefore it will be 100% full after 2F*5*1000=2.77hours. If i apply 10v across it, it will take 2.77h to charge up to 10v and if i apply 220v it will take the same amount of time to charge up to 220v? What happens if i apply a constant current for example 2A, how fast does the capacitor will charge? And what happens if i still apply 2A current after the maximum charge that can store?

So if i change myself the current through the capacitor time constant is not working anymore?

Apr 5, 2008
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Last edited: May 10, 2014
3. ### MrChips Moderator

Oct 2, 2009
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It's all in the math.

When you charge a capacitor from a constant voltage source Vs through a series resistor R, as the capacitor voltage Vc increases and the voltage across the resistor (Vs - Vc) decreases.

The current through the resistor also decreases, I = (Vs - Vc)/R
at an inverse exponential rate.

With a constant current source, the voltage on the capacitor Vc increases at a constant rate over time. For the current source to remain constant, the voltage Vs of the current source must also increase in order to keep (Vs - Vc) constant.

The charge Q on a capacitor is given by the formula Q = C x V, where the charge Q is the current I x time t seconds

Hence the rate at which the voltage increases is V/t = I/C

With I = 2A and C = 2F, V/t = 1 volt/s

In this case there is no equivalent concept of time constant.

There is no maximum charge on the capacitor except what happens when the capacitor is destroyed by over voltage.

If the voltage of the current source can keep on increasing, the voltage on the capacitor will keep rising until the capacitor violently self destructs.

Last edited: May 10, 2014