RC time constant

Hello,

I have a fairly simple question. If a capacitor is being charged through a resistor the time it takes to reach full charge or for the cap to reach the battery voltage is 5*RC. If two caps are in series...and that is charged through a resistor will the time for it to reach full charge/battery voltage be 10*RC?

I may have my theory completely wrong. Would really appreciate some help.

Thanks,

George L.

in a single R-C circuit, the voltage on the capacitor varies as exp(-t/R/C) but when two RCs are cascaded, the load on the first R-C modifies its voltage so the math gets complicated. At any rate, by 10RC, the 2nd cap is nearly fully charged.

George L. said:

Hello,

I have a fairly simple question. If a capacitor is being charged through a resistor the time it takes to reach full charge or for the cap to reach the battery voltage is 5*RC. If two caps are in series...and that is charged through a resistor will the time for it to reach full charge/battery voltage be 10*RC?

I may have my theory completely wrong. Would really appreciate some help.

Click to expand...

There are some cases where time is roughly calculated by: R * C
and there are some cases where time is roughly calculated by 2 * pi * R * C, where pi is the value when you press the PIE symbol on your calculator.

I think that 2 * pi * R * C is mainly used for high-frequency, and R * C is used for low frequencies. I'm not sure about this.

If you want to get more precise, the basic equation for time is: 0.69 * R * C. I use 0.69 because I think that waiting (R * C) seconds will charge capacitor C through Resistor R, but the capacitor will be charged about 69%.

I don't know if 69% is wrong, but the range is between 60% and 70%

George L. said:

Hello,

I have a fairly simple question. If a capacitor is being charged through a resistor the time it takes to reach full charge or for the cap to reach the battery voltage is 5*RC. If two caps are in series...and that is charged through a resistor will the time for it to reach full charge/battery voltage be 10*RC?

I may have my theory completely wrong. Would really appreciate some help.

Thanks,

George L.

Click to expand...

A capacitor will charge to a little over 99% of final value (the battery voltage) in 5 time constants (5*RC). If you put two caps (C1 and C2) in series, but still have only one resistor, the resulting capacitance is

C=C1*C2/(C1+C2)
As you can calculate, if C1=C2, then the resulting capacitance is half the value of the individual caps.

Calculate the new time constant using this capacitance.

Mstechca doesn't know what he's talking about.

He sure gets pi in his face a lot.... It was amusing to read though. I hope hes not really a "tech."