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.

Attached Images

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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%

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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.

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Ron

 

hyedenny

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