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I have read in some books that to determine the equivalent time constant of charging of a capacitor in any RC circuit containing either only one capacitor and any number of resistors arranged in any way or any number of capacitors arranged in any way and one resistor, we follow the follow the following steps: 1)Short circuit all the batteries. 2)In case of many resistors, the product of equivalent resistance across the capacitor and the capacitance of the capacitor will give the time constant of the circuit. 3)In case of many capacitors, the product of equivalent capacitance across the resistance and the resistance of the resistor will give the equivanent time constant. I have given a lot of thought on this but I am not able to figure out why that method works. Can someone please help? Can't this method be somehow extended to the general case containing any number of capacitors and resistors?

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    $\begingroup$ Why does the method work? Perhaps looking up Thevenin theorem would help. $\endgroup$
    – Elendil
    Apr 8 '20 at 4:27
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@SKTG

Can't this method be somehow extended to the general case containing any number of capacitors and resistors?

Yes, but you need to be very careful in reducing all parallel paths of common components to serial paths containing one resistive, one capacitive and one inductive component (if all are present).

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