The voltage would divide according to the capacitance values (assuming 0 leakage in both caps). Using the formula C=Q/V, where C is capacitance in Farads, Q is charge in Coulombs, and V is voltage (in volts, of course!), Q necessarily has to be the same for both caps in a series circuit once both are charged. Therefore, Q = C1/V1 = C2/V2, and the magic of algebra yields V1/V2 = C1/C2.
73, -Larry/NE1S Jim Candela wrote: > It would be interesting to see how the voltages would imbalance if the > capacitors were poorly matched in terms of capacitance (assume equal leakage > currents).... Any thoughts? > > Regards, > Jim Candela > WD5JKO
