The voltage across capacitors in series is inversely proportional to their capacitance.
This is because:
- they are all partaking of the same current, so charge is evenly distributed between them.
- Q=CV so C=Q/V so capacitance = charge / voltage
- so farads can be thought of as 'coulombs per volt'
- so if two capacitors in series each hold 2C of charge, and cap A is rated 1F and cap B is rated 2F, the 2C in A keep its terminals 2V apart, but the 2C in B only keeps its terminals 1V apart.
- cap A, rated with half the capacitance of B, thus has twice the voltage across it. TICK
Thanks again to Grob for pointing this out where Cook, and Art of Electronics, didn't.
(I understand capacitance better if I think of it as 'charge sensitivity'. Higher capacitance means lower voltage between the terminals after being filled with the same amount of charge.)
Anti-nuisance lawsuit warning: The purpose of these notes is to remind me, Zoegond, of stuff or to help me work stuff out. They may contain mistakes.
Quick
- ($a, $b....) = unpack("A2A7...", $packed)
- push( array, list )
