basically a capacitor will charge to the input DC level however it will mathematically never happen since capacitors charge at a certain rate the voltage drop across a capacitor will follow the R C time constant or 63% of the applied voltage for a unit time.
AnswerIn the case of an a.c. supply, yes, there will be a voltage drop across a capacitor. In the case of an 'ideal' capacitor, this will be the product of the load current and the capacitive reactance of the capacitor.
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It's not.
The voltage drop across an LED varies, but is typically around 2 volts.
Depending on the circuit, 63% of the available voltage.
(a) what is the total capacitance of this arrangement (B) the charge stored on each capacitor (C) the voltage across the 50 micro farad capacitor and the energy stored in it. 20v and 20+30+50 micro farad
Voltage across all parallel capacitor's is same i.e. it is equal to supply voltage, it can be measured using digital volt meter (any high input impedance volt meter). When capacitors are in series; voltage drop depends on charge stored in the capacitor. it can be given by the formula V x V = 2 / (joules x capacitance). This voltage can also be measured using digital volt meter.