Q:Sent by Tanh, VLSI engineer
What will be the voltage level between the two capacitors? The Vcc = 10v DC.
Hint from Hitequest:
Q1= V1*C1 //ammount of charge on C1
Q2= V2*C2 //ammount of charge on C2
Q = Q1 = Q2 //all caps in series have the same charge
(V-V2)*C1=V2*C2
V2=V*C1/(C1+C2)
V2=10*20/50=4V //the result
Hector M.:
there is an error on the question with
the 2 capacitors in series (20uf and 30uf)....
The solution is 6 volts instead of 4 volts. The voltage
divider formula is:
V=(C2.Vcc)/(C1+C2)
Marvin:
My comment is related to the question sent by Tanh. As
shown in the figure the required voltage is the voltage on
the capacitor C2 (The voltage is always measured between
two points). The response shown is correct.
Hector M comment is wrong - The voltage divider for the
capacitors is V2= V*C1/(C1+C2) which obviously is not as
shown by him.
His formula is an extension of the resistor
voltage divider, V2=V*R2/(R1+R2)where R1 and R2 are the
resistors replacing the capacitors in a resistor network.
anonym:
Tanh's problem about two capacitors in series is under-specified. If the
capacitors start out discharged, then they will charge up as indicated since the
current is the same through the circuit and therefore q1 = q2.
However, if the capacitors have some initial charge, that changes everything.
One can imagine a stable condition in which, for instance, V1 = 9V and V2 =
1V.
The net charge across each capacitor equals its own voltage, so no current
flows.
