Question on capacitance?

Question

Hi all! I've got a question on capacitance.
For resistors in series, the rule is that the higher the resistance of an appliance, the higher the potential difference across it.
What about capacitors in series? Is it the higher the capacitance of a capacitor, the lower the potential difference across it?

Answer 1

This topic is foreign to me, however, I may have a site with the answer: http://lectureonline.cl.msu.edu/~mmp/kap19/RR480app.htm
The page says:
In this applet, you can select different configurations of capacitors in parallel and/or in series, change the capacitances of the individual capacitors, and the applet will calculate the equivalent capacitance of the network. You can practice applying the equivalent capacitance formulas in each case, and check with what the applet gives.

Answer 2

'Let's make sure what you are asking: By potential, I assume, you mean the voltage. And that voltage is DC, so impedance has nothing to do with it here.
Unlike a resistor, a capacitor only "conducts" electricity when it is "empty".( = uncharged). So there are 2 possible scenarios:
a) During the initial moment of charging an "empty" capacitor: The potential is actually close to zero (short circuit), and rising in a reverse-exponential curve to the original value of the potential (voltage).
b) If the capacitor is charged, the potential across the capacitor equals the original potential.
BOTH scenarios are true no matter what value of capacitors you have, only the time it takes from empty to charged state is longer: The higher the capacitor value is (dependent on total circuit resistance, including internal resistance of the potential's source ( = power supply), the longer it takes to charge it.

Answer 3

impedence of a capacitor
Z=[R^2+(1/2pi f c)^2]^1/2 in ohms, where
r=resistance in ohms
f=frequency cycles/sec
c=capacitance in farads
we observe that capacitance is in the denominator. So larger the capacitance smaller the impedence.