[ilovemath]

here are the MS points :

Capacitor stores charge/charges up (1)
(If voltage is constant) capacitor doesn’t discharge (1)


I don't understand how the answer makes sense. The question asks how placing a capacitor causes the power source to produce a constant voltage. but how does putting the capacitor there suddenly make the power source decide its voltage?

[doomhalo]

This is called smoothing. Because the DC is changing, with the addition of a capacitor which has a voltage which decays slowly over time, by the time the voltage from the power supply has increased back to it's high point, the capacitor hasn't discharged much. So roughly, the voltage stays constant.

http://www.kpsec.freeuk.com/powersup.htm#smoothing

[ilovemath]

(Original post by doomhalo)
This is called smoothing. Because the DC is changing, with the addition of a capacitor which has a voltage which decays slowly over time, by the time the voltage from the power supply has increased back to it's high point, the capacitor hasn't discharged much. So roughly, the voltage stays constant.

http://www.kpsec.freeuk.com/powersup.htm#smoothing

sorry i don't understand how this works why does the power supply depend on the capacitor.

[ilovemath]

BUMP*


OK so I get the first bit on the ms

Capacitor stores charge/charges up (1)
(If voltage is constant) capacitor doesn’t discharge (1)

but not the second bit...."if voltage is constant" BUT IT ISN'T
please could someone explain

thanks

[Foghorn Leghorn]

By the looks of it, the supply is an ac supply that has been fed through a rectifier to create a direct current in the circuit, hence the voltage increasing and decreasing between 0 and 6 but not dropping below 0. If this is the case then every time there is a positive charge it will charge the capacitor, when the charge drops due to the initial ac supply then the capacitor discharges and bridges the gap between the peak voltages.

[Stonebridge]

Look here
http://www.thestudentroom.co.uk/show...php?p=35040118