Stupid conceptual question regarding capacitors

Discussion in 'Homework Help' started by hitmen, Oct 8, 2008.

1. hitmen Thread Starter Active Member

Sep 21, 2008
159
0
I have been thought that for a capacitor connected to a load in series ,

C d(vc)/dt + Vc(t)/R = 0

I don't understand why this is so. According to the passive sign convention, anything absorbing power is assigned a positive value while that supplying power is assigned a negative value.

So shouldnt the equation be NEGATIVE C d(Vc)/dt PLUS Vc(t)/R = 0 since the capacitor is supplying power while the resistor is absorbing power?

Also wouldnt the current be added twice for this equation I am unable to grasp this concept. Can anyone explain or refer me to the appropriate source?

2. mik3 Senior Member

Feb 4, 2008
4,846
70
I think you confuse series with parallel circuits.

If you have a battery and a capacitor in series with a resistor and connect them across the battery you have this equation according to KVL:

Vo=Vc+IR

where
Vo=battery voltage
Vc=voltage across the capacitor
I=current through the circuit
R=resistance

if you substitute I=C*(dVc/dt) you get

Vo=Vc+RC*(dVc/dt)

this is a differential equation and you can solve it to get the formula about the voltage across the capacitor varying with time.

3. neon9 Member

Oct 8, 2008
15
0
first of all sign are applied in electron flow positive to negative not hole flow whereby the flow of holes flow the other way. a capacitor opposes flow of electrons basicaly it stores them as potential. so when you apply sudden voltage to a cap the voltage will rise exponentialy as dv/dt

4. hitmen Thread Starter Active Member

Sep 21, 2008
159
0
You mentioned that Vo = Vc + IR

How do you know whether to to put a positive or negative before Vc?

Also, when a capacitor is connected solely to a resistor,

is the equation Vc +IR = 0 or -Vc+IR =0

I know that the first equation is right but I dunno why. Isnt the capacitor supposed to be acting as a source in this case? 5. mik3 Senior Member

Feb 4, 2008
4,846
70
You assume the polarities initially and after you have done the calculation you will see which polarities should be in the other direction. They will appear as negative on your answer.