how does one capacitor charge another capacitor?

Status
Not open for further replies.

pugongying

Member level 3
Joined
Apr 2, 2006
Messages
64
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,288
Visit site
Activity points
2,110
C1 has been connected to a voltage source V1 for a long time,at time 0,the switches switch the connection from V1 to C2,i.e. V1 is cutoff,and the two capacitor are parallel connected,what is the voltage waveform on C2?

(all the capacitors and voltage source has one node connected to groud)
 

The charge on the C1 is C1V1 before switch. After switch, the charge will share between C1 and C2. The voltage V2 will be C1V1/(C1+C2)
 

Thanks! I agree with the initial value and final value,but how about the waveform? Is it linear some exponential or something?
 

pugongying said:
Thanks! I agree with the initial value and final value,but how about the waveform? Is it linear some exponential or something?

I think it's exponential
 

If switch resistivity is equal to zero the voltage will be changed by step.
 

the total charge should be kept as a constant.
 

hi
i think if the circuit has zero resistivity, the charged capacitor acts as a voltage source and the 2nd capacitor will charge with a exponential function that initial and final value are as said before. but because the zero resistivity, the exponential wave has RC=0 !! and this is a step function exactly.

the exponential function : v(t)=A(1-e^(-t/RC))*u(t)
RC=0 => v(t)=A(1-e^(-t/0))*u(t)
=> v(t)=A(1-e^(-∞))*u(t)=Au(t) :: Step function
 

Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…