### Welcome to EDAboard.com

#### Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Status
Not open for further replies.

#### kishore680

##### Member level 5
Initially 0.5uf is uncharged whereas 1uf is charged with 10v... what wud be the voltage across c2 i.e 0.5uf??? after some time... They are in parallel . so voltage should be equal? And then discharge through 10k ?..

#### vinodstanur

When you connect the charged capacitor (1uF) to the circuit, instantly the voltage across it will decrease. The .5uF capacitor will try to get charged from the 1uF capacitor.
Now since there is a 10K resistor, practically the final voltage across the capacitors will be zero because it discharges the two capacitors. (after some time)

#### jeffrey samuel

there will be decreasing sinusoids of voltage curve in real time as the CAP voltage is dissipated across the resistor

#### albert22

##### Full Member level 6
I agree with vinodstanur.
When you connect the charged capacitor to the uncharged capacitor, part of the charge will be transffered to uncharged capacitor. Theoretically the current will not have limit (infinite) because the cap will present a short. In practice it will be limited by the ESR. Infinite current will charge instantly the .5uF. You will be able to calculate the resultant voltage taking in account that the total charge of the 1 uf cap will be contained now in one cap of 1.5uF. Neglecting the current on the R. This produces a step in the voltage. Then the 1.5uF capacitance will discharge exponentially with a RC constant of 1.5uF x 10Kohms to give 0v (asymptotically)
Calculating the curve of voltage can be done by solving a differential equation.
There is no way in that can be sinusoids involved unless an inductance is involved. Real caps and resistors do have inductance but I think you are presented with ideal components ( No ESR no inductance, etc)
Please post the final solution if you ever confirm it.

another subject, ( as per vinodstanur status line):
I disagree that all waveforms are composed of different sinusoids. I think that although any waveform can be mathematically described by fourier series, they are generated by physical and electrical phenomenon that usually not involve sine waves. As the one which are studying here. Applying a mathematical abstraction not necessarily means that the real wolrd behaves the same way. But who knows what really happens ? We only have models.

Regards

#### jeffrey samuel

I agree with vinodstanur.

another subject, ( as per vinodstanur status line):
I disagree that all waveforms are composed of different sinusoids. I think that although any waveform can be mathematically described by fourier series, they are generated by physical and electrical phenomenon that usually not involve sine waves. As the one which are studying here. Applying a mathematical abstraction not necessarily means that the real wolrd behaves the same way. But who knows what really happens ? We only have models.

Regards
If you are able to write any signal then when you frequency analysis of it it generally becomes a sinusoid (fourier transform)

this is true in most cases

Status
Not open for further replies.