0
\$\begingroup\$

I am watching this video where the guy deducts the voltage across the capacitor during the transient phase.

enter image description here

There is an integral, in the middle of the equations. The guy integrates that and there is no constant of integration after that.

This is not just this guy, I am not aware of any of the capacitor/inductor formulas where the tutor considers them.

Why is that? Why can you simply ignore the constant of integration for all these equations?

\$\endgroup\$
2
\$\begingroup\$

I believe that the constant of integration is taken care of here: -

enter image description here

\$V_0\$ is the initial condition of the capacitor: -

enter image description here

And, when the left-hand integral is resolved \$V_0\$ is properly handled. For the right-hand integral having "0" as a limit means there won't be a "k" factor.

\$\endgroup\$
  • \$\begingroup\$ are you telling me that if the integral has a range, like \$ \int_0^t \$ or \$ \int_{Q_0}^Q \$ is it ok to ignore the constant of integration? \$\endgroup\$ – SpaceDog Aug 8 '18 at 14:42
  • 1
    \$\begingroup\$ When the integral gets resolved and the limits applied you get an (U+K)-(L+K) situation where U is the upper limit and L is the lower limit. K (the constant) cancels but it was a long time ago that I learned this LOL. \$\endgroup\$ – Andy aka Aug 8 '18 at 15:29
  • \$\begingroup\$ brilliant, thanks. What you said gave me a lead to research the topic and you are right. When the integral is defined the constant of integration is cancelled and the limit situations are handled. THANKS, I LOVE YOU! \$\endgroup\$ – SpaceDog Aug 8 '18 at 15:46
  • 1
    \$\begingroup\$ Easy now chap LOL \$\endgroup\$ – Andy aka Aug 8 '18 at 15:49
  • 1
    \$\begingroup\$ haha 😃!!!!!!!!!!!!! \$\endgroup\$ – SpaceDog Aug 8 '18 at 15:54
0
\$\begingroup\$

At t=0 the switch is open and thus current flowing is zero.

The usual assumption in such example cases is that circuit is in steady state when t<0 which means the voltage across the capacitor is also zero.

As a result the constant on integration, the initial conditions, are zero.

I agree it is a bit sloppy as it isn't always the case and sometimes the dynamics of the circuit and the bias point does result in the constant of integration being non-zero

\$\endgroup\$
  • \$\begingroup\$ Yes I agree. It appears that the solution is doing a log of assumptions. The problem is that this solution is the most complete I have found. If you know some document online, book or something where all these conditions are considered, for RL, RC and RLC, please let me know. Thanks. \$\endgroup\$ – SpaceDog Aug 8 '18 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.