How Does Closing a Switch Affect Capacitor Charge Over Time?

[thunderhadron]

a thundermistake!

lol

 

[thunderhadron]

Hi friends,

There is a problem after integrating with proper limits its going in this manner,

∫dq/ (CV - q) = ∫ dt/ RC

Left limit ==> CV/4 to q &

Right limit ==> 0 to t

ln[(CV - q)/ CV] = - t/RC

Placing the limits,

ln[(CV - q)/ CV] - ln [ (CV - CV/4) / CV] = -t / RC

since, ln a - ln b = ln (a/b)

hence,

ln [(CV - q) / (3CV/4) ] = -t / RC

==> (CV - q) / (3CV/4) = e^{-t / RC}

==> (CV - q) = (3CV/4) . e^{-t / RC}

==> q = CV - (3CV/4) . e^{-t / RC}

==> q = CV ( 1 - 3/4 . e^{-t / RC} )

Now it'll distribute in both as,

q = CV/2 . ( 1 - 3/4 . e^{-t / RC} )


Still it is not the answer as given, q = CV / 2 . ( 1 - 1/2 . e^{-t / RC} )

 

[tiny-tim]

ln[(CV - q)/ CV] - ln [ (CV - CV/4) / CV] = -t / RC

no, you're getting confused …

your q is the charge on both capacitors combined, so at t = 0 that's q = CV/2

(and i really really REALLY think you should stop using this "limits" method, use standard integration as in my previous post )

 

[thunderhadron]

Oh My dear tim,
I got the correct answer. And I've understand the problem completely now. And that is only by din't of you & Yukoel.

Cordially Thank you very much guys.

I appreciate your help.:!)