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Capacitor in parallel

Why the answer is B? The voltage of capacitor should increase or decrease exponentially.Screenshot 2020-05-28 at 11.06.08 AM.png
Original post by ialfighter2020
Why the answer is B? The voltage of capacitor should increase or decrease exponentially.Screenshot 2020-05-28 at 11.06.08 AM.png


Slightly "trick" question, in the real world the behaviour should be more like D (but with a steady state section afterwards), exponential charge and discharge up to the limit of Vc = V.

HOWEVER, the power supply is described as "ideal" and has negligible internal resistance. The time constant for exponential charging T = R(int) * C which for R=0 is also 0, hence instantaneous charging to Vc = V and then steady state from then onwards. A bit unphysical as there is always some internal resistance and the power supply has to do some work to charge the cap.
Original post by Mr Wednesday
Slightly "trick" question, in the real world the behaviour should be more like D (but with a steady state section afterwards), exponential charge and discharge up to the limit of Vc = V.

HOWEVER, the power supply is described as "ideal" and has negligible internal resistance. The time constant for exponential charging T = R(int) * C which for R=0 is also 0, hence instantaneous charging to Vc = V and then steady state from then onwards. A bit unphysical as there is always some internal resistance and the power supply has to do some work to charge the cap.

Closer inspection of D rules that answer out immediately; the correct discharge curve as the switch is opened would be exponential decay with an asymptote trajectory approaching zero at t = infinity. i.e. it only achieves zero at infinity.
(edited 3 years ago)
Original post by uberteknik
Closer inspection of D rules that answer out immediately; the correct discharge curve as the switch is opened would be exponential decay with an asymptote trajectory approaching zero at t = infinity. i.e. it only achieves zero at infinity.

Is there any explanation behind this phenomenon (i.e. the voltage never drops to zero). I know that capacitor follows exponential decay but it seems strange to me that not all the charge would be discharged.
(edited 3 years ago)
Original post by ialfighter2020
Is there any explanation behind this phenomenon (i.e. the voltage never drops to zero). I know that capacitor follows exponential decay but it seems strange to me that not all the charge would be discharged.


The charge and discharge behaviour of an IDEAL capacitor via a resistor is nicely modelled with a 1st order differential equation, where the rate of charge is a function of the voltage difference between the source and the voltage at a specific time. As it charges up, the rate of charge goes down. If you solve this differential equation (that's a nice 1st year university problem) you get an exponential solution where V = Vin [1- exp (t/RC)] which approaches Vin, but only gets there for infinitely long times.

Now in the real world, “other stuff” gets in the way of this "infinite time, perfect solution", capacitors leak slightly and are a little bit inductive. They act as aerials and pick up a bit of signal from radio waves and also charge from the air. A really big capacitor can charge itself enough from this to kill you, which is why you seem them stored with shorting bars on. More explanations at the link below.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html
Original post by Mr Wednesday
The charge and discharge behaviour of an IDEAL capacitor via a resistor is nicely modelled with a 1st order differential equation, where the rate of charge is a function of the voltage difference between the source and the voltage at a specific time. As it charges up, the rate of charge goes down. If you solve this differential equation (that's a nice 1st year university problem) you get an exponential solution where V = Vin [1- exp (t/RC)] which approaches Vin, but only gets there for infinitely long times.

Now in the real world, “other stuff” gets in the way of this "infinite time, perfect solution", capacitors leak slightly and are a little bit inductive. They act as aerials and pick up a bit of signal from radio waves and also charge from the air. A really big capacitor can charge itself enough from this to kill you, which is why you seem them stored with shorting bars on. More explanations at the link below.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html

Ok, got it! Thanks a lot for all your help!

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