Charging - discharging simple problem

thunderhadron

Yes and out of it there is right cap which is totally uncharged before the closing the switch. so the circuit will look like the same ckt as you showed me.
And the P.D. across the right capacitor will become V/2 due to battery. M I going wrong?

tiny-tim

yes the voltage across the two capacitors will be the same so long as the switch is closed (and will be the same as across the "second" resistor)

so what are the charges on each capacitor when the switch is first closed?

thunderhadron

As the voltage across the two capacitors and the second resistor is same as V/2 then the charges on both the capacitors according the relation Q = CV are CV/2 & CV/2
???

tiny-tim

no, if the charge on the left capacitor was CV/2 immediately before closing the switch (and 0 on the right capacitor), how can it be CV/2 on both capacitors (total of CV) immediately after?

the total charge has to be the same … impulsive current cannot flow through resistors

(the maximum possible charge, ie the relation Q = CV, is irrelevant)

thunderhadron

Do u mean to say that the charge on both will become CV/4 ?

tiny-tim

yes … it's the same total charge, and the capacitances are the same, so it will be equally distributed

does that make the rest of the problem clear?

thunderhadron

OK.. I am getting this. But please tell me one thing: At the time of closing the the switch the right cap is connected with left capacitor hence the charge distribute equally on both. Bit the right cap is also connected to the battery at this instance so what is the role of battery at the moment?

tiny-tim

at that moment (between "immediately before" and "immediately after" the switch closes), the battery and the resistors have no role …

the battery and the resistors need time to do anything, but the charge rearranges itself "instantaneously" …

comparing it with mechanics: the switch gives an impulsive force, and non-impulsive forces (such as gravity, or the battery) have no role

thunderhadron

But some times when I solve some problems of charging- discharging, take the example of a simple circuit when a cap and a resistor are connected in parallel and a battery is connected with this circuit in parallel.
Many times the problem says at the instance of closing the switch : Current will have two ways on the junction of resistance and cap. but the current will flow through capa at that time, it acts like zero resistance or superconducting wire at t = 0 but the time goes on the cap stores charge by leaps and bounce and resistance of it keeps on increasing.
And after a very long time when it gets fully charged it no charge flow through it, it acts like
R→∞.
Is this wrong, so please tell me what is the actual concept but mind one thing I've solved ample number of problem keeping this concept in my mind, If you want I can show you some examples.

NascentOxygen

That's correct. The capacitor offers very low impedance under those circumstances. But if there is a resistance in series with the capacitor. then that resistor limits the current and hampers the capacitor in doing what it might otherwise do. Only where the effective series resistance is very low can the capacitor branch act like a "short circuit". You have to examine the circuit very closely to see which resistances are in the relevant charge/discharge current paths and under which circumstances.

tiny-tim

i don't understand that
you do know that no current ever flows through a capacitor?

talking about the "resistance" of a capacitor is just an analogy

a capacitor does not have a resistance (R), it has a reactance

ie it is not Ohmic, with I proportional to V, it is reactive, with I proportional to dV/dt
(you mean "leaps and bounds"?)

no, the charge flows smoothly …

it starts fast, and ends slow, but the change is smooth

immediately after you close the switch, the charge on the capacitor is still zero, but the rate of charge is non-zero

thunderhadron

I am sending you an img.

Well I mean to say that at the instance of closing the ckt the current has two ways at point c, one from capacitor link and another from resistor through wire cd. So the resistance sorry capacitive reactance of the topmost capacitor is very less at this time hence the charge will flow form capacitor.

thunderhadron

Friend please make me understand some according to this fig. Is this fig is correct. Will the ckt look like this?

NascentOxygen

What do you mean by "the topmost capacitor"?

Yukoel

Hello thunderharon,
Look at the rectangular loop formed by the right and left capacitor.Apply Kirchhoff's Voltage rule.The capacitances being same they ought have the same charge initially.The left capacitor achieved steady state in the time before switch was closed ,true but now it has to achieve the steady state once again in accordance with Kirchhoff's rules.
regards
Yukoel

tiny-tim

ok, i understand now …

you're talking about problems with a resistor and a capacitor (or more than one
capacitor) in parallel, so that the current (or charge) can flow both ways

then yes, you're correct (apart from the "leaps and bounds"): the current (which before the switch closed was flowing entirely towards the resistor) will start flowing entirely towards the capacitor, but will then flow more and more towards the resistor until finally it flows entirely towards the resistor

(as if the capacitor started with zero resistance, and ended with infinite resistance)

thunderhadron

::Yukoel:: So like "tiny-tim" was telling me Its P.D. will become V/4 now so as the P.D. of the right capacitor. Is this correct?