Capacitors , why can't series combination be used here?

[K Dhiraj Bhak]

Homework Statement

Two capacitors A and B with capacitance 3 μF and 2 μF are charged to potential difference of 100 V and 180 V respectively. These two capacitors are connected to an uncharged capacitor of 2 μF using a switch as shown in the diagram. Determine the final charge on each capacitor when the switch is closed. ( please refer the figure (attachment))...

Homework Equations

Q=CV

The Attempt at a Solution

the actual method is by charge conservation and Kirchhoff Loop rule.. ..but i tried a different method and terribly failed! BEFORE THE SWITCH IS ON, i tried to find the net equivalent capacitance of the 2 capacitors lying in the bottom,and then,WHEN SWITCH IS ON ,by assuming that a charge (q') flows to the uncharged capacitor from the NET charged capacitor till the capacitors are at the same potentials.But i ended up with the wrong answer...so my method fails...but why?...i think i went wrong in the first step itself...i feel i shouldn't have found the net capacitance using series formula...but why not? where did i go wrong actually? please help. in brief, why can't we apply series combination formula in an open circuit??

 

[NascentOxygen]

Welcome to the Physics Forums.

I can see how you determined your equivalent single capacitance, but how did you determine what initial charge it should have so as to give an equivalent outcome here?

 

[K Dhiraj Bhak]

i added the charge contained by the individual capactors(3μF and 2F)..ie-->(300 μC + 360 μC)=660μC

 

[K Dhiraj Bhak]

Since the 2 capacitors are isolated from the other capacitor ( isolated b'cos the switch is off) , their initial charges remain constant as long as the switch is NOT ON. that's why i combined their charges into the equivalent capacitor...am i correct?

 

[K Dhiraj Bhak]

same question can be posed like this also...

Two capacitors A and B with capacities 3 mF and 2 mF are charged to a potential
difference of 100 V and 180 V respectively. The plates of the capacitors are
connected as shown in the figure with one wire from each capacitor free. The
upper plate of A is positive and that of B is negative. An uncharged 2 mF
capacitor C with lead wires falls on the free ends to complete the circuit.
Calculate
(i) the final charge on the three capacitors, and
(ii) the amount of electrostatic energy stored in the system before and after
the completion of the circuit..
...this question is same as the original one,,but this one explains much better

 

[NascentOxygen]

Since the 2 capacitors are isolated from the other capacitor ( isolated b'cos the switch is off) , their initial charges remain constant as long as the switch is NOT ON. that's why i combined their charges into the equivalent capacitor...am i correct?

Try working backwards. Determine the initial charge needed on the single equivalent capacitor so that it transfers the same charge to the 2μF capacitor as do the series pair in this question.

 

[K Dhiraj Bhak]

so...why am I not allowed to combine the capacitors here??...is there any rule that capacitors cannot be combined in an open circuit??..if yes , why?

 

[NascentOxygen]

so...why am I not allowed to combine the capacitors here??...is there any rule that capacitors cannot be combined in an open circuit??..if yes , why?

Certainly you can combine them. Regardless of all else, if they are in series then it's always true that 1/C = 1/C1 + 1/C2

That gives you the capacitance that behaves equivalently, but you have yet to determine the initial charge it should have so as to lead to identical circuit current & voltage when the switch gets closed.

 

[sophiecentaur]

so...why am I not allowed to combine the capacitors here??...is there any rule that capacitors cannot be combined in an open circuit??..if yes , why?

Apologies for repeating some of what's gone before but I think this is relevant (or just put another way).
The point here, I think, is that the two charged capacitors will have different charges on them (from Q=CV, they will have 300 and 360C, respectively). If you took the two capacitors (uncharged) and connected them across a 280V supply, what charge would the series arrangement have? The simple series equivalent capacitor formula assumes that the charges are the same on each. I think this is the crux of your problem.
You need to set up some simultaneous equations here, I think - not too hard though.

 

[DrZoidberg]

You can just combine A and B into one 1.2µF capacitor charged to 280V.
So what happens when you connect 1.2µF in parallel to 2µF?
Calculate the resulting voltage (280V/3.2µF * 1.2µF) and from that you can then calculate all the other values.
To calculate the final charges on A and B however you have to take into account the fact that they had different amount of charge on them. The difference in charge has to remain constant. i.e. Ca (charge of A) - Cb = constant

 

[K Dhiraj Bhak]

.
The point here, I think, is that the two charged capacitors have different charges on them (from Q=CV, they will have 300 and 360C, respectively).The simple series equivalent capacitor formula assumes that the charges are the same on each.

i agree that the 2 capacitors have different charges,,but when they are connected (as shown before the switch is ON) , won't the charges flow and redistribute so as to make the charges equal on both capacitors,,,,so that we can apply the series combination formula for these 2 charged capacitors,,(now that they have equal charges..) ...?

 

[sophiecentaur]

i agree that the 2 capacitors have different charges,,but when they are connected (as shown before the switch is ON) , won't the charges flow and redistribute so as to make the charges equal on both capacitors,,,,so that we can apply the series combination formula for these 2 charged capacitors,,(now that they have equal charges..) ...?

The end result must be that Q=CV must apply to all three Capacitors. Imagine that capacitors were uncharged and then connected across the 'supply' (Capacitor A). Charge will flow around the circuit until the PDs across each Capacitor add to zero. This is a modification of Kirchoff 2, in effect. Where the supply is not a fixed emf but follows V=Q/C.
The point i am making is that the simple series C calculation is not appropriate until it has all settled down and you will have to solve it for equilibrium and not when you switch on.
There is a sting in the tail here because you will usually arrive at a situation where Energy Conservation is not followed. Total stored Energy(initial) may not equal stored Energy (final). You have to allow some to be lost in a finite resistance, to be radiated or accept a constant oscillation due to the Inductance of the loop of circuit. Like the well known charge sharing situation when capacitors are connected ' in parallel' - sometimes mis-named as a paradox.

 

[NascentOxygen]

i agree that the 2 capacitors have different charges,,but when they are connected (as shown before the switch is ON) , won't the charges flow and redistribute so as to make the charges equal on both capacitors

No, charges cannot flow anywhere before the switch is closed. Whenever charge flows onto one plate of a capacitor, at that same moment and at the same rate, charge must flow from the other plate and into an external circuit, to close the loop. But with the switch OPEN there is no path out so charge will not flow into either capacitor.

This property makes it seem that transient current does flow through a capacitor, because we see a current flowing into one lead of the capacitor while an exactly equal current flows out of the other lead!

so that we can apply the series combination formula for these 2 charged capacitors,,(now that they have equal charges..) ...?

The two capacitors here start off with unequal charges, and because the two are connected in series, they will always retain that charge difference, there being no way here that one of the capacitors could be given more charge (or have some removed) without the other capacitor experiencing an identical change in its charge.

Hint: when the uncharged capacitor is switched in and accepts a charge Q, each of the other two capacitors loses an identical amount of charge Q, the result of a brief transient current flowing around the closed loop.

 

[K Dhiraj Bhak]

...OK , got it, thanks a lot people.