Compare the two cases in terms of energy dissipated while charging a capacitor...?

Question

First Case:

A capacitor (capacitance C) is fully charged using a 15 volt battery. Let the energy dissipated in the process be E1.

Second Case:

A capacitor (capacitance C) is connected to a 15 V battery but some how allowed to charge only till 5 V and then disconnected. Immediately it is reconnected and charged to 10 V from previous 5 V. The process is redone (disconnected and reconnected to charge it to 15 V from 10 V.) Let the total Energy dissipated in this case be E2.

Question:

Compare E1 and E2 (as E1= nE2.) The connecting wires are ideal and so are the capacitors and the batteries.

Please explain, in detail, how are the two cases different (if they are.)

Apart from mathematical equations, logical conclusions will also be welcomed.

Thankyou.

Answer 1

Mike has a point. It's a BS question. All you can say is that E1 = E2, whatever value that may be. With ideal everything (no resistance anywhere) it would all be over in zero time and the current would be a mathematical impulse function of infinite amplitude. There would be zero dissipation in both cases, thus E1 = E2 = 0.

However, we have to take them as they come, which in this case means assuming nonzero times (so the charging interval can "some how" be broken up), finite current, dissipation, etc. During the infinitesimal disconnected periods, nothing happens except for someone's feverish activity opening and closing a circuit. Time doesn't pass, there's no change in Vb, Vc or Q. Thus the charging curves of the second case are just sections of the charging curve of the first case, meaning no difference in energy loss. E1 = E2 again.

Answer 2

hi..

I hope u r aware of the charging/discharging eqns for a capacitor..

charging..

q=q(max)[1-e^-t/T]

when t approarches infinity..the capacitor is charged completely...

In case 1:

The capacitor is charged completely...so q=15C(using Q=CV)

E1=1/2*Q*V=1/2*15C*15=225/2*C^2;

In case 2:

Capacitor is connected to the 15V battery..but when the potential difference between the two plates is 5V capacitor is disconnected...

so,Q=C*V=5C,

E21=25/2*C^2,

Now it is reconnected...

since,while calculating charge on a capacitor we calculate the potential difference between the plates...

V=10V,in this case,

E22=100/2*C^2,

& similarly..

E23=225/2*C^2,

E2=E21+E22+E23;

E2=350/2*C^2,

so I hope this solves your query...

actually..

in the second case...you have to consider the potential difference between the plates...

so most of the people will take 'V' as (10-5)V..which I presume is wrong...

If you think Iam wrong...plz contact me..

cheers:)

Answer 3

First logical approach: The two cases are different because one takes slightly longer than the other.  The reason it takes longer is that there are gaps when no current flows.  The same amount of charge is transferred , the losses are 0 in each case ( no energy dissipated in battery, no time dependent charge recovery process in battery) and the starting and end conditions are identical  so E1=E2 and it all takes more time. - but they say "immediately it is reconnected" so we know that this is neither logical nor even faintly connected to reality - it's just another badly expressed physics problem designed to make you think about something, possibly associated with short current spikes.  What could it be? Do they want you to think about self-inductance in 'ideal wiring' ? em radiation from the set up?  Talk about it in a little mini-essay and show that you have spent some time trying to figure out what they might be trying to get at.

There is a different logical aproach which starts from the premise that everything is ideal and in an ideal world any answer you give would be right but I doubt that you will get away with that one.