Kirchoff's laws and capacitance

[Praveen1901]

Homework Statement

When steady state is reached by keeping switch in position 1, then switch 'S' is transferred to position 2, then:
(A) Charge on 3F capacitor is 30C after S is transferred from position 1 to 2.
(B)Charge on 3F capacitor is 650/19 C after S is transferred from 1 to 2.
(C) Work done by battery 380 J after switch S is transferred from 1 to 2.
(D) Work done by battery is 80 J after S is transferred from 1 to 2.

Relevant Equations

Q=CV

Equivalent capacitance before and after remains the same.

Now the 10F capacitor (which was initially connected in parallel with 20F) would have 30 C charge. Hence an additional 20C must have been supplied to it. The only path which may supply the charge is through battery. However this leads to 6F capacitor having an additional charge of 20C as well. But in steady state the charge on it should have been the same as equivalent capacitance remained the same before and after

I'm unable to apply kirchhoffs law correctly here.

 

[BvU]

The only path which may supply the charge is through battery

Is only 'somewhat correct': capacitors can exchange charge to achieve a new 'equilibrium', i.e. voltage.
Look at the charges on the top right 10 and 20 F capacitors when S is opened.

 

[DaveE]

Sorry, I found the problem statement too confusing to follow your reasoning. I'm not saying it is wrong, I'm saying I'm too lazy to fill in all of the gaps in your description. Also I think there are typos in your table, everything is after 1 → 2.

But in steady state the charge on it should have been the same as equivalent capacitance remained the same before and after

This doesn't make sense to me. There are 2 steady states, 1 before, 1 after, so that confuses me. Also, the reasoning that since the capacitance didn't change the charge shouldn't change baffles me.

I know I'm not helping, but you will get better answer from people if you ask your questions more carefully. I may be willing to help solve problems, but I am not very motivated to work on the puzzle "what was the question?". Give us all of the information, for example.

Finally, any time you have a transient problem with capacitors in parallel (or inductors in series) you are treading on thin ice, physics-wise. There are some inherent paradoxes since your lumped element model isn't physically realizable, parasitics can not be ignored. For example: what is the peak current when the switch is closed? It's infinite, that's not realistic. Avoid any questions about what happens when the switch changes, only deal with before and after. For this circuit, stick with conservation of charge, stay away from energy conservation, it will confuse you.

I am not saying this problem isn't good for some instructional purposes, depending on exactly how it is presented, but be careful.

 

[DaveE]

A more clear example of the parallel capacitor paradox may help, or it may make you more confused, yet a bit smarter anyway:

I'm too lazy to draw a schematic, so I'll describe it. Given a single loop circuit with three elements all in series; two are capacitors (1 F each), and a perfect switch. There are no other components, like resistors.

- Initially the switch is open, C1 has 2 V across it, C2 has 0 V across it.
- Initially, C1 has 2 C of charge (Q=CV) and 2J of energy (E=CV²/2). C2 has 0 C and 0 J.
- After the switch has closed and the charge has been redistributed (conservation of charge), each capacitor will have 1 V, 1 C, and 0.5 J of energy. The total charge in the circuit remains the same at 2 C, however, the total stored energy has decreased from 2 J to 1 J.

You could also work this problem assuming energy is conserved and show that charge isn't.

Also, don't assume the problem lies in the switch, it is perfect. I could have describe a transient problem with no switch and the result would be the same.

Ultimately the answer is that the question wasn't realistic. We ignored the processes at work when charge is moved, which is a much more advanced subject.