Can a charged capacitor act like a smaller EMF battery in a circuit?

[clavin]

the circuit is as shown.

http://img186.imageshack.us/img186/6627/circuitg.jpg

now i have to find the current from the battery. now the book says since the capacitor is charged its branch will no more have any current and we can neglect that branch. now my qs is. after the capacitor is charged won't it work like a smaller emf battery and try to oppose the current of the main source in other branches(even if for a small time since the capacitor would discharge.) the reason i am asking this is that i am trying to take analogy of a charged capacitor with a battery(keeping the discharging point apart). also is it possible to find the voltage across that capacitor.

 

[rl.bhat]

Hi clavin, welcome to PF.
In battery, there is no stored charge. It acts like a pump, which circulates the charge in the circuit. In the capacitor charge is stored. During discharge through a resistance the current will flow.

 

[tiny-tim]

Welcome to PF!

Hi clavin ! Welcome to PF!

Yes, Kirchhoff's rules still work even when the currrent is zero …

in this case, using Kirchhoff on the outer circuit, we find the current is 1/4 A.

On the "lower" circuit, we find total potential drop across the two resistors is 6*1/4 V + 4*0 V = 1.5 V, which must be the "emf" across the capacitor.

On the "upper" circuit, we find total potential drop across the two resistors is 2*1/4 V + 4*0 V = 0.5 V, which must be the "emf" across the capacitor and battery combined, again making it 2 V - 0.5V = 1.5 V across the capacitor.

 

[clavin]

thank u rlbhatt
and well thnx for the solution tim
but i don't exactly want the solution
i want to know why you have taken the assumption that the current in middle branch is zero
or why a charged capacitor works as a broken wire
and lastly if i replace the capacitor with a similar emf battery
i.e. 1.5 v will the current in all the resistors still be the same(in my case i am getting the current thru outer circuit to be 11/42amp)

 

[tiny-tim]

Hi clavin!

i want to know why you have taken the assumption that the current in middle branch is zero or why a charged capacitor works as a broken wire

Because that's the definition of a capacitor being fully charged …

"fully charged" means that no current is flowing.

and lastly if i replace the capacitor with a similar emf battery
i.e. 1.5 v will the current in all the resistors still be the same(in my case i am getting the current thru outer circuit to be 11/42amp)

Yes, the currents will all be the same, including the zero current along the middle.

(how did you get 11/42 ? )

 

[clavin]

ok maybe i am getting it a bit
i know now this is getting annoying but can u answer one more circuit problem
lets say i have a 5 v battery and a 3 micro farad capacitor and a resiter 2 ohm all in parallel to each other
now i believe i have to think as if no current passes thru the 3 micro farad capacitor in steady state
so the current across the 2 ohm resistor is 2.5 amp
right?
now i change the capacitor with a battery of 5 v now the current thru the 2.5 ohm resistor is 0. right?
well first i wud like to know if i am right
and if i am then y in this case does it yield different results?
since in ur last post u said that the current will be the same i am believing that i am thinking in the right direction
but its the above example which is confusing me
thnx

 

[rl.bhat]

now i change the capacitor with a battery of 5 v now the current through the 2.5 ohm resistor is 0. right?
From where did get 2.5 ohm resistor? If you mean it is 2 ohm, the current through it is 2.5 A. not zero.
While charging the capacitor, the charges never cross the plates of the capacitor. Due to the applied emf, electrons are stripped from one plate and transferred to the other plate. This process continues until the potential difference across the capacitor is equal to the applied emf. If you connect a resistance in series with the capacitor, time taken to reach the maximum voltage will be more. Rate of flow of charge dq/dt can be treated as the current flowing towards the plates of the capacitor. During the process of charging the current through the resistance in parallel to the capacitor will increase gradually and reaches maximum when the capacitor is fully charged.

 

[tiny-tim]

Hi clavin!

… now i change the capacitor with a battery of 5 v now the current thru the 2.5 ohm resistor is 0. right?

Depends which way round this battery is.

(I'm assuming you mean a circuit shaped like your original diagram. You must also have additional resistors in series with both the capacitor and the battery, otherwise the current will be infinite. I'll assume those additional resistors have negligible but non-zero resistance. )

If one battery faces to the left, and the other to the right, then Kirchhoff's rules tell us that the current around the whole "B" will be zero.

But if both batteries face to the right, then it's like having two batteries in parallel in a piece of electrical equipment … the voltage is the same as with one battery, the current is the same, and the only difference is that the batteries will last twice as long.

It's like having a turbine operated by water from two pipes leading from two reservoirs at the same height … all that matters is the pressure (the "head"), which corresponds to the voltage, and it's the same pressure whether there's one reservoir or two.