Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have a conceptual question.

The circuit is shown below. Let us assume that the capacitor shown in the figure is of 2.2 micro-farads having rated level of 5 volts and we decide to charge it up to 2 volts so that we don't smoke it.

Circuit - enter link description here

If we connect $R_1$ and $R_2$ in the circuit for charging and discharging respectively, how will the circuit work?

The red loop shows the charging circuit and the green loop shows the discharging circuit.

The circuit is as shown until the capacitor is charged and then the switch S is closed.

Will the circuit discharge instantaneously?

If yes, when will the capacitor be re-charged again? Will be it charged after the long exponential charge decay of the capacitor or will it start to charge itself after it has a certain amount of charge left?

Will this process of charging and discharging be continuous if the switch S remains in closed position?

Let me start with some basics. The red loop circuit will charge with $R_1\times C$ time constant. So the charging will be quick because of low $R_1$. It will be 22 microseconds.

After the capacitor has been charged to 2V (max given by power supply), we close switch S and then the discharge process will start. As expected it will also be quick and similar to charging time constant.

Will this capacitor charge again? If yes, when? And can it be controlled?

share|improve this question
2  
I think this should be migrated to electronics.stackexchange.com –  daaxix Jan 10 '13 at 5:26
add comment

1 Answer

For starters (in case you overlooked it), note that the capacitor will not discharge completely: The stationary state for the closed circuit has a continuous current flowing through the corrent branches with the resistors, so the point A will have a finite, non zero $V_A$ and the capacitor will experience a finite voltage. The higher the value of $R_1$ is compared to the value of $R_2$, the more the capacitor will discharge.

The capacitor can be recharged, but it will not happen spontaneously if the switch keeps closed. You would have to open it again so that the capacitor may recharge again.

I hope this explanation has been useful, I'm trying to stay conceptual, as the OP requested.

share|improve this answer
    
Thanks, I made a circuit just like above and I am going to plug the input and output to a DAQ to see what is going on here. I will discuss my findings later tomorrow. I have a circuit where the switch opens and closes for 10's of microseconds. –  xsystem1 Jan 10 '13 at 1:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.