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 –
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?
Best 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.