Lets say I have 2 capacitors P and Q connected to a 9V supply. Across P there's a resistor connected in parallel with the switch open (off position). When I turn on the battery they both fully charge. When I switch on the battery but also simultaneously switch on the switch to the resistor across P, does P discharge (even though its connected to both the power supply and the resistor) ?

From a question, the answer is that it does discharge. Why ? Does the resistor take priority over the power supply ? Or is it just that the question forgot to mention that the power supply is then turned off ?

My question more clearly is this : when a charged capacitor is connected in parallel to both a power supply and a resistor does the capacitor discharge ?

Diagram for reference:

enter image description here

  • $\begingroup$ You should provide a diagram because your circuit is unclear sounds like you are charging capacitors without a resistor in series with the battery, which is a no no $\endgroup$
    – Bob D
    Mar 21 at 15:42
  • $\begingroup$ ok will do but why cant the resistor not be in series with the battery? $\endgroup$ Mar 21 at 15:50
  • $\begingroup$ In real life the capacitor can blow up $\endgroup$
    – Bob D
    Mar 21 at 15:52
  • $\begingroup$ wait so in the diagram i just attached the resistor is ok? $\endgroup$ Mar 21 at 15:55
  • $\begingroup$ Charging should take place with the switch initially in position Y so that the resistor can limit the charging current. In the X position the the only thing limiting the initial charging current would be the internal resistance of the battery and the connecting wires. Too high a current can damage capacitors $\endgroup$
    – Bob D
    Mar 21 at 16:02

2 Answers 2


If you have the switch set to the Y position you will get an initial current through the resistor, but this will decay exponentially with time and given enough time the current in the resistor will be zero.

If the current in the resistor is zero then the voltage across the resistor is also zero, and since the capacitor P is in parallel with the resistor the voltage across the capacitor P must also be zero.

And if the voltage across the lower capacitor is zero it is discharged. So the answer to your question is that yes with the switch set to Y the capacitor P will be discharged.

  • $\begingroup$ wait but why does the resistor decay exponentially with time?.. am i missing something? is it always the case that when a resistor is parallel to the capacitor even though theres a power supply, this happens? $\endgroup$ Mar 22 at 13:40
  • $\begingroup$ @Safayousif the voltage across the resistor would change in quite a complicated way immediately after the power supply was turned on, but at longer times it would decay exponentially to zero. $\endgroup$ Mar 22 at 16:06

Think about removing the capacitor P from your circuit. You have just capacitor Q and resistor R in series (with switch closed). What happens?
As the capacitor charges up the current falls, the voltage across the resistor drops, eventually to zero. Capacitor Q has 9v across it.
Now connect capacitor P in parallel with R, what happens? Nothing. The voltage across R is zero so the voltage across P is zero.
This is what happens long term if you begin with the setup as you have it.

If you begin with P and Q both charged, there is no current. Then close the switch to the resistor, the voltage across P will fall as current flows through R. The voltage across Q will gradually increase to 9V.
Remember that in a DC circuit current only flows through a capacitor as it's charging. Once it is charged to power supply voltage there's no more current flow.

  • $\begingroup$ okay i think that cleared up the concept for me thank you. i didnt realise current was zero with a charged capacitor $\endgroup$ Mar 23 at 7:49
  • 1
    $\begingroup$ @Safayousif That is what one means by "charged capacitor". Note that this doesn't mean the capacitor can't physically have more charge; it just means that no more charge will be added for the applied potential. If no more charge is being added, then of course there is no more current $\endgroup$ Mar 25 at 16:28

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