1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Discharging a capacitor through a voltmeter

  1. Nov 22, 2009 #1
    so here are my questions:

    Does a single capacitor take longer to discharge than two capacitor is series? Why? Would the voltage also be lower when discharging in series?
    In discharging a capacitor, if a resistor is not a part of the circuit, does it discharge through the voltmeter's resistance? <<<please explain this...i'm kinda confused. Would it affect the rate at which it discharges?:confused:
    Last edited: Nov 22, 2009
  2. jcsd
  3. Nov 22, 2009 #2


    Staff: Mentor

    Do you recall the formula for the time constant (discharging time) of a RC circuit? How does the time change as you change capacitance and resistance?
  4. Nov 22, 2009 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Look at the expression for caps connected in series? What does it tell you?

    Voltmeters have a resistance which is usually specified as ohms/volt. That means it depends upon the scale used. A capacitor cannot know what is providing the load, only that there is a path of conduction between the plates, so the resistance of a voltmeter is the same as any other resistance connected between the plates.
  5. Nov 22, 2009 #4


    User Avatar
    Gold Member

    Hi and welcome to PF!
    I'll give a little try.
    It depends on their capacitance. Say you have a capacitor whose capacitance is C. It would discharge at about 63% after a time equal to RC where R is the resistor of the circuit. After a time equal to 5RC we usually consider the capacitor to be discharged, although it is not totally discharged (around 99.3%).
    Now you put another capacitor with a capacitance worth C in series with the first capacitor. Their capacitance won't add up, instead [tex]\frac{1}{C'}=\frac{1}{C}+\frac{1}{C}=\frac{2}{C}\Rightarrow C'=\frac{C}{2}[/tex]. Hence the capacitance of the circuit is halved by adding another capacitor to the circuit, if it has the same capacitance than the first capacitor.
    Use the formula [tex]Q=CV[/tex] to check it out yourself.
    What is the resistance of an ideal voltmeter?
  6. Nov 22, 2009 #5
    That depends on the capacitance of all of these capacitors.
    You can have two capacitors in series that exactly equal the capacitance of another capacitor. Depending on their capacitance.
    Depending on the capacitance is how fast the capacitor will discharge. The smaller the capacitance, the less force there is holding all of the charge together, meaning it will discharge faster.

    That depends on the voltage of the individual capacitors that you happened to put in series. To figure out the total voltage, just add them all up, as if they were batteries.


    The rate at which a capacitor discharges is proportional to the capacitance * resistance.
    Higher resistance = Longer discharge rate.
    Higher capacitance = Longer discharge rate.

    The reason for the longer discharge rate with the resistor is that with a higher resistance the charge that was accumulated in the capacitor cannot come out of the capacitor as quickly.

    Hope that answers your questions.
  7. Nov 22, 2009 #6
    Thank you very much...you've all been so helpful. cleared up alot for me. Thanks
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook