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!

Battery and capacitor in series

  1. Jul 24, 2013 #1
    In 1. (see attached), a is attached to ground, so has a voltage of 0 V. The voltage at b is -1 V, so the voltage across the capacitor (Vb-Va) is -1 V.

    My understanding is so rusty... What happens when a is now wired to the positive end of a 1.5 V battery, and the negative end of the battery is wired to the ground, as in 2? Specifically, what are the new voltages at a and b, and what is the new voltage drop across the capacitor? Is it still -1 V, or has it changed?
     

    Attached Files:

  2. jcsd
  3. Jul 25, 2013 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The voltage (potential) is referred to the ground, where the potential is taken zero. The voltage (potential) difference is 1.5 V across the terminals of the battery. One terminal is connected to the ground. What is the voltage (electric potential) of the other terminal connected to a?
    The voltage (potential) of point b is kept -1 V with respect to the ground. Does it change if something is connected to an other point ?

    ehild
     
  4. Jul 25, 2013 #3
    Hi ehild, thanks so much for your help. I'm embarrassed how rusty I am at this!

    In the second configuration, a is now at 1.5 V (relative to ground). Doesn't that mean that b is now 1.5-1 = 0.5 V relative to ground? So the voltage difference across the capacitor is still -1 V (b-a)? Is it correct to say that changing configuration from 1. to 2. had no effect on the voltage across the capacitor?
     
  5. Jul 25, 2013 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    NO. B is still -1V with respect to the ground:Vb=-1 V. You would be right if the potential difference across the capacitor would be unchanged. But that is the question. The potential of b is unchanged, and the potential difference across the capacitor is Vb-Va.

    ehild
     
  6. Jul 25, 2013 #5
    Hi ehild - sorry to be pedantic, but I just want to make sure I have this right.

    The voltage at b does not change. Vb=-1V in both situations.
    The voltage at a does change. Va=0v in situation 1., and Va=1.5V in situation 2.
    Therefore the voltage across the capacitor, Vb-Va, changes. Vb-Va=-1V in situation 1. to Vb-Va=-2.5V in situation 2.

    Is this correct? Many thanks.
     
  7. Jul 25, 2013 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is correct.

    ehild
     
  8. Jul 25, 2013 #7
    There's something I don't understand...

    If the potential across the capacitor changes, then the charge on the capacitor would have to change. But b is floating, so there's no source of charge for b (no way to get current to it). So how does this work? Specifically, how does the charge on the b side of the capacitor change?

    Again many thanks.
     
  9. Jul 25, 2013 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The text of the problem is not clear. You can understand it in the way that b is floating, (it was not stated in the OP), in that case the charge of the capacitor and the potential difference across the capacitor remain the same, and the potential of b increases by 1.5 V, so it becomes 0.5 V.

    The other possibility is that the potential of b originates from an external source, for example, it is connected to the negative pole of a battery, with the other pole connected to the ground. That makes a closed circuit, current can flow to recharge the capacitor causing a different voltage across it.

    I understood the problem in the second way. Did you copy the problem text exactly as it was written?


    ehild
     
  10. Jul 25, 2013 #9
    Hi - sorry for the confusion! b is indeed floating. I (think) I set this in the circuit diagram, but I didn't explicitly state it in the question. The question is worded by me, so all confusion is my fault!

    I understand what happens now. Thanks very much - super useful!
     
  11. Jul 25, 2013 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    If you said that a charged capacitor is connected to point a it would have been more clear. Anyway, I am pleased that you understood the concept of potential and potential difference.


    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Battery and capacitor in series
  1. Battery in series (Replies: 1)

Loading...