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!

Homework Help: Two capacitors (different capacitance and innitial charge) in a series circuit

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Two capacitors C1 and C2 and a resistor R are all connected in series. At t=0, a
    charge Q1 resides on C1 and a charge Q2 on C2 (Q1 > Q2; C1 < C2), and the positive plate
    on C2 is connected to the negative plate on C1.

    a) In a first experiment, the positive plate on C2 is connected to the negative plate on
    C1. Derive an expression for the time dependence of the current I(t) which flows
    in the circuit. Indicate the direction of this current on a circuit diagram, and
    include the sign of the charges on the plates on each of the capacitors at t=0.

    b) The same circuit is used for a second experiment, with the same initial charges,
    except that in this case the negative plate of C2 is connected to the negative plate
    on C1. All three components are otherwise still connected in series as before.
    Derive an expression for I(t), and indicate the direction of the current on a second
    diagram (again indicate the sign of initial charge on each capacitor on this

    c) Hence calculate the energy dissipated in the resistor in each case.

    2. Relevant equations
    V = IR

    V=Q/C for a capacitor

    3. The attempt at a solution

    ok, I'm stuck on the first part. firstly, the voltage gain across the two capacitors is:

    [itex]V = \frac{Q_{1}(t)}{C_{1}} + \frac{Q_{2}(t)}{C_{2}}[/itex]

    And since V = IR, we have:

    [itex] I = \frac{1}{R}(\frac{Q_{1}(t)}{C_{1}} + \frac{Q_{2}(t)}{C_{2}})[/itex]

    and the capacitors must each lose the same amount of charge per unit time:

    [itex]\frac{dQ_{1}(t)}{dt} = \frac{dQ_{2}(t)}{dt} = -I [/itex]

    the current is negative since the charges are decreasing. So, we have:

    [itex]\frac{dQ_{1}(t)}{dt} = \frac{dQ_{2}(t)}{dt} = \frac{-1}{R}(\frac{Q_{1}(t)}{C_{1}} + \frac{Q_{2}(t)}{c_{2}})[/itex]

    And this is where I'm stuck. I'm not entirely sure how to go about evaluating this for Q1 and Q2.
    Last edited: Sep 28, 2011
  2. jcsd
  3. Sep 28, 2011 #2


    User Avatar

    Staff: Mentor

    You have the initial potential that the resistor will see: the sum of the two capacitor voltages as you've shown. What initial current must flow then?

    The circuit consists of a resistor and two capacitors in series. What's the equivalent capacitance? How about the circuit's time constant?

    What can you do with an initial current and the time constant?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook