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Very simple RC-circuit

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data

    C = 1 uF
    E = 12 V
    Total R in circuit = 100 Ω
    Q = 0 before the breaker closes

    Problem A:
    Derive an expression for the voltage across the capacitor and the current through the circuit as a function of time after the breaker closes.

    2. Relevant equations

    This is my first problem. I do not know any relavent equations.

    3. The attempt at a solution

    I have trouble starting this exercise, but from what I have found on Wikipedia, the capacitor needs 5τ to fully charge, and 1 τ = R times C.

    From what I understand, the voltage across the capacitor will decrease until the capacitor is fully charged.
    But that comes in conflict with my understanding about that a capacitor will have a voltage across when fully charged (maybe only when there is no other voltage sources present?)

    Please kick me in the right direction here, so that I can get going :)
  2. jcsd
  3. Oct 15, 2011 #2


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    Staff: Mentor

    You've been looking at the discharge curve showing how the voltage across an initially charged capacitor decays down to zero as it discharges through a resistor.

    Capacitors can also charge up, gaining voltage, as current flows from a source through the resistor and onto the capacitor. This curve looks like the first one only flipped vertically.


    By the wording of the given problem they want you to derive the mathematical expression for the charging capacitor. It will involve writing and then solving a differential equation.

    Attached Files:

  4. Oct 15, 2011 #3
    Is the current curve exactly opposite as the voltage curve?
  5. Oct 15, 2011 #4


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    Staff: Mentor

    What do you mean by "exactly opposite"?

    The current is flowing in different directions (with respect to the capacitor) in each case. But the eventual steady state condition for both charging and discharging is zero current (capacitor all charged up --> no more current; capacitor fully discharged --> no more current).
  6. Oct 15, 2011 #5
    Hm, I need to play around with my calculator for a while, thanks for pointing things out.

    Be back shortly :)
  7. Oct 15, 2011 #6

    I am taking introductory Physics and need practice and help with circuits. Can I post my attempt at the solution?

    Thank you
  8. Oct 15, 2011 #7
    Yes, please do
  9. Oct 15, 2011 #8
    Part 2 of the exercise is to calcuate the charge of the capacitor when it's done charging.

    I found a few relevant equations for that:

    [tex]q_c = C \times u_c(t)[/tex]
    where q = time varying charge, C = capacitance and u = time varying voltage across capacitor

    [tex]u_c = E \times (1 - e^( \frac{-t}{\tau}) )[/tex]
    Where E = applied voltage (12V) and τ = time constant = RC

    So, I merged those to equations and set the time to 5τ as that is supposed to be the time until the capacitor is fully carged.

    This is how it looked like:
    [tex]u_c = C \times E \times (1 - \frac{-5\tau}{\tau})[/tex]

    [tex]u_c = 10^-6 F \times 12 V \times (1 - e^(\frac{-(5 \times 100Ω \times 10^(-6))}{100Ω \times 10^(-6)}))[/tex]
    [tex]= 0.0000119191 = 11.91 μC[/tex]
    But the answer should be 12.0
  10. Oct 15, 2011 #9


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    Staff: Mentor

    [itex]5\tau[/itex] is the approximate time when most of the "excitement" is over for a circuit with a time constant of [itex]\tau[/itex]. It's a handy rule of thumb, but it is only an approximation. The final value that the equation is heading for can be found by taking the limit as t → ∞.

    Another way to look at it is, when the capacitor is fully charged and the current goes to zero there can be no voltage drop across the resistor (Ohm's law -- no current no potential drop). To have no potential difference across the resistor, the potential on the capacitor must be the same as the potential of the source.
  11. Oct 15, 2011 #10
    ah, there's still hope then!

    Maybe the remaining 0.09 uC appears between 5t and eternety :)
  12. Oct 15, 2011 #11
    Potential across charged capacitor is Q/C.
    The current through charged capacitor is zero.

    Applying Kirchhoff's loop rule

    ε - q/C - IR = 0

    ε - q/C - (dq/dt) R = 0

    ( dq / dt ) R = ε - q/C

    ( dq / dt ) R = (εC - q) / C

    dq / (-εC + q) = -dt / ( RC )

    [itex]\int_0^q[/itex] dq / (-εC + q) = [itex]\int_0^t[/itex] -dt / ( RC )

    ln (-εC + q) = -t / (RC)

    by definition of the natural log:

    -εC + q = e^(-t / (RC) )

    q(t) = e^(-t / (RC) ) + εC

    I = -1/ (RC) e^ (-t / (RC) )

    ΔV (t) = q(t) / C

    In a circuit consisting of battery and capacitor, after fully charging, the voltage between the capacitor equals the voltage between the terminals of the battery. In a circuit consisting of a battery, capacitor and a resistor, after fully charging, is the voltage across the capacitor equal to the voltage across the terminals of the battery?
  13. Oct 15, 2011 #12
    wow JosephK.

    This is way beyond me. I lost you at line number 5 or something.
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