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Kirchhoff loop rule and Ohm's Law

  1. Feb 11, 2008 #1
    1. The problem statement, all variables and given/known data

    To understand the behavior of the current and voltage in a simple R-C circuit

    A capacitor with capacitance C is initially charged with charge q0. At time t=0 a resistor with resistance R is connected across the capacitor

    PART A
    Use the Kirchhoff loop rule and Ohm's law to express the voltage across the capacitor V(t) in terms of the current I(t) flowing through the circuit

    PART B

    We would like to use the relation V(t) = I(t)R to find the voltage and current in the circuit as functions of time. To do so, we use the fact that current can be expressed in terms of the voltage. This will produce a differential equation relating the voltage V(t) to its derivative. Rewrite the right-hand side of this relation, replacing I(t) with an expression involving the time derivative of the voltage.

    [​IMG]

    2. Relevant equations



    3. The attempt at a solution

    For PART A I got the answer to be V(t) = I(t)R, but I'm having problems with the PART B. I'm not sure what the "time derivative of the voltage" is. Any suggestions?
     
  2. jcsd
  3. Feb 11, 2008 #2
    Remember that C=Q/V, and dQ/dt=I(t)

    I need to stop here if the problem was to actually FIND the differential equation, because any more and I'd pretty much finish it >_> Take a stab keeping in mind what I typed
     
  4. Feb 11, 2008 #3
    The only thing that comes to mind is:

    [tex]V(t) = \frac{dQ}{dt}R[/tex]

    but I don't think that's the right answer.
     
  5. Feb 11, 2008 #4
    Ok more prodding
    Q=C*V, take the derivative of both sides, dQ/dt=C*dV/dt (because C is constant)

    oh look an expression for dQ/dt that involves the time derivative of V....
     
  6. Feb 11, 2008 #5
    Power wouldn't happen to be the same as dV(t)/dt?
     
  7. Feb 11, 2008 #6
    No

    All part B wants is V(t)=some stuff involving dV/dt

    right now you have V(t)=some stuff involving I(t)

    You have seen how dQ/dt = somet stuff involving dV/dt

    dQ/dt = ? *pokes you with cattle prod*
     
  8. Feb 11, 2008 #7
    dQ/dt=C*dV/dt

    therefore dV/dt = (dQ/dt)/C

    I(t) = dQ/dt = V(t)/R

    dV/dt = (V(t)/R)/C

    C(dV/dt) = V(t)/R

    V(t) = RC(dV/dt)

    If this is it, I'm going to kick myself....
     
  9. Feb 11, 2008 #8
    That's it

    You could've also saved some steps and algebra and just gone from I=C*dV/dt, and plugged that into V=IR and gotten V=RC*dV/dt

    Someday soon you might do a similar problem with LC circuits, and even take it a step further to RLC circuits, just don't forget the basic procedure you used here, just different things will be derivatives and different things will be constants
     
  10. Feb 11, 2008 #9
    For some reason the correct answer is V(t) = -RC(dV/dt)
     
  11. Feb 11, 2008 #10
    Yah that negative slips in there somewhere because the capacitor is discharging

    So charge is flowing out of the capacitor so...-dQ/dt=C*dV/dt?
     
  12. Feb 11, 2008 #11
    So another question unlocked itself after you finish this one:

    Now solve the differential equation V(t) = -RC(dV/dt) for the initial conditions given in the problem introduction to find the voltage as a function of time for any time t.

    Express your answer in terms of q, R, C, and t.

    so I know:

    V(t) = -RC(dV/dt) and dv/dt = (dQ/dt)/C = I(t)/C

    V(t) = -RC(I(t)/C)
    V(t) = -R(I(t)) and I know Q = I(t)*t => I(t) = q/t
    V(t) = -R(q/t), but I can't seem to get the C in there.

    Any suggestions?
     
  13. Feb 12, 2008 #12
    I'm not sure where you were going starting with "so I know", it's separation of variables

    So you have V=-RC*dV/dt, separate them to get dV/V=-RC*dt

    Remember RC is a constant(it's got a special name that I forget too)integrate both sides

    You'll have a natural log on the left, and just -RCt + C on the right, use your initial conditions to find C, then exponentiate both sides(raise both sides as powers of e)remembering that e^(ln(anything))=anything
     
    Last edited: Feb 12, 2008
  14. Feb 12, 2008 #13
    Bit rusty on my integration (havent dont it in about a year or so)

    so the integral of v/dv would be ln|v+C|.

    and the "+C" is just a constant.
     
    Last edited: Feb 12, 2008
  15. Feb 12, 2008 #14
    You'd usually just stick the constant on the other side

    I messed up rearranging the equation fyi, it's dV/V=-dt/RC, but 1/RC is still a constant so it's not problem. The absolute value on the natural log is to avoid a domain error, but you usually just ignore it since you're gonna get rid of the ln anyways, so you end up with ln(V)=-t/RC+C, and it's not "just" a constant, you have initial conditions you have to utilize to determine it, basically at t=0 V=such and such, so you can solve for C
     
  16. Feb 12, 2008 #15
    At time t = 0, V would have to be 0 since voltage is a measure of the pressure under which electricity flows, and at time t = 0, there isnt any flow of electricity. There is also no current at time t = 0, so I = 0 as well.

    so C = q

    but that's also wrong. *sigh*
     
    Last edited: Feb 12, 2008
  17. Feb 12, 2008 #16
    There's still a V at t=0

    if there wasn't then electricity would never flow
     
  18. Feb 12, 2008 #17
    so if there's voltage and charge, then C = VQ

    I apologize if it's wrong, but I'm next to useless when it comes to Circuits
     
  19. Feb 12, 2008 #18
    Right, at t=0 the V=C/Qo

    that's the initial voltage across the capacitor and hence the circuit
     
  20. Feb 12, 2008 #19
    I submitted e^((-t/RC)+V(t)Q), but it says the answer does not depend on the variable v

    I think it's e^((-t/RC)+tQ) instead
     
  21. Feb 12, 2008 #20
    You didn't find c right(lower case c=the constant from integration, upper case is capacitance)

    You plug in t=0, then plug in C/Qo for V, and solve for c
     
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