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Differential equation for an LCR circuit

  1. Oct 9, 2014 #1
    1. The problem statement, all variables and given/known data
    LCR series circuit.PNG
    Find the differential equation that Q(t) satisfies.

    2. Relevent equations

    Kirchoffs loop law and voltage across a capacitor, resistor and inductor.

    3. The attempt at a solution

    So I'm thinking, by Kirchoffs voltage rule, that the sum of the voltages in this circuit must add to zero. So the voltage across the capacitor will be equal to the sum of the voltage drops of the other two elements:

    [itex] \frac{Q}{C} = \frac{dQ}{dt}R + \frac{d^2Q}{dt^2}L [/itex]

    A friend of mine, who is always right, tells me that the differential equation should be;

    [itex] \frac{Q}{C} + \frac{dQ}{dt}R + \frac{d^2Q}{dt^2}L = 0 [/itex]

    I know my DE is wrong, but why is mine wrong, and his right?

    Thank you.
     
    Last edited: Oct 9, 2014
  2. jcsd
  3. Oct 9, 2014 #2

    Mark44

    Staff: Mentor

    Note: Fixed the LaTeX in your two equations. Each one was missing a }.
    From what you wrote, "the sum of the voltages in this circuit must add to zero."
    Your equation can be rewritten as
    ##\frac{Q}{C} - \frac{dQ}{dt}R - \frac{d^2Q}{dt^2}L = 0##, so you are subtracting the voltages across the resistor and coil from the voltage across the capacitor.
     
  4. Oct 9, 2014 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You are right, that the potential drop across the resistor is RI and across the inductor is LdI/dt. But I = - dQ/dt as Q decreases as the capacitor is discharged.

    ehild
     
  5. Oct 9, 2014 #4
    Thank you Mark44. Okay that makes sense, but then in this question why can we write;
    ## \varepsilon = \frac{Q}{C} + R\frac{dQ}{dt} ## but we cannot write ## \varepsilon + \frac{Q}{C} + R\frac{dQ}{dt} = 0 ## ?
    blblblblblblbl.PNG
    blablablalbbal.PNG
     
  6. Oct 9, 2014 #5
    Ahhhhh, I see. Thank you ehild. However, as naive as my above question may be, i would still appreciate an answer (whenever anyone is ready).
     
  7. Oct 9, 2014 #6

    Mark44

    Staff: Mentor

    ##\varepsilon## in the first equation will have the opposite sign of the same symbol in the second equation.

    For a much simpler example, 5 = 3 + 2, but it's not true that 5 + 3 + 2 = 0. However, -5 + 3 + 2 = 0 is true.
     
  8. Oct 9, 2014 #7
    OK, but why is the sign the opposite in the 2nd equation if by kirchoffs law, the sum of the voltages is zero? I know these are all quite stupid questions, i would just like to build a strong intuitiveness for this sort of stuff.
     
  9. Oct 9, 2014 #8

    Mark44

    Staff: Mentor

    If you add three numbers together, and get zero, then at least one of the numbers has to be negative. Is that what you're struggling with?
     
  10. Oct 9, 2014 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Instead of "voltages" work with potentials. Still discussing the first problem, assign zero potential to the bottom wire and assuming that the upper plate of the charged capacitor is positive, follow the current round the circuit and determine the potential at the nodes.

    So you have Q/C potential at the upper plate of the capacitor. The current flows clockwise. The potential drops by IR across the resistor, so it is Q/C-IR between the resistor and inductor. It drops again, by LdI/dt, on the inductor, and you arrived back to zero potential. Q/C-IR-LdI/dt = 0. And I=-dQ/dt, as the current flows away from the capacitor, decreasing the charge.

    In the second problem, let be the negative terminal of the battery at zero potential. The upper terminal is positive, and its potential is E. The left plate of the capacitor is positive, the right plate is negative, the potential drops by Q/C from left to right across the capacitor.So the potential is E-Q/C on the right plate. Then the current flows through the inductor and causes potential drop LdI/dt from right to left. At the left end of the inductor, the potential is E-Q/C-LdI/t and you are back at the negative terminal of the battery, where the potential is zero. So
    E-Q/C-LdI/dt=0.
    In that circuit, the charge is increased by the current, so I=dQ/dt.

    ehild
     
  11. Oct 10, 2014 #10
    Thanks alot ehild and mark 44, I understand it alot better now!
     
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