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: Differential Difficulties in an RL Circuit Problem

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data

    An inductor with L = 50 mH is in series with a resistor of R = 180 ohms. At t = 0, a potential difference of 45 V is suddenly applied across the series circuit. At what rate is the current increasing after 1.2 milliseconds?

    2. Relevant equations

    V = IR + Emf
    Emf = -L(dI/dt)
    I = dQ/dt

    3. The attempt at a solution

    My first instinct was to try and find the maximum current, or when the rate of change of the current is 0, but I'm not sure that doing that would accomplish anything. I've been puzzling over how to relate the concepts with differential equations, but I've gotten pretty stuck there, too. Any points in the right direction would be greatly appreciated! Thank you!
  2. jcsd
  3. Dec 9, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    First of all, you need to construct the DE which describes this circuit and solve it to obtain I(t). Once you do that, then you can calculate dI/dt for t = 1.2 ms.
  4. Dec 9, 2013 #3
    Okay! But I'm kind of lost there, too - what information do I need to construct the differential equation?
  5. Dec 9, 2013 #4
    KCL would be a good start.
  6. Dec 9, 2013 #5
    I don't know anything about KCL...

    After doing a little research, though, it looks like maybe Q(t) = L*V(1-e^(tR/L)): if this is true, then it's just a matter of derivatives. I'm not sure it's right, though (I cobbled it together from Wikipedia and Hyperphysics) - would knowing the KCL help?

    (the formula comes from a similar one which I learned for RC circuits, in which Q(t) = C*V(1-e^(t/RC)), so I'm not so sure that it's as simple as switching the time constant)
  7. Dec 9, 2013 #6
    I don't remember the solutions, its quite easy to find them. I meant Kirchoff's laws, you should have heard of them.
  8. Dec 9, 2013 #7
    Ah! Yes, okay - I think that's already been essentially completed. The circuit is a single loop in series, so V = IR + Emf, like I put above.
  9. Dec 9, 2013 #8
    Yes. :)

    So what is the D.E (differential equation) you get?
  10. Dec 9, 2013 #9
    V = R(dQ/dt) - L(dI/dt)

    The problem is I don't know the equation describing Q - I've found a candidate, but I'm not terribly confident about it.
  11. Dec 9, 2013 #10
    Why minus? :confused:
  12. Dec 9, 2013 #11
    The back Emf opposes the direction of current flow, so it's negative until the current starts switching directions, is my understanding
  13. Dec 9, 2013 #12
    Well, I am not sure how to explain it as I am a student myself, not an expert. :)

    Yes, the back emf opposes the direction of current flow. As you state, it's negative which would mean,


    Last edited: Dec 9, 2013
  14. Dec 9, 2013 #13


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    What is Q supposed to represent in the circuit? The charge of what? The answer is there is no element in the circuit with a charge Q, so it doesn't make sense to express any quantity in terms of this ill-defined variable.

    The current I is apparently the current through the inductor. How is that related to the current through the resistor?
  15. Dec 9, 2013 #14


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    From a problem-solving standpoint, you should follow this convention: Assume a direction of current flow for each element (R, L, and C). Label the end where the current enters with a + and the end where the current leaves with a -. As you traverse a loop, if you go from + to - across an element, you subtract the potential difference. If you go from - to +, you add the potential difference. With this sign convention, you have the relations
    v &= iR \\
    i &= C\frac{dv}{dt} \\
    v &= L\frac{di}{dt}
    \end{align*} where v is the potential difference across the element and i is the current flowing through the element.

    (Batteries follow the opposite sign convention because they are sources. The current flows out through the + side.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted