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Circuit with two resistances and inductance

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data
    http://img534.imageshack.us/img534/5788/gimpjl.png [Broken]

    I have this circuit and I want to know [itex]i_1[/itex] and [itex]i_2[/itex] (currents in [itex]R_1[/itex] and [itex]R_2[/itex]).
    [itex]\varepsilon[/itex] is electromotive force and L is inductance.

    I'd like to know if the following system of equations is correct and if I can get [itex]i_1[/itex] and [itex]i_2[/itex].

    2. The attempt at a solution

    \begin{cases} \varepsilon=R_1(i_1+i_2)+R_2i_2 \\ \varepsilon-L\frac{d}{dt}i_1=R_1(i_1+i_2) \end{cases}

    I think I'm wrong, because i1 is the current in L, not in R1, but I must get current in R1!
    How can I correct the system of equations?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 21, 2013 #2


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    Staff: Mentor

    Do you have to solve the differential equation for the circuit, or can you take advantage of the known properties of first order circuits (initial conditions at t = 0+; steady state conditions; the forms of their transient solutions) to write the solution?

    As a general hint/suggestion, note that if you happened to have an expression for the potential where R1 and R2 (and L) meet, then it would be a simple matter to write expressions for the two currents that you're looking for.
  4. Jan 21, 2013 #3

    Thank you for your answer!! My main problem is writing the equations of the system. Can you help me?
  5. Jan 21, 2013 #4


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    Staff: Mentor

    The set of equations that you've already written will allow you to find a differential equation for the current through the inductor (i1 in your equations). If you solve for this current you can then go back and find i2, and then find the current through R1 since its the sum of the other two. The algebra may get tedious.

    I think that if I were to do this problem I'd consider starting with a nodal equation and find the potential at the one independent node (so only one equation to deal with). This would involve an integral equation rather than a differential equation, but a quick differentiation would reduce it to familiar form. Solving it, having the expression for the node voltage would let me easily find the two currents. Also, I think I'd simplify the circuit first: a Norton equivalent would get rid of one of the resistances.
  6. Jan 21, 2013 #5
    thank you so much!!! :)
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