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Find the current through the inductor

  1. Feb 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Find a differential equation for current through the inductor and then solve it. The battery carries voltage ##V_0##
    (The switch is closed at t=0s)

    circuit.png

    2. Relevant equations
    ##v=\frac{q}{C}## (Capacitor)
    ##v=L\frac{di}{dt}## (Inductor)
    ##v=IR## (Resistor)

    3. The attempt at a solution

    From KVL around the first square I get ##0=V_0-iR_1-L\frac{di}{dt}##

    Now I have a couple questions. First off is ##i## in this case actually the current flowing through the inductor or is it the current flowing through the circuit. 2nd, how would I go about solving this differential equation?
     
  2. jcsd
  3. Feb 20, 2016 #2

    cnh1995

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    It is the current through the inductor. Also, current through R1 is not same as current through L.
     
  4. Feb 20, 2016 #3
    Is this because ##L## is in parallel with ##R_2##?
     
  5. Feb 20, 2016 #4

    cnh1995

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  6. Feb 20, 2016 #5

    cnh1995

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    During the transient, current through L and R1 are different. But in the steady state, since the inductor acts as a short, R2 is shorted and current flows only through R1 and L.
     
  7. Feb 20, 2016 #6
    Ok, I originally thought I would just assign a new variable for the currents but then I obtain the equation ##V_0-iR_1-L\frac{di_p}{dt}=0## but this doesn't look too promising so I think it's better to use the junction rule and then assign current ##i## through ##R_1##, current ##i_1## through ##L## and current ##i_2## through ##R_2##, then from the junction rule ##i=i_1+i_2\Longrightarrow i_1=i-i_2##, which gives $$V_0-iR-L\frac{d(i-i_2)}{dt}=0$$. Unfortunately I can't see where to go from here either.
     
  8. Feb 20, 2016 #7

    cnh1995

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    You'll need to frame a DE with one variable only. You can use the two loops to find another relation between i1 and i2 and substitute in place of i2.
    For example, if total current is i, you can write iR1+LdiL/dt=Vo
    And
    LdiL/dt=iR2R2.
    Also, i=iL+iR2
    You can play with these three equations and set a DE in iL.
     
  9. Feb 21, 2016 #8
    Okay so combining equations:

    $$V_0=iR_1+L\frac{di_1}{dt}=(i_1+i_2)R_1+L\frac{di_1}{dt}=i_1R_1+i_2R_1+L\frac{di_1}{dt}=i_1R_1+L\frac{R_1}{R_2}\frac{di_1}{dt}+L\frac{di_1}{dt} \\
    V_0=i_1R_1+\frac{di_1}{dt}(L\frac{R_1}{R_2}+L)\Longrightarrow (V_0-i_1R_1)dt=(L\frac{R_1}{R_2}+L)di_1\\ \frac{dt}{L\frac{R_1}{R_2}+L}=\frac{di_1}{V_0-i_1R_1}
    $$

    Let ##D=L\frac{R_1}{R_2}+L## then
    $$
    \int \frac{dt}{D}=\int \frac{di_1}{V_0-i_1R_1}\Longrightarrow \frac{t}{D}+C=-\ln(V_0-i_1R_1)/R_1 \\
    i_1=\frac{V_0-e^{-tR_1/D-C}}{R_1}
    $$

    Is there a way to show that the constant C=0 in this case? I'm pretty sure this is the case since it makes sense that the current through it as ##t\to \infty## would equal ##V_0/R_1##
     
  10. Feb 21, 2016 #9

    cnh1995

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    Initially, i1=0. You can use definite integration here. Take limits for current as 0 to i and limits for time as 0 to t. This way, you can have a genralized relationship between i and t. Then you can put t= to find the final inductor current.
     
  11. Feb 21, 2016 #10

    cnh1995

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    Well, here it should be in the form
    i(t)=Vo(1-e-t/T)/R1
     
  12. Feb 21, 2016 #11
    Okay using definite integrals with limits ##0\to t## and ##0\to i_1## we end up with
    $$
    \frac{t}{L\frac{R_1}{R_2}+L}=\frac{\ln(V_0)-\ln(V_0-i_1R_1)}{R_1}
    $$
    After further simplification: $$
    i_1R_1=V_0-e^{\ln(V_0)-\frac{R_1}{t}{D}}=V_0-V_0e^{-\frac{R_1t}{D}}\Longrightarrow i_1=\frac{V_0(1-e^{-\frac{R_1t}{D}})}{R_1}$$

    I just have one last question, how come the inductor acts as a short in the steady state?
     
  13. Feb 21, 2016 #12

    cnh1995

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    In the steady state i.e. at t=∞, there is no back emf in the circuit, hence no voltage drop across the inductor. You can see the final current is V/R1 which flows only through R1 and L, bypassing R2. In steady state, you can replace the inductor by a wire.
     
  14. Feb 21, 2016 #13

    gneill

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    Out of curiosity, since you're studying first order RL and RC circuits now, surely you must have been introduced to Thevenin and Norton equivalents? Would it not have been simpler to reduce the resistor network and source to a Thevenin or Norton equivalent first?

    upload_2016-2-21_8-48-22.png
     
  15. Feb 21, 2016 #14
    I found ##V_{th}=iR_2## and ##R_{th}=R_1+R_2## which when plugged into the differential equation obtained from KVL: $$iR_2=i(R_1+R_2)+L\frac{di}{dt}
    $$
    Solving this equation results in
    $$
    i=\frac{e^{-\frac{tR_1}{L}}}{R_1}$$
    Which doesn't seem to match up with my previous result. Did I calculate ##V_{th}## and ##R_{th}## incorrectly?
     
  16. Feb 21, 2016 #15

    gneill

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    Your Thevenin model is not correct. What is "i"? There should be no "i" in the Thevenin model, just the original voltage source value and the resistors.
     
  17. Feb 21, 2016 #16
    Ok I believe I found the correct values now although I haven't done the integration yet, I got ##V_{th}=\frac{R_2}{R_1+R_2}V_0## and ##R_{th}=R_1+R_2##
     
  18. Feb 21, 2016 #17

    gneill

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    Your Thevenin resistance is still incorrect. How do you find the Thevenin resistance?
     
  19. Feb 21, 2016 #18
    Removing the voltage sources and then calculating the resistance across the terminals. Is my mistake that ##R_1## and ##R_2## are in parallel so ##R_{th}=\frac{R_1R_2}{R_1+R_2}##
     
  20. Feb 21, 2016 #19

    gneill

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    Yup.
     
  21. Feb 21, 2016 #20
    Thanks, I managed to get the same answer now.
     
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