Differential equations

Discussion in 'Homework Help' started by MrRockchip, May 25, 2010.

  1. MrRockchip

    Thread Starter New Member

    May 25, 2010
    18
    0
    Please, tell me, what are the Correct differential equations
    for calculating the electric current on inductivity ? :confused:

    http://myimgs.net/images/zynyf.jpg

    I've created some equations, but at the end (after solving them)
    I get the wrong result - so, they were wrong. :(
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,152
    1,793
    Voltage, Vr, across a resistor R, according to Ohm's Law, is "I*R"
    Voltage, Vl across an inductor L is L*(di/dt)

    The result of writing KVL around a loop should be a first order equation. with a step function representing the closing or opening of the switch. Does that help?
     
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  3. MrRockchip

    Thread Starter New Member

    May 25, 2010
    18
    0
    For inductor:

    i(-0)=E/R;

    L(di/dt)=E / ((R*R)/R+R);
    L(di/dt)=2*E/R;
    (di/dt) = 2*E/(R*L)
    Because E,R,L=const., after integrating :

    i(t)=(2*E/(R*L))*t

    Something is wrong here, because i(t) can't be infinite;
    it's obvious that i(+inf)=E/R . :(
     
    Last edited: May 26, 2010
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Here's a start ... note that R2 only determines the value of I_{0}

    L\frac{di}{dt}+R_{1}i=E

    \frac{di}{(\frac{E}{R_{1}}-i)}=\frac{R_{1}}{L}dt

    \int^{i}_{I_{0}}\frac{di}{(\frac{E}{R_{1}}-i)}=\int^{t}_{0}\frac{R_{1}}{L}dt
     
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  5. MrRockchip

    Thread Starter New Member

    May 25, 2010
    18
    0
    Thank you, Tnk! :)

    Here's my solution:

    [​IMG]

    We know that [​IMG]

    So, the final answer is [​IMG]

    It seems to be correct, because
    [​IMG]

    Am i right ? ;)
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Actually

    I_{0}=\frac{E}{R_{1}||R_{2}}

    You happened to have

    I_{0}=\frac{2E}{R_{1}}

    "correct" only because R1=R2 in the problem statement.

    Otherwise you appear to have got a handle on the problem solution.
     
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