Stuck on finding the natural response

Discussion in 'Homework Help' started by p75213, Nov 22, 2011.

  1. p75213

    Thread Starter Member

    May 24, 2011
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    Hello,
    I am attempting this problem but have got stuck on finding the natural response to the circuit at t>0.
    When the independent current source is turned of we get a short circuit from the top to bottom. I am struggling to come up with the appropriate KCL and KVL equations.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    No we don't. Presumably you mean turned on....?

    Through which elements does the current flow at t=0+ when the switch first closes? The source will adopt whatever voltage is required to match its terminal load conditions.

    Through which elements does the current flow at t=∞?
     
  3. p75213

    Thread Starter Member

    May 24, 2011
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    At t=0+ i(0+)=i(0-)=0
    At t=\infty i=2
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    And - at t=0+ all of the 2A flows into the left hand branch.
     
  5. p75213

    Thread Starter Member

    May 24, 2011
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    For the LHS I used v(t)=v(\infty)+(v(0)-v(\infty))e^{-t/RC}
    And for the RHS i(t)=i(\infty)+(i(0)-i(\infty))e^{-tR/L}

    The result is the same as the answer given however in both cases I get the exponent = -2t rather than -5t.
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The solution is somewhat convoluted ...

    Denote the current in the left branch as i1(t) and the voltage at the top node as v1(t)

    v_1(t)=10i_1+20\int i_1 dt=4i+2\frac{di}{dt}

    also

    i=2-i_1 & \frac{di}{dt}=-\frac{di_1}{dt}

    v_1(t)=10i_1+20\int i_1 dt=4(2-i_1)-2\frac{di_1}{dt}

    leading to

    10i_1+20\int i_1 dt=4(2-i_1)-2\frac{di_1}{dt}

    and

    \frac{di_1}{dt}+7i_1+10\int i_1 dt=4

    \frac{d^2i_1}{dt^2}+7\frac{di_1}{dt}+10 i_1=0

    Initial conditions: i1(0)=2, i(0)=0, v1(0)=20

    The real roots of the CE are -5 and -2

    So

    i_1(t)=K_1e^{-2t}+K_2e^{-5t}

    Using i1(0)=2

    K_1+K_2=2

    Using v1(0)=20 and i1 as defined above, along with the relationship

    v_1(t)=8-4i_1-2\frac{di_1}{dt}

    The values of K1 & K2 can then be found and so on ...

    As it turns out the value of K1=0 which accounts for the existence of the exponential solution being in -5t only.

    This problem is far more readily solved by the use of Laplace transforms.
     
  7. p75213

    Thread Starter Member

    May 24, 2011
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    Your right - it is convoluted. Haven't got to Laplace Transforms yet. That's towards the end of the book. I notice Khan Academy has included them in the differentials section. This is a good free learning resource for anybody interested.
     
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