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Using nodal analysis with Laplace transform find “v” for “t > 0”

Founded initial conditions:

i(0-)=i(0+)=0A

v(0-)=v(0+)=0V

i(infinite)=3A

v(infinite)=4*3A=12V

My nodal analysis with Laplace transform equations are:

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v1 – node where current source flows into circuit

The Laplace of “v” is in the second row, but it seems to be an error in my equations – the answer in book isn’t equal to reverse transform of second row! May anyone help to properly write that equations?

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I see two errors. #1; a switch in series will NOT stop a current source. You must use a switch in parallel with the source and open it at t=0 to enable. #2; the Laplace transform of the suddenly applied 3A source is 3/S. This results in Vc=60/[(s)(S+5)] from which Vc(t)=12(1-e^(-5t)). Note that the final value of Vc is 12volts as expected.

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I get this set of nodal equations:

$$\begin{align*} \frac{V_1}{10\:\Omega}+\frac{V_1}{4\:\Omega}&=\frac{V_C}{10\:\Omega}+\frac{V_L}{4\:\Omega}+3\:\textrm{A}\\\\\frac{V_C}{10\:\Omega}+50\:\textrm{mF}\cdot\frac{\textrm{d}\:V_C}{\textrm{d}\:t}&=\frac{V_1}{10\:\Omega}\\\\\frac{V_L}{4\:\Omega}+\frac{1}{2\:\textrm{H}}\cdot\int V_L~\textrm{d}\:t&=\frac{V_1}{4\:\Omega} \end{align*}$$


Which results in this Laplace matrix:

$$\begin{array}{ccc} \left[\begin{array}{ccc} \left(\frac{1}{10}+\frac{1}{4}\right)&\left(\frac{-1}{10}\right)&\left(\frac{-1}{4}\right)\\\\\left(\frac{-1}{10}\right)&\left(\frac{1}{10}+\frac{s}{20}\right)&\left(0\right)\\\\\left(\frac{-1}{4}\right)&\left(0\right)&\left(\frac{1}{4}+\frac{1}{2\: s}\right) \end{array}\right]&\left[\begin{array}{ccc}V_1\left(s\right)\\\\\\V_C\left(s\right)\\\\\\V_L\left(s\right)\end{array}\right]=&\left[\begin{array}{ccc}3\\\\\\0\\\\\\0\end{array}\right]\end{array}$$


Which yields:

$$\begin{align*} V_1\left(s\right) &=\frac{30s+60}{s+5} \\\\V_C\left(s\right)&=\frac{60}{s+5}\\\\V_L\left(s\right)&=\frac{30 s}{s+5} \end{align*}$$


I get the same results. I haven't bothered with the initial conditions, though. Are you taking the initial conditions into account?

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  • \$\begingroup\$ It seems to be I don’t understand how to take the initial conditions into account with Laplace transform. May you help me with it? The inverse transform gives Vc = 60*e^(-5*t), but it is not the correct answer! What problem is I may not found! \$\endgroup\$
    – MaxMil
    Dec 19 '16 at 15:56
  • \$\begingroup\$ I think my equations must be corrected to use initial conditions! Anyone may help with that? \$\endgroup\$
    – MaxMil
    Dec 20 '16 at 6:53

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