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I have two equations and I'm trying to solve $V_2$. The equations are: $$\frac{V_1}{sL}+\frac{V_1-V_2}{R}=I_s,\\ \frac{V_2-V_1}{R}+\frac{V_2}{1/sC} = 0$$

If I solve the second equation with respect to $V_1$ I get: $V_1=V_2 (sRC+1)$

It all goes down hill when I sub $V_1$ into first equation. I end up getting $$\frac{I_s}{\frac{1}{sL}+\frac{sC}{R}} $$

However the correct answer is: $$\frac{s I_s}{C[s^2+\frac{R}{L}s +\frac{1}{LC}]}$$

Can someone help me walk through this step by step to figure out why I am getting the wrong answer.

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Where does $V_b$ appear in any of the above? What are you trying to solve for? –  copper.hat Oct 24 '12 at 16:30
    
Sorry it was $V_2$ –  Nick Oct 24 '12 at 16:36
    
I find maxima (wxmaxima) useful for symbolic calculations. A bit cumbersome, but free. –  copper.hat Oct 24 '12 at 16:47

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When I plug $V_1=V_2(sRC+1)$ and $V_1-V_2=V_2sRC$ into $$\frac{V_1}{sL}+\frac{V_1-V_2}{R}=I_s$$ I get $$ \begin{align} V_2 \left(\frac{sRC+1}{sL}+sC\right)&=I_s\\ V_2 \frac{sRC+1+s^2LC}{sL} &= I_s \end{align} $$ which means $$V_2= \frac{I_ssL}{sRC+1+s^2LC}.$$ For some reason, in the answer from your book they continued to get $$V_2= \frac{I_ssL}{sRC+1+s^2LC} = \frac{I_ssL}{LC \left(\frac{sR}L + \frac1{LC} + s^2\right)} = \frac{I_ss}{C \left(\frac{sR}L + \frac1{LC} + s^2\right)}.$$

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The terms $\frac{R}{L}$ (frequency of sorts) and $\frac{1}{LC}$ (square of resonant frequency) have physical interpretations for electrical engineers... –  copper.hat Oct 24 '12 at 16:42
    
I see. So that's probably the reason why they did not leave that it that form but continued until they had there those expressions. –  Martin Sleziak Oct 24 '12 at 16:43
    
Yes this is algebra for an output voltage using the laplace transform. –  Nick Oct 24 '12 at 16:49

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