0
$\begingroup$

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.

  • 1
    Where does $V_b$ appear in any of the above? What are you trying to solve for?2012-10-24
  • 0
    Sorry it was $V_2$2012-10-24
  • 0
    I find maxima (wxmaxima) useful for symbolic calculations. A bit cumbersome, but free.2012-10-24

1 Answers 1

1

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 $$V_2= \frac{I_ssL}{sRC+1+s^2LC} = \frac{I_ssL}{LC \left(\frac{sR}L + \frac1{LC} + s^2\right)} = \frac{I_ss}{C \left(\frac{sR}L + \frac1{LC} + s^2\right)}.$$

  • 2
    The terms $\frac{R}{L}$ (frequency of sorts) and $\frac{1}{LC}$ (square of resonant frequency) have physical interpretations for electrical engineers...2012-10-24
  • 0
    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.2012-10-24
  • 0
    Yes this is algebra for an output voltage using the laplace transform.2012-10-24