Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.