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I am trying to solve a set of non linear equation but using the function nsolve but mathematica is not giving me proper answer. The code is given below.

h = 0.8;
\[Epsilon]r = 2.94;
ae = 5;

The Equations are given below.

Equation Number 1....

1/we = (120*\[Pi])/(ae (w/h + 1.393 + 0.667*Log[w/h + 1.44]));

Equation Number 2....

1/we = (4.38/
ae)*(E^((-0.627*\[Epsilon]r)/(((\[Epsilon]r + 1)/
  2) + (\[Epsilon]r - 1)/2 + 1/Sqrt[(1 + (12*h)/w)])));

Nsolve for solving both equations for "we" and "w" simultaneously.

NSolve[{1/we == (120*\[Pi])/(
ae (w/h + 1.393 + 0.667*Log[w/h + 1.44])), 
1/we == 4.38/
ae*(E^((-0.627*\[Epsilon]r)/(((\[Epsilon]r + 1)/
   2) + (\[Epsilon]r - 1)/2 + 1/Sqrt[(1 + (12*h)/w)])))}, {we, 
w}, Reals]

I am not getting any solution for the above equations using NSOLve.

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  • $\begingroup$ NSolve deals primarily with linear and polynomial equations. In your case you might have more luck using e.g. FindRoot with an adequate estimate of the solution values. $\endgroup$ – MarcoB Jun 8 '15 at 5:02
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    $\begingroup$ Using {{w, 100}, {we, 2}} for the starting values should do it with FindRoot as recommended by MarcoB. $\endgroup$ – JimB Jun 8 '15 at 5:17
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    $\begingroup$ @JimBaldwin , I used the Find Root Method with the code given below .FindRoot[{1/we == (120*\[Pi])/( ae (w/h + 1.393 + 0.667*Log[w/h + 1.44])), 1/we == (4.38/ ae)*(E^((-0.627*\[Epsilon]r)/(((\[Epsilon]r + 1)/ 2) + (\[Epsilon]r - 1)/2 + 1/Sqrt[(1 + (12*h)/w)])))}, {{w, 100}, {we, 2}}] But that gives me some dimension related error . $\endgroup$ – AK K Khan Jun 8 '15 at 6:47
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Using the starting value {w,100} and setting both sides equal (we can ignore the reciprocal 1/we) and then solve for w first. Apologies to Jim and MarcoB, this is a rip answer to clean things up.

h = 0.8;
\[Epsilon]r = 2.94;
ae = 5;

eqn = (120*\[Pi])/(ae (w/h + 1.393 + 0.667*Log[w/h + 1.44])) == 
(4.38/ae)*(E^((-0.627*\[Epsilon]r)/(((\[Epsilon]r + 1)/
        2) + (\[Epsilon]r - 1)/2 + 1/Sqrt[(1 + (12*h)/w)])));

FindRoot[eqn, {w, 100}] (* Gives  w -> 106.761 *)
1/(eqn[[1]] /. w -> 106.761) (* Implies we is 1.83182 *)
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  • $\begingroup$ @Histograms...Thanks for your answer..This was really helpful. $\endgroup$ – AK K Khan Jun 8 '15 at 8:26
  • $\begingroup$ No need to apologize. Thank you for writing this up! $\endgroup$ – MarcoB Jun 8 '15 at 14:03

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