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Re: Complex and Solve
*To*: mathgroup at smc.vnet.net
*Subject*: [mg125162] Re: Complex and Solve
*From*: Howard Lovatt <howard.lovatt at gmail.com>
*Date*: Sat, 25 Feb 2012 01:56:24 -0500 (EST)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <201202211115.GAA08330@smc.vnet.net>
Thanks Bob,
That is even better. I am somewhat surprised that you need to manually
specify $Assumptions and that since this is global that Solve doesn't take
this into account anyway.
But this solution is much better than my original, so thanks again,
-- Howard.
On 23 February 2012 23:07, Bob Hanlon <hanlonr357 at gmail.com> wrote:
> Actually I should have checked, with version 8 Solve can handle
> inequalities so either Reduce or Solve works here.
>
> $Assumptions = rfe > 0 && l > 0;
>
> r = 1;
> \[Omega] = 300;
> vrms = 2;
> irms = 1/2;
> \[Phi] = 30 Degree;
>
> impedance = r + 1/(1/rfe + 1/(I \[Omega] l)) // FullSimplify;
>
> voltage = vrms (Cos[\[Phi]] + I Sin[\[Phi]]) // Simplify;
>
> Solve[{irms == voltage/impedance, $Assumptions}, {rfe,
> l}][[1]] // FullSimplify
>
> {rfe -> (1/11)*(-7 + 30*Sqrt[3]), l -> (1/600)*(17 - 4*Sqrt[3])}
>
> Alternatively,
>
> Solve[{irms == voltage/impedance, Im[l] == 0, Im[rfe] == 0}, {rfe,
> l}][[1]] // FullSimplify
>
> {rfe -> (1/11)*(-7 + 30*Sqrt[3]), l -> (1/600)*(17 - 4*Sqrt[3])}
>
>
> Bob Hanlon
>
> On Thu, Feb 23, 2012 at 5:46 AM, <howard.lovatt at gmail.com> wrote:
> > Thanks Bob,
> >
> > Your solution is much better than mine. I had no idea that Reduce could
> be used like you showed.
> >
> > Bizarre that Solve and Reduce behave so differently!
> >
> > Thanks again,
> >
> > -- Howard.
> >
>
>
>
> --
> Bob Hanlon
>
--
-- Howard.
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