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There are two oscillating circuits (figure) with capacitors of equal capacitances. How must inductances and active resistances of the coils be interrelated forthe frequencies and damping of free oscillations in both circuits to be equal ? The mutual inductance of coils in the left circuit is negligible. |
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Answer» Solution :We have `L_(1)dot(I_(1))+R_(1)I_(1)=L_(2)dot(I_(2))+R_(2)I_(2)` `=-(int Idt)/(C)` `I=I_(1)+I_(2)` Then differentiating we have the equations `L_(1)C ddot(I_(1))+R_(1)Cdot(I_(1))+(I_(1)+I_(2))=0` ` L_(2)Cddot(I_(2))+ R_(2)Cdot(I_(2)+(I_(1)+I_(2))=0` Look for a solution `I_(1)=A_(1)e^(alphat), I_(2)=A_(2)e^(alphat)` Then ` (1+ alpha^(2)L_(1)C+ alpha R_(1)C)A_(1)+A_(2)=0` `A_(1)=(1+alpha^(2)L_(2)C+alpha R_(2)C)_(2)=0` This SET of simultaneous equation has a nontrivial solution only if `(a+alpha^(2)L_(1)C+alphaR_(1)underset.C)(1+alpha^(2)L_(2)C+alphaR_(2)C)=1` or `alpha^(3)+ alpha^(2)(L_(1)R_(2)+ L_(2)R_(1))/( L_(1)L_(2))+alpha(L_(1)+L_(2)+R_(1)R_(2)C)/(L_(1)L_(2)C)+(R_(1)+R_(2))/( L_(1)L_(2)C)=0` This cubic equation has one real root which we ignore and two complex conjugate roots. We require the condition that this pair of complex conjugate roots is identical with roots of the equation `alpha^(2) LC=alphaRC + 1 =0` The general solution of this problem is not easy . We look for special cases. If `R_(1)=R_(2)=0`, then `R=0`, and `L=(L_(1)L_(2))/(L_(1)+L_(2)).` If `L_(1)=L_(2)=0, `then `L=0` and `R=R_(1)R_(2)//(R_(1)+R_(2))`. These are the quoted solution but they are MISLEADING. We shall give the solution for small `R_(1), R_(2)`. Thenwe put `alpha=-beta+iomega` when `beta` is small We get `(1- omega^(2)L_(1)C-2ibetaomegaL_(1)C-betacancel(R_(1))C+iomegaR_(1)C)` `(1- omega^(2)L_(2)C-2ibetaomegaL_(2)C-betacancel(R_(2))C+iomegaR_(2)C)=1` `(` we neglect `beta^(2)& beta R_(1), betaR_(2))`. Then `(1-omega^(2)L_(1)C)(1-omega^(2)L_(2)C)=1implies omega^(2)=(L_(1)+L_(2))/(L_(1)L_(1)C)` This is identical with `omega^(2)=(1)/(LC)` if `L=(L_(1)L_(2))/(L_(1)+L_(2))`. ALSO`(2 beta L_(1)-R_(2))(1- omega^(2)L_(2)C)+(2 betaL_(2)-R_(2))(1-omega^(2)L_(2)C)=0` This gives `beta=(R)/(2L)=(R_(1)L_(2)^(2)+R_(2)L_(1)^(2))/(2L_(1)L_(2)(L_(1)+L_(2)))implies R=(R_(1)L_(2)^(2)+R_(2)L_(1)^(2))/((L_(1)+L_(2))^(2))` 
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