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Question-174975




Question Number 174975 by sciencestudent last updated on 15/Aug/22
Answered by MikeH last updated on 15/Aug/22
R_1 +R_2  = 690 or x+y = 690.....(1)  ((xy)/(x+y)) = 150  ⇒ ((xy)/(690)) = 150 ⇒ xy = 103500   from (1) x = 690−y  ⇒ y(690−y)= 103500  ⇒ y^2 −690y +103500 =0  y = 220.4 Ω or y = 469.6 Ω  so R_1  = 220.4Ω and R_2  = 469.6Ω
$${R}_{\mathrm{1}} +{R}_{\mathrm{2}} \:=\:\mathrm{690}\:\mathrm{or}\:{x}+{y}\:=\:\mathrm{690}…..\left(\mathrm{1}\right) \\ $$$$\frac{{xy}}{{x}+{y}}\:=\:\mathrm{150} \\ $$$$\Rightarrow\:\frac{{xy}}{\mathrm{690}}\:=\:\mathrm{150}\:\Rightarrow\:{xy}\:=\:\mathrm{103500}\: \\ $$$$\mathrm{from}\:\left(\mathrm{1}\right)\:{x}\:=\:\mathrm{690}−{y} \\ $$$$\Rightarrow\:{y}\left(\mathrm{690}−{y}\right)=\:\mathrm{103500} \\ $$$$\Rightarrow\:{y}^{\mathrm{2}} −\mathrm{690}{y}\:+\mathrm{103500}\:=\mathrm{0} \\ $$$${y}\:=\:\mathrm{220}.\mathrm{4}\:\Omega\:\mathrm{or}\:{y}\:=\:\mathrm{469}.\mathrm{6}\:\Omega \\ $$$$\mathrm{so}\:{R}_{\mathrm{1}} \:=\:\mathrm{220}.\mathrm{4}\Omega\:\mathrm{and}\:{R}_{\mathrm{2}} \:=\:\mathrm{469}.\mathrm{6}\Omega \\ $$

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