Question-34617

Question Number 34617 by ajfour last updated on 09/May/18

Commented by candre last updated on 09/May/18

$${r}_{{th}} ={R}_{\mathrm{1}} \parallel{R}_{\mathrm{3}} +{R}_{\mathrm{2}} \parallel{R}_{\mathrm{4}} \\ $$$${V}_{{th}} =\left(\frac{{R}_{\mathrm{3}} }{{R}_{\mathrm{1}} +{R}_{\mathrm{3}} }−\frac{{R}_{\mathrm{4}} }{{R}_{\mathrm{2}} +{R}_{\mathrm{4}} }\right)\varepsilon \\ $$$${I}=\frac{{V}_{{th}} }{{r}_{{th}} } \\ $$

Commented by ajfour last updated on 09/May/18

$${Find}\:{current}\:{I}. \\ $$

Answered by ajfour last updated on 09/May/18

$${I}={i}_{\mathrm{1}} −{i}_{\mathrm{3}} \\ $$$$\:\:\:{i}\:=\frac{\varepsilon}{\left(\frac{{R}_{\mathrm{1}} {R}_{\mathrm{2}} }{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }+\frac{{R}_{\mathrm{3}} {R}_{\mathrm{4}} }{{R}_{\mathrm{3}} +{R}_{\mathrm{4}} }\right)} \\ $$$${i}_{\mathrm{1}} ={i}\left(\frac{{R}_{\mathrm{2}} }{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }\right)\:\:\:;\:\:{i}_{\mathrm{3}} ={i}\left(\frac{{R}_{\mathrm{4}} }{{R}_{\mathrm{3}} +{R}_{\mathrm{4}} }\right) \\ $$$${I}=\frac{\varepsilon}{\left(\frac{{R}_{\mathrm{1}} {R}_{\mathrm{2}} }{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }+\frac{{R}_{\mathrm{3}} {R}_{\mathrm{4}} }{{R}_{\mathrm{3}} +{R}_{\mathrm{4}} }\right)}\left[\frac{{R}_{\mathrm{2}} }{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }−\frac{{R}_{\mathrm{4}} }{{R}_{\mathrm{3}} +{R}_{\mathrm{4}} }\right] \\ $$$${I}=\:\frac{\varepsilon\left({R}_{\mathrm{2}} {R}_{\mathrm{3}} −{R}_{\mathrm{1}} {R}_{\mathrm{4}} \right)}{{R}_{\mathrm{1}} {R}_{\mathrm{2}} \left({R}_{\mathrm{3}} +{R}_{\mathrm{4}} \right)+{R}_{\mathrm{3}} {R}_{\mathrm{4}} \left({R}_{\mathrm{1}} +{R}_{\mathrm{2}} \right)}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/May/18

$${for}\:{balanced}\:{wheaston}\:{bridge}\:{that}\:{means} \\ $$$${when}\:\frac{{R}_{\mathrm{1}} }{{R}_{\mathrm{2}} }=\frac{{R}_{\mathrm{3}} }{{R}_{\mathrm{4}_{} } }\:\:\:{I}=\mathrm{0} \\ $$

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