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_1 −i_3 i =(ε/((((R_1 R_2 )/(R_1 +R_2 ))+((R_3 R_4 )/(R_3 +R_4 ))))) i_1 =i((R_2 /(R_1 +R_2 ))) ; i_3 =i((R_4 /(R_3 +R_4 ))) I=(ε/((((R_1 R_2 )/(R_1 +R_2 ))+((R_3 R_4 )/(R_3 +R_4 )))))[(R_2 /(R_1 +R_2 ))−(R_4 /(R_3 +R_4 ))] I= ((ε(R_2 R_3 −R_1 R_4 ))/(R_1 R_2 (R_3 +R_4 )+R_3 R_4 (R_1 +R_2 ))) .](https://www.tinkutara.com/question/Q34624.png)
$${I}={i}_{\mathrm{1}} −{i}_{\mathrm{3}} \\ $$$$\:\:\:{i}\:=\frac{\varepsilon}{\left(\frac{{R}_{\mathrm{1}} {R}_{\mathrm{2}} }{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }+\frac{{R}_{\mathrm{3}} {R}_{\mathrm{4}} }{{R}_{\mathrm{3}} +{R}_{\mathrm{4}} }\right)} \\ $$$${i}_{\mathrm{1}} ={i}\left(\frac{{R}_{\mathrm{2}} }{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }\right)\:\:\:;\:\:{i}_{\mathrm{3}} ={i}\left(\frac{{R}_{\mathrm{4}} }{{R}_{\mathrm{3}} +{R}_{\mathrm{4}} }\right) \\ $$$${I}=\frac{\varepsilon}{\left(\frac{{R}_{\mathrm{1}} {R}_{\mathrm{2}} }{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }+\frac{{R}_{\mathrm{3}} {R}_{\mathrm{4}} }{{R}_{\mathrm{3}} +{R}_{\mathrm{4}} }\right)}\left[\frac{{R}_{\mathrm{2}} }{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }−\frac{{R}_{\mathrm{4}} }{{R}_{\mathrm{3}} +{R}_{\mathrm{4}} }\right] \\ $$$${I}=\:\frac{\varepsilon\left({R}_{\mathrm{2}} {R}_{\mathrm{3}} −{R}_{\mathrm{1}} {R}_{\mathrm{4}} \right)}{{R}_{\mathrm{1}} {R}_{\mathrm{2}} \left({R}_{\mathrm{3}} +{R}_{\mathrm{4}} \right)+{R}_{\mathrm{3}} {R}_{\mathrm{4}} \left({R}_{\mathrm{1}} +{R}_{\mathrm{2}} \right)}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/May/18
$${for}\:{balanced}\:{wheaston}\:{bridge}\:{that}\:{means} \\ $$$${when}\:\frac{{R}_{\mathrm{1}} }{{R}_{\mathrm{2}} }=\frac{{R}_{\mathrm{3}} }{{R}_{\mathrm{4}_{} } }\:\:\:{I}=\mathrm{0} \\ $$