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

Commented by candre last updated on 09/May/18

r_(th) =R_1 ∥R_3 +R_2 ∥R_4   V_(th) =((R_3 /(R_1 +R_3 ))−(R_4 /(R_2 +R_4 )))ε  I=(V_(th) /r_(th) )

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

Commented by ajfour last updated on 09/May/18

Find current I.

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

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 ))) .

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

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

for balanced wheaston bridge that means  when (R_1 /R_2 )=(R_3 /R_4_  )   I=0

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

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