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Given , epsilon-L(di)/(dt)=iR, find the value ofi at any time t in terms of constantepsilon, L and R. At t=0 , i=0 . |
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Answer» Solution :We have , `epsilon-L (DI)/(dt)=iR` or `(epsilon -iR)=L (di)/(dt)` or `(di)/((epsilon-iR))=(dt)/(L) "" ` ...(i) Integrating both sides of the equation (i), we get `INT (di)/((epsilon-iR))= int (dt)/(L)` Here LIMIT of time varies from 0 to t and corresponding limits of i varies from 0 to i. ` :. "" int_(0)^(i)(di)/((epsilon-iR))=int _(0)^(t) (dt)/(L)` For integration of LHS, substitute ` epsilon-iR=z` Also `"" (d)/(di)(epsilon-iR)=(DZ)/(di)` or `(0-R)=(dz)/(di)` `:. "" di=(dz)/((-R))` and ` int _(0)^(t)(di)/((epsilon-iR)) = int _(0)^(i) (dzl(-R))/(z)` ` =((1)/(-R)) int_(0)^(i) (dz)/(z)` `=((1)/(-R))|ln z|_(0)^(i)` `=((1)/(-R))|ln (epsilon-iR)|_(0)^(i)` `=((1)/(-R)){ln(epsilon-iR)-ln(epsilon-0)}` `=((1)/(-R)) ln ((epsilon-iR))/(epsilon)" " ` ...(ii) and RHS` int_(0)^(t) (dt)/(L)=((1)/L))|t|_(0)^(t)=(1)/(L)(t-0)` `=(t)/(L)` From equation (i) and (ii) , we have `(-(1)/(R)) ln ((epsilon-iR)/(epsilon))=(t)/(L)` or ` " " ln((epsilon-iR)/(epsilon))=-(R)/(L)t` or ` ((epsilon-iR)/(epsilon))=e^(-(Rt)/(L))` or `" " i=(epsilon)/(R)(1-e^(-(tR)/(L)))` |
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