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A resistance of 100 ohms and inductance of 0.5 henries are connected in series With a battery of 20 volts. Find the current at any instant if the relation between L,R,E is L `(di)/(dt)+Ri=E.`

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#### Solution

L`(di)/(dt)+Ri=E.`

`therefore (di)/(dt)+(Ri)/L=E/L`

Solution is given by ,

`i,e^(int(R/L)dt=inte^(int(R/L)dt)E/Ldt+c`

`therefore"i".e^((Rt)/L)=(Ee^((Rt)/L))/R+c`

At t = 0, i = 0 `thereforec=E/R`

`therefore "i".e^((Rt)/L)=(Ee^((Rt)/L))/R+(-E)/R`

`therefore "i" = E/R(1-e^(-(Rt)/L))`

For given condition R = 100, L = 0.5, E = 20

`therefore i=0.2(1-e^(-200t))`

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