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In case of series R-L circuit excited by DC supply (V) derive equation for transient current IL with initial conditions.
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$consider\ series\ R-L\ circuit\ with\ initial\ condition\ i\left(0^-\right)=^{'}i^{'}\ A\ switch\ is\ closed\ at\ time\ t=0$ Writing  KVL equation for t\gt0 $Ri\left(s\right)+Lsi\left(s\right)=\dfrac{v}{s}+Li\left(0^-\right) $ $\therefore{}i\left(s\right)=\dfrac{\dfrac{V}{L}}{s\left(s+\dfrac{R}{L}\right)}+\dfrac{Li\left(0^-\right)}{L\left(s+\dfrac{R}{L}\right)}=\dfrac{\dfrac{V}{L}}{s\left(s+\dfrac{R}{L}\right)}+\dfrac{i\left(0^-\right)}{\left(S+\dfrac{R}{L}\right)} $   $i\left(s\right)=\dfrac{A}{s}+\dfrac{B}{s+\dfrac{R}{L}}+\dfrac{i\left(0^-\right)}{s+\dfrac{R}{L}} $ $A=\dfrac{s\times{}\dfrac{V}{L}}{s\left(s+\dfrac{R}{L}\right)}\left\vert{}\ s=0=\dfrac{V}{R};\ \ \ B=\dfrac{\left(s+\dfrac{R}{L}\right)\times{}\dfrac{V}{L}\ }{s\left(s+\dfrac{R}{L}\right)}\right\vert{}\ s=\left(-\dfrac{R}{L}\right)=\left(-\dfrac{V}{R}\right)$ $\therefore{}I\left(s\right)=\dfrac{\dfrac{V}{R}}{S}+\dfrac{\dfrac{-V}{R}}{s+\dfrac{R}{L}}+\dfrac{i\left(0^-\right)}{s+\dfrac{R}{L}} $ Taking Inverse Laplace transform \(I\left(t\right)=\dfrac{V}{R}-\dfrac{V}{R}e^{-\dfrac{R}{L}t}+i\left(0^-\right)e^{-\dfrac{R}{L}t}\ \ \ \ for\ t\gt0\) \(\therefore{}I\left(t\right)=\dfrac{V}{R}-e^{-\dfrac{R}{L}t}\left[\dfrac{V}{R}-i\left(0^-\right)\right]\ \ for\ t\gt0\) Applying …

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