If $I = 0$ at $t = 0$ find i.

Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : 2013

$\dfrac {di}{dt}+\dfrac RLi=\dfrac EL\sin wt \ $ Which is same as that of $$\dfrac {di}{dt}+ pi= Q \ $$

Here $P=\dfrac RL$ and $Q=\dfrac EL\sin wt \\ \therefore I.f = e^{\int p\space dt}=e^{\int \frac RL dt}=e^{\frac {Rt}l} \\ $

Solution is

$i.If=\int Q\times If\space dt \\ i\times e^{\frac {Rt}L}=\int \dfrac EL\sin wt \space e^{\frac {Rt}L} \\ =\dfrac EL\int e^{\frac {Rt}L}\sin wt\space dt + C\\ $

We have direct formula for this type of integration

$ie^{\frac {Rt}L}=\dfrac EL\dfrac {e^{\frac {Rt}L}}{(\dfrac RL)^2+w^2}\Bigg[\dfrac RL\sin wt- w\cos wt\Bigg] + c\\ ie^{\frac {Rt}L}=\dfrac EL e^{\frac {Rt}L}\times \dfrac {L^2}{R^2+w^2L^2}\Bigg[\dfrac {R\sin wt-wL\cos wt}L\Bigg]+c \\ ie^{\frac {Rt}L}=\dfrac {Ee^{\frac {Rt}L}}{R^2+w^2L^2}(R\sin wt-wL\cos wt)+ c\\ \therefore i=\dfrac E{R^2+w^2L^2} (R\sin wt-wL\cos wt) +c.e^{\frac {Rt}L} \\ AT \space t=0 , i=0 \\ 0=\dfrac E{R^2+w^2L^2}[R\times 0-wL\times 1]+ C\\ \therefore c=\dfrac {EwL}{R^2+w^2L^2}\\ \therefore i=\dfrac E{R^2+w^2L^2}[R\sin wt-wL\cos wt] +\dfrac {Ewl}{R^2+w^2L^2}e^{-\frac {Rt}L}$