$\dfrac {RdQ}{dt}+\dfrac QC=V $ $ If\space Q=0 \space at \space t=0 $ $show \space that$ $ i=\dfrac VRe^{-1/RC},i=\dfrac {dQ}{dt}$

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

Marks : 6

Year : 2015

$$\dfrac {RdQ}{dt}+\dfrac QC=V $$ $$ \therefore \dfrac {dQ}{dt}+\dfrac 1R\dfrac QC=\dfrac VR$$

Which is similar to

$\dfrac {dQ}{dt}+p'Q=Q'\\ where \space p'=\dfrac 1{RC} \space and \space Q'=\dfrac VR\\ I.f =e^{\int p'dt}\\ =e^{\int \dfrac 1{RC}dt}\\=e^{t/RC}\\ \text{Solution is}\\ Q.If=\int Q'If\space dt\\ Q.e^{t/RC}=\int \dfrac VRe^{t/RC}dt\\ =\dfrac VR\dfrac {e^{t/RC}}{1/RC} + c\\ Q.e^{t/RC}=VCe^{t/RC}+ c\\ \therefore Q=VC+ ce^{-t/RC}\\ \text{It is given that at } t = 0, Q = 0\\ 0=VC+c\\ C=-VC\\ Q=VC-VCe^{-t/RC}$

Differentiating w.r.t t we get

$i=\dfrac {dQ}{dt}=0-VCe^{-t/RC}\times \Big(-\dfrac 1{RC}\Big) \\ i=\dfrac VR e^{-t/RC}$