Question Number 55725 by Tinkutara last updated on 03/Mar/19
Answered by ajfour last updated on 03/Mar/19
$$\:\:\:{short}-{circuited}\:{case} \\ $$$$\:\:\frac{{LdI}}{{dt}}=−{IR} \\ $$$$\Rightarrow\:\int_{{I}_{\mathrm{0}} } ^{\:{I}} \frac{{dI}}{{I}}\:=\:−\frac{{R}}{{L}}\int_{\mathrm{0}} ^{\:\:{t}} {dt} \\ $$$$\Rightarrow\:\:{I}\:=\:{I}_{\mathrm{0}} {e}^{−{Rt}/{L}} \\ $$$$\:\:\Rightarrow\:\:{q}_{{s}} ={I}_{\mathrm{0}} \int_{\mathrm{0}} ^{\:\infty} {e}^{−{Rt}/{L}} {dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:{I}_{\mathrm{0}} \left(\frac{{L}}{{R}}\right) \\ $$$${had}\:{the}\:{circuit}\:{not}\:{been}\:{short} \\ $$$${circuited}\:{and}\:{current}\:{had} \\ $$$${remained}\:{constant},\:{then}\:{in} \\ $$$${one}\:{time}\:{constant}\:{charge}\:{that} \\ $$$${would}\:{have}\:{passed}\:{is} \\ $$$$\:\:\:\:\:\:\:{q}_{{c}} \:=\:{I}_{\mathrm{0}} \left(\frac{{L}}{{R}}\right)\:\:\:\:\:\:\left({the}\:{same}\right). \\ $$
Commented by Tinkutara last updated on 03/Mar/19
But before short circuiting the total charge would be 1/e that written in last line? Then the two will not be equal.