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Question-55725




Question Number 55725 by Tinkutara last updated on 03/Mar/19
Answered by ajfour last updated on 03/Mar/19
   short-circuited case    ((LdI)/dt)=−IR  ⇒ ∫_I_0  ^( I) (dI/I) = −(R/L)∫_0 ^(  t) dt  ⇒  I = I_0 e^(−Rt/L)     ⇒  q_s =I_0 ∫_0 ^( ∞) e^(−Rt/L) dt             = I_0 ((L/R))  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_0 ((L/R))      (the same).
$$\:\:\:{short}-{circuited}\:{case} \\ $$$$\:\:\frac{{LdI}}{{dt}}=−{IR} \\ $$$$\Rightarrow\:\int_{{I}_{\mathrm{0}} } ^{\:{I}} \frac{{dI}}{{I}}\:=\:−\frac{{R}}{{L}}\int_{\mathrm{0}} ^{\:\:{t}} {dt} \\ $$$$\Rightarrow\:\:{I}\:=\:{I}_{\mathrm{0}} {e}^{−{Rt}/{L}} \\ $$$$\:\:\Rightarrow\:\:{q}_{{s}} ={I}_{\mathrm{0}} \int_{\mathrm{0}} ^{\:\infty} {e}^{−{Rt}/{L}} {dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:{I}_{\mathrm{0}} \left(\frac{{L}}{{R}}\right) \\ $$$${had}\:{the}\:{circuit}\:{not}\:{been}\:{short} \\ $$$${circuited}\:{and}\:{current}\:{had} \\ $$$${remained}\:{constant},\:{then}\:{in} \\ $$$${one}\:{time}\:{constant}\:{charge}\:{that} \\ $$$${would}\:{have}\:{passed}\:{is} \\ $$$$\:\:\:\:\:\:\:{q}_{{c}} \:=\:{I}_{\mathrm{0}} \left(\frac{{L}}{{R}}\right)\:\:\:\:\:\:\left({the}\:{same}\right). \\ $$
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.

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