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Question Number 55725 by Tinkutara last updated on 03/Mar/19

	Answered by ajfour last updated on 03/Mar/19

	$s h o r t - c i r c u i t e d c a s e$ $\frac{L d I}{d t} = - I R$ $\Rightarrow \int_{I_{0}}^{I} \frac{d I}{I} = - \frac{R}{L} \int_{0}^{t} d t$ $\Rightarrow I = I_{0} e^{- R t / L}$ $\Rightarrow q_{s} = I_{0} \int_{0}^{\infty} e^{- R t / L} d t$ $= I_{0} (\frac{L}{R})$ $h a d t h e c i r c u i t n o t b e e n s h o r t$ $c i r c u i t e d a n d c u r r e n t h a d$ $r e m a i n e d c o n s t a n t, t h e n i n$ $o n e t i m e c o n s t a n t c h a r g e t h a t$ $w o u l d h a v e p a s s e d i s$ $q_{c} = I_{0} (\frac{L}{R}) (t h e s a m e) .$
	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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