Question-55725

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.