Question-35496

Question Number 35496 by ajfour last updated on 19/May/18

Commented by ajfour last updated on 19/May/18

S o l u t i o n t o Q . 35493

Answered by ajfour last updated on 19/May/18

(x + z) R + z R = y R = ξ \dots (i)

\Rightarrow x R + 2 z R = ξ

\frac{q}{C} = z R \dots (i i)

u s i n g (i) i n (i i) :

x R + \frac{2 q}{C} = ξ a n d a s x = \frac{d q}{d t}

\frac{d q}{d t} = \frac{1}{R} (ξ - \frac{2 q}{C})

\Rightarrow \int_{0}^{q} \frac{d q}{ξ - \frac{2 q}{C}} = \int_{0}^{t} \frac{d t}{R}

\Rightarrow \ln (ξ - \frac{2 q}{C}) ∣_{0}^{q} = \frac{- 2 t}{R C}

o r \frac{ξ - \frac{2 q}{C}}{ξ} = e^{- \frac{2 t}{R C}}

\Rightarrow q = \frac{C ξ}{2} (1 - e^{- \frac{2 t}{R C}}) .

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