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

Commented by ajfour last updated on 19/May/18

$$SolutiontoQ.35493$$

Answered by ajfour last updated on 19/May/18

$$(x+z)R+zR=yR=\xi ...\left(i\right)$$$$\Rightarrow xR+2zR=\xi$$$$\frac{q}{C}=zR...\left(ii\right)$$$$using\left(i\right)in\left(ii\right):$$$$xR+\frac{2q}{C}=\xi andasx=\frac{dq}{dt}$$$$\frac{dq}{dt}=\frac{1}{R}(\xi -\frac{2q}{C})$$$$\Rightarrow {\int }_0^q\frac{dq}{\xi -\frac{2q}{C}}={\int }_0^t\frac{dt}{R}$$$$\Rightarrow \ln (\xi -\frac{2q}{C}){\mid }_0^q=\frac{-2t}{RC}$$$$or\frac{\xi -\frac{2q}{C}}{\xi }=e^{-\frac{2t}{RC}}$$$$\Rightarrow q=\frac{C\xi }{2}(1-e^{-\frac{2t}{RC}}).$$

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