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

Commented by ajfour last updated on 19/May/18

Solution to Q.35493

$${Solution}\:{to}\:{Q}.\mathrm{35493} \\ $$

Answered by ajfour last updated on 19/May/18

(x+z)R+zR=yR= ξ      ...(i)  ⇒   xR+2zR=ξ  (q/C)=zR     ...(ii)  using (i) in (ii):  xR+((2q)/C) = ξ    and   as  x=(dq/dt)        (dq/dt) = (1/R)(ξ−((2q)/C))  ⇒    ∫_0 ^(  q)  (dq/(ξ−((2q)/C))) = ∫_0 ^(  t)  (dt/R)  ⇒    ln (ξ−((2q)/C))∣_0 ^q = ((−2t)/(RC))  or        ((ξ−((2q)/C))/ξ) = e^(−((2t)/(RC)))   ⇒    q = ((Cξ)/2)(1−e^(−((2t)/(RC))) ) .

$$\left({x}+{z}\right){R}+{zR}={yR}=\:\xi\:\:\:\:\:\:...\left({i}\right) \\ $$$$\Rightarrow\:\:\:{xR}+\mathrm{2}{zR}=\xi \\ $$$$\frac{{q}}{{C}}={zR}\:\:\:\:\:...\left({ii}\right) \\ $$$${using}\:\left({i}\right)\:{in}\:\left({ii}\right): \\ $$$${xR}+\frac{\mathrm{2}{q}}{{C}}\:=\:\xi\:\:\:\:{and}\:\:\:{as}\:\:{x}=\frac{{dq}}{{dt}} \\ $$$$\:\:\:\:\:\:\frac{{dq}}{{dt}}\:=\:\frac{\mathrm{1}}{{R}}\left(\xi−\frac{\mathrm{2}{q}}{{C}}\right) \\ $$$$\Rightarrow\:\:\:\:\int_{\mathrm{0}} ^{\:\:{q}} \:\frac{{dq}}{\xi−\frac{\mathrm{2}{q}}{{C}}}\:=\:\int_{\mathrm{0}} ^{\:\:{t}} \:\frac{{dt}}{{R}} \\ $$$$\Rightarrow\:\:\:\:\mathrm{ln}\:\left(\xi−\frac{\mathrm{2}{q}}{{C}}\right)\mid_{\mathrm{0}} ^{{q}} =\:\frac{−\mathrm{2}{t}}{{RC}} \\ $$$${or}\:\:\:\:\:\:\:\:\frac{\xi−\frac{\mathrm{2}{q}}{{C}}}{\xi}\:=\:{e}^{−\frac{\mathrm{2}{t}}{{RC}}} \\ $$$$\Rightarrow\:\:\:\:{q}\:=\:\frac{{C}\xi}{\mathrm{2}}\left(\mathrm{1}−{e}^{−\frac{\mathrm{2}{t}}{{RC}}} \right)\:. \\ $$

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