A capacitor of capacitance $$C$$ is connected to a battery of emf $$\varepsilon $$ at $$t=0$$ through a resistance $$R$$. Find the maximum rate at which energy is stored in the capacitor. When does the rate has this maximum value?
Let at any time $$t, \ q = EC (1 - e^{-t/CR})$$
$$E=$$ Energy stored $$= \dfrac{q^2}{2c} = \dfrac{E^2C^2}{2c} ( 1- e^{-t/CR})^2 = \dfrac{E^2C}{2} ( 1 - e^{-t/CR})^2$$
$$R=$$ rate of energy stored = $$\dfrac{dE}{dt} = \dfrac{-E^2C^2}{2} \left( \dfrac{-2}{RC}\right) ( 1 - e^{-t/RC})e^{-t/RC} = \dfrac{E^2}{CR}.e^{-t/RC} ( 1 - e^{-t/CR})$$
$$\dfrac{dR}{dt} = \dfrac{E^2}{2R} \left[\dfrac{-1}{RC} e^{-t/CR} . ( 1 - e^{-t/CR}) + (-) . e^{-t/CR}{ (-1 /RC)} . e^{-t/CR}\right]$$
$$\dfrac{dR}{dt}=\dfrac{E^2}{2R} \left( \dfrac{-e^{-t/CR}}{RC} + \dfrac{e^{-2t/CR}}{RC} + \dfrac{1}{RC} . e^{-2t/CR}\right) = \dfrac{E^2}{2R} \left( \dfrac{2}{RC} . e^{-2t/CR} - \dfrac{e^{-t/CR}}{RC}\right)$$ ...(1)
For $$R_{max} dR / dt = 0 \Rightarrow 2.e^{-t/RC} - 1 = 0 \Rightarrow e^{-t/CR} = 1/2$$
$$\Rightarrow -t/RC = - ln^2 \Rightarrow t = RC \ in \ 2$$
$$\therefore$$ Putting $$t = RC \ ln \ 2$$ in equation (1) We get $$\dfrac{dR}{dt} = \dfrac{E^2}{4R}$$
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