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A capacitor of capacitance C is connecte...

A capacitor of capacitance C is connected to a battery of emf `(epsilon)`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?

Text Solution

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Let at any time,
` q = epsilon c ( 1- e^(-t/rc))`
` E = energy stored `
` = q^2/2c = (epsilon^2 C^2/2C)(1- e^(-t/rc))^2`
` = (1 + e^(-t/rc))^2`
` r = rate of energy stored `
` = (depsilon/dT) = (-epsilon^2 C /2) dr/dt `
` = (1-e^(-t/rc))`
` dr/dt e^(-t/rc)= (depsilon/2r) (-1/rC^2) (e^(-t/rc)(1-e^(-t/rc))`
` epsilon^2/2r = [(e^(-t/rc))(1-e^(-t/rc)) + (-).(e^(-t/rc)) (-1/rc)(e^(-t/rc))]`
` epsilon/2r = (epsilon/2r + epsilon/2r + 1/2r (e^(-2t/cr)))`
` = (2/rC)e^(-2t/cr) epsilon/rC)`
` For r_max' dr/dt = 0 `
` rArr 2. (e^(-t/rc) = 0`
` rArr e^(-t/rc)= 1/2 `
` rArr = -t/rc = -In 2 rArr t= rc In 2 `
` Putting t = In 2 in equation (i) `
` we get, dr/dt = epsilon^2/4r`.
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