A capacitor $C_{1} (= C)$ is charged to a potential difference $ε$ is connected to a charging circuit by changing the switch S as shown. Assume the instant of switching as t=0. Capacitor $C_{2}$ is, initially uncharged, then the charge on $C_{2} (= C)$ is changing according to equation; $q_{2} = Q_{2} (1 - e^{- t / δ}) .$ Where, $δ$ is called time constant and $Q_{2}$ is the steady state charge on $C_{2}$ . Let $Q_{2} = n_{2} (C ε)$ and $δ = n_{1} (R C)$ .Find $(\frac{n_{2}}{n_{1}})$ .

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Detailed Solution

After switching, let charge q flows from $3 ε$ , then

KVL:

$I R + \frac{(C_{1} ε + q)}{C_{1}} + \frac{q}{C_{2}} = 3 ε = 0 I R + q (\frac{1}{C_{1}} + \frac{1}{C_{2}}) = 2 ε (\frac{d q}{d t}) R + q (\frac{1}{C_{e q}}) = 2 ε (\frac{d q}{d t}) R = (2 ε - \frac{q}{C_{e q}}) \int_{0}^{q} \frac{d p}{(2 ε - \frac{q}{C_{e q}})} = \frac{1}{R} \int_{0}^{t} d t - I n (\frac{2 ε - \frac{q}{C_{e q}}}{2 ε - 0}) = \frac{1}{R} t (1 - \frac{q}{2 ε C_{e q}}) = e^{t / R C_{e q}} q = 2 ε C_{e q} (1 - e^{- t / R C_{e q}}) Q_{2} = 2 ε C_{e q} = 2 ε (\frac{C}{2}) = C ε, n_{2} = 1 δ = R C_{e q} = R (\frac{C}{2}) = \frac{1}{2} R C, n_{1} = \frac{1}{2} (\frac{n_{2}}{n_{1}}) = 2$