A capacitor of capacity 5F is charged by a battery of 2 volt and then connected with 5 uncharged capacitors and 5 resistors with the help of a switch ‘S’ as shown in the figure. At t = 0, switch is closed. Choose the correct option(s).

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Charge on capacitor X at time t is given by $\frac{10}{9} (1 - e^{- \frac{27}{160} t})$

Current in resistor Y at time t is given by $\frac{1}{4} (e^{- \frac{27}{160} t})$

Total energy dissipated in the circuit is $\frac{50}{9} J$

Total energy dissipated in the circuit is $\frac{40}{9} J$

answer is A, B, D.

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Detailed Solution

At time 't'

$- \frac{q}{4} - i \times \frac{8}{3} + \frac{10 - q}{5} = 0 \int_{0}^{q} \frac{\frac{d q}{9}}{2 - \frac{9}{20} q} = \int_{0}^{t} \frac{3}{8} d t q = \frac{40}{9} [1 - e^{- \frac{27}{160} t}] U_{i} = \frac{{(10)}^{2}}{2 \times 5} U_{f} = \frac{{(40 / 9)}^{2}}{2 \times 4} + \frac{{(50 / 9)}^{2}}{2 \times 5} Δ U = U_{i} - U_{f} = \frac{40}{9} J$