1.

In precious question the switch is kept closed for a long time and then assume it is opened at t = 0. Write the expression of charge on the capacitor as a function of time.

Answer»

`q = 2CV E^(- t//3RC)`
`q = CV e^(- t//3RC)`
`q = CV e^(- t//2RC)`
`q = CV e^(- 2t//3RC)`

Solution :After the CAPACITOR is FULLY CHARGED then the amount of charge with it is Q=CV and it will be initial charge on capacitor when the switch is opened. When it is opened then the upper loop makes a discharging circuit and its time constant becomes 3RC. Using the expression of charge for discharging `q= q_(0) e^(-t//tau)` we can write the following expression. `q= CVe^(-t//3RC)` Hence OPTION (b) is correct


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