A capacitor of capacity `1muF` is connected in a closed series circuit with a resistance or `10^(7)` ohms, an open key and a cell of 2 V with negligible internal resistance :
(i) When the key is switched on at time t=0, find,
(a) The time constant for the circuit.
(b) The charge on the capacitor at steady state.
(c) Time taken to deposit charge equal to half of charge that will deposit at steady state.
(ii) If after completely charging the capacitor, the cell is shorted by zero resistance at time t=0, find the charge on the capacitor at t=50 s. (Given : `e^(-5)=6.73xx10^(-3), ln^(2)=0.693`)

Answer

`(i) (a) 10s (b) 2 mu C " "(c) 10 In 2 = 6.93 sec`.
`(ii) q=(2e^(-5))mu C=1.348xx10^(-8) C`