(a)Find the current I supplied by the battery in the network shown in figure in steady state. (b)find the charge on the capacitor.

Explanation

(a)Once the capacitor is charged, no current will go through it and hence the current through the middle branch of the circuit is zero in steady state. The `(4Omega)`resistor will have no current in it and may be omitted for current analysis. The`(2Omega) and (6Omega)`resistors are, therefore, connected in series and hence
`i-(2V)/(2Omega+6Omega)=0.25A.`
(br)(b) The potential drop across the `(6Omega)`resistor is `(6Omegaxx0.25A=1.5V)`.As there is no current in the (4Omega) resistor, there is no potential drop across it.The potential difference across the capacitor is,therefore `1.5V`.The charge on this capacitor is
`Q=CV=2(mu)Fxx1.5V=3(mu)C.`