Consider the circuit shown in fig 5.174 where a battery of emf 4V and a capacitor of capacitance 1 `muF` is connected to a combination of resistacnes. Find out the stedy-state current in the battery.

Explanation

Let us consider I to be the steady-state current through
the cirucuit. In the steady-state condition , current I is constant
in the circuit and the capacitor offers infinite resistance. So the
resistance `10Omega` becomues ineffective in the circuit. So in this
case, the euqivalent resistance of resistors `2Omega and 4Omega` that are
connected in parallel is
`(1)/(R_P) = (1)/(2) +(1)/(4) = (3)/(4) or R_P = (4)/(3)Omega`
Therefore, the total resistance of the circuit is
`(4)/(3) +(2)/(3) +(2)/(3) = (6)/(3) = 2Omega`
Hence, the steady state current in the circuit is `4/2 = 2A.`