Calculate the steadystate current in the `2Omega` resistor shown in the circuit in the figure. The internal resistance of the battery is negligible and the capacitance of the condenser C is `0.2 muF`.

Answer

0.9

Explanation

In steady state situation no current will flow through the capacitor `2 Omega and 3Omega` are in parallel.
`R=(2xx3)/(2+3)=1.2Omega ,` Net Current through the battry `i=6/(1.2+2.8)=1.5A`
The current will distribute in inverse ratio of their risistance in ` 2Omega and 3Omega`
`therefore (i_2)/(i_3)=3/2 ori_2=(3/(3+2))(1.5)=0.9 A`