In the circuit shown in the figure, find the total resistance of the circuit and the current in the arm CD.

Explanation

At a steady of capacitor in branch CE, the current will not pass. So, the resistor EF will not be considered. Resistance in branches BC and CD are in series. So, their equivalent resistance will be
`R. = 3+3 = 6Omega`
Net resistance across branch AD: `R_(AD) = (3R.)/(3+ R.) = (3 xx 6)/(3+6) = 2Omega`
Total resistance, `R_("eqvt") = 2+ 3= 5Omega`
Net current in this network will be.
`I= (E )/(R_("eqvt")) = (15)/(5) = 3A`