Calculate the equivalent of the network across the points `A and B` shown in

Explanation

The given network, shown in (Fig. 3.13) (a), can be represented as shown in (Fig. 3.13) (b) The resultant of `4 Omega and 8 Omega` resistances, which are in series in the arm `EDC`, is givn by
`R_1 = 4 Omega + 8 Omega = 12 Omega`
Similarly, the resultant of `4 Omega and 8 Omega` in the EFC is given by
`R_2 = 4 Omega + 8 Omega = 12 Omega`
The equivslent circuit of `R_1 and R_2` which are in parallel is shown in (Fig. 3.13)( c)
If `R` is the equivalent resistance between the points `A and B`, then
`(1)/(R ) = (1)/(R_1) + (1)/(R_2) = (1)/(12) + (1)/(12) = (2)/(12) = (1)/(6)`
or `R = 6 Omega`.
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