Calculate the equivalent resistances between the points `A and B` of the circuits shown in (fig. 2.33)(b).
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Answer

(a) `2.4 Omega`
(b) `4.8 Omega`

Explanation

In (Fig. 3.22) (a), total resistance in the path `ACB`, i.e., `R_1 = 8.5 Omega + 3.5 Omega = 12 Omega`
Since `R_1 and R_2` are in parallel, the effective resistance between the points `A and B`, i.e.,
`R = (R_1 R_2)/(R_1 + R_2) = (3 xx 12)/(3 + 12) Omega = 2.4 Omega`
In (Fig. 3.22)(b), total resistance in the path `ADCB`, i.e., `R_1 = (8 + 9 + 7) Omega = 24 Omega`
Further, `R_1 (=24 Omega) and 6 Omega` are in parallel between the points `A and B`.
Thus, effective resistance between `A and B`, i.e., `R = (24 xx 6)/(24 + 6) Omega = 4.8 Omega`.