For the combination of resistors shown in (fig.)
(a) find the equivalent resistance between
(i) C and D and
(ii) A and B.
.

Explanation

(i) The resistors `R_2,R_3 and R_4` are in series. Their equivalent resistance `R_s` is given by
`R_s = 3 Omega + 3 Omega + 3 Omega`
=`9 Omega`
Further, `R_s (= 9 Omega) and R_5(= 3 Omega)` are in parallel, (Fig. 3.17) (b) and their equivalent resistance is given by
`R_p = ("product of two resistances")/("sum of two resistances")`
=`(9 xx 3)/(9 + 3) omega = (27)/(12) Omega= 2.25 Omega`
Thus, the equivalent resistance between C and D is `2.25 Omega`.
(ii) Now `R_1 (= 3 Omega), R_p (=2.25 Omega)`
and `R_6 (=3 Omega)`
are in series, (Fig. 3.17).
Thus,equivalent resistance between A and B, (Fig. 3.17)(d),
=`R_1 + R_p + R_6`
=`3 Omega + 2.25 Omega + 3 Omega`
=`8.25 Omega`
.