Calculate the equivalent resistance between the points `A and B` in the circuits shown in (Fig. 3.19)
(a) and (Fig. 3.19) (b).
.

Answer

(a) `4 Omega`
(b) `6 Omega`.

Explanation

(a) Two resistances `(6 Omega,6 Omega)` in the upper arm are in series and their resultant resistance `= 6 Omega + 6 Omega = 12 Omega`. Further, `12 Omega and 6 Omega` (in the lower arm) are in parallel. The resultant resistance of `12 Omega and 6 Omega` in parallel is `(12 xx 6)/(12 + 6) Omega = 4 Omega`.
(b) The resultant resistance of `6 Omega,6 Omega` which are in parallel in the first loop is `(6 xx 6)/(6 + 6) Omega = 3 Omega`. Similarly, the resultant resistance in the second loop is also `3 Omega`. Since the two loops are in series, the total resistance between A and B is `3 Omega + 3 Omega = 6 Omega`.