Find equivalent resistance of the network in fig. Between points (i) A and B and (ii) A and C.

Explanation

(i) The `10Omega` and `30Omega` resistors are connected in parallel between points A and B. The equivalent resistance between A and B is
`R_(1)=(10xx30)/(10+30)ohm=7.5Omega`
(ii) The resistance `R_(1)` is connected in series with resistor of `7.5Omega` hence the equivalent resistance between point A and C is `R_(2)=(R_(1)+7.5)ohm=(7.5+7.5)ohm=15Omega`.