Question

Calculate equivalent resistance in the following cases:

Answer 1

(i) Resistance P and Q are in series = 3 Ω + 3 Ω = 6 Ω

Resistance of 6 Ω segment and 3 Ω are parallel.

∴ `1/"R" = 1/6 + 1/3 = (1 + 2)/6 = 3/6` or R = 2Ω

(ii) Resistance of the arm PS = 2Ω +2Ω = 4Ω (in series)

Resistance of the arm QR = 2Ω + 2Ω = 4Ω (in series)

Resistance of the arm PQ, QR and RS are in series

= (4 + 4 + 4)Ω = 12Ω

Now resistance of the arm RS and arms (PQ + QR + RS) are in parallel.

`1/"R" = 1/4 + 1/12`

`= (3 + 1)/12 = 4/12`

or R = 3Ω