When a potential difference of `2 V` is applied across the ends of a wire of `5 m` length, a current of `1 A` is found to flow through it. Calculate
(i) the resistance per unit length of the wire,
(ii) the resistance of `2 m` length of the wire.
(iii) the resistance across the ends of the wire if it doubled on itself.

Answer

(i) `0.4 Omega//m`
(ii) `0.8 Omega`
(iii) `0.5 Omega`.

Explanation

(i) `R = (V)/(I) = (2 V)/(1 A) = (2 V)/(1 A) = 2 Omega`. Resistance//unit length `= (2 Omega)/(5 m) = 0.4 Omega//m`
(ii) Resistance of `2 m` length of the wire `= 0.4 xx2 = 0.8 Omega`
(iii) When the wire is doubled on itself, it will be equivalent to two resistances (each of value `1 Omega`) parallel to each other. The resultant resistance across the ends of this folded wire `= (1 Omega)/(2) = 0.5 Omega`.