A wire of 15 `Omega` resistance is gradually stretched to double its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 3.0 volt battery. Find the current drawn from the battery.

Explanation

Here R = 15 `Omega`. When the wire is gradually stretched to double its original length, its cross-section area is reduced to half of its original value so that the volume remains same. Hence l = 21 and A. = A/2.
` therefore ` New resistance of wire `R. = (rho l.)/(A.) = (rho . (2l))/((A/2)) = 4 (rho l)/(A) = 4R = 4 xx 15 = 60 Omega`
On cutting the wire into 2 equal parts, resistance of each part `R_1 = R_2= (R.)/(2) = 30 Omega`
On joining the two parts in parallel, equivalent resistance
`R_(eq) = (R_1 R_2)/(R_1 + R_2) = (30 xx 30)/(30 + 30) = 15 Omega`
If emf of battery be `epsi` = 3.0 V, then current drawn from the battery
`I = (epsi)/(R_(eq)) = (3.0 V)/(15Omega) = 0.2 A`