a wire of 15 ohm resistance is gradually stretched to double its original length. it is then cut into two equal parts. these parts are connected in parallel across a 3 V cell. find the current drawn from the cell
Let R0, l0, A0 be the original resistance, length and area of the wire.
Let R, l, A be the original resistance, length and area of the stretched wire. Since the length is doubled ; l =2l0
When the wire is stretched its Volume remains the same.
Therefore, V= Al = A0l0
A(2l0) = A0l0
R/R0 = lA0/l0A
R/R0 = 2l0 A0/ (l0 (A0/2)) = 4
R = 4R0 = 4x15 = 60Ω
When the resistance of wire R=60Ω is cut into two equal parts, the length becomes half but area remains the same. Hence resistance of each part is R1= R/2 = 60/2 = 30Ω
When these two wires of resistance 30Ω are connected in parallel across a 3V battery, the equivalent resistance Rp = R1/2 = 30/2 =15 Ω
(1/Rp = (1/ 30) +( 1/ 30) = 2/30 = 1/15
Rp =15Ω )
Therefore the current through the circuit is V/Rp = 3/15 = 0.2A