Question

A wire of 15 Ω resistances 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 300 volt battery. Find the

current drawn from the battery.​

Answer 1

Explanation:

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

A=A0/2

Resistance R=ρl/A

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