**Change of Resistance in Wires by Stretching**

**Change of Resistance in Wires by Stretching**

We shall understand with an example how the resistance of a wire changes with stretching its length.

**Example-1 –**Change of Electrical Resistance with wire stretching

### The resistance of L length wire is R. If half of its length stretch such as the total length becomes 2L what will be the final resistance?

The resistance of L length wire is R.

The resistance of L/2 length wire is R/2

Now the other L/2 length of the wire is stretched to length X to have full length of conductor 2L

L/2 +X = 2L

X= 3/2 L

The length of the stretched conductor is 1.5 L.

Length of conductor L/2 is stretched up to 1.5 L, the length of the stretch conductor is now 3 times of the original length hence the diameter of the conductor is d/3. The resistance of the stretched conductor.

**R1= ⍴ x L/A **

**= ⍴ x L/πd ^{2} **

**= 3 /(1/9)**

**R1= 27 times of the original resistance of the half length wire**

R1= 27 x R/2 =13.5 R

The resistance of conductor after stretching is

= R/2 + 13.5 R=

= 14 R

**The new resistance of the wire after stretching its half length is 14 times of the original resistance of conductor.**

**Example-2 **

**What is the new resistance of a wire if it is stretched to twice its original length. Original resistance is 10 ohm. **

After stretching of wire

Length of wire L_{1} = 2 L

New Area A_{1} = A/2

In above conditions,the volume of the wire remain the same.

New Resistance R1 = ρ L_{1} /A_{1}

= ρ x 2L /A/2

= ρ x 4L /A

R1 = 4 ρ L /A

= 4 R = 4 x 10

R1 = 40Ω

New resistance will be 4 times of the original resistance

**Example-3 **

### A wire has a resistance of 30 ohms. We stretch the wire by 10% of its original length, what will be the new resistance?

Let the original length and area of a wire is L and A respectively.

The resistance of wire R = ρ L /A

After 10 % stretching the length and area of wire is L_{1} and A_{1}

L_{1} = 1.1 L

The volume must remain the same after stretching

A_{1} = (1/1.1 )A = 0.909 A

The new resistance of wire

R_{1} = ρ L_{1/A1}

= ρ (1.1L)/ (0.909 A)

R_{1} = 1.21 ρ L /A = 1.21R

= 1.21 x 30

** R _{1} = 36.3 Ω**

The above illustrative examples shows how the resistance of a wire changes with stretching its length.