A 12.1 Ω metal wire is cut into three equal pieces which are then connected side by side to form a new wire (with length one-third the original length). What is the resistance of this new wire? Answer in units of Ω.

The resistance R, of a wire is directly proportional to the length (L) of the wire and inversely proportional to the cross sectional area (A) of the wire as follows;

R ∝ [tex]\frac{L}{A}[/tex]

R = p [tex]\frac{L}{A}[/tex] -------------------(i)

Where;

p = proportionality constant = resistivity of the wire

Now;

let the resistance of the wire before being cut be R₀ = 12.1Ω

Hence, equation (i) becomes;

R₀ = p [tex]\frac{L}{A}[/tex] -----------------------(ii)

Moving forward;

Since the wire is cut into three equal pieces, then the resistance in each of the wires are the same. Also, the length of each wire is one-third of the original length i.e [tex]\frac{L}{3}[/tex]

let the resistance of the each piece of cut wire be Rₓ

Hence, equation (i) becomes;

Rₓ = p [tex]\frac{L}{3A}[/tex] ---------------------------(iii)

Comparing equations (ii) and (iii), equation (iii) can be re-written as;

Rₓ = [tex]\frac{R_{0} }{3}[/tex] ------------------(iv)

Therefore, each of the pieces of wire will have a resistance of Rₓ = [tex]\frac{R_{0} }{3}[/tex].

Now, since the pieces of wire are connected side by side to form a new wire, then the wires are connected in parallel to one another. Therefore, the resistance (Rₙ) of the new wire is found by finding the effective resistance of the individual resistances (Rₓ) of the three wires which are connected in parallel as follows;

[tex]\frac{1}{R_{n}}[/tex] = [tex]\frac{1}{R_{x} }[/tex] + [tex]\frac{1}{R_{x} }[/tex] + [tex]\frac{1}{R_{x} }[/tex]

[tex]\frac{1}{R_{n}}[/tex] = [tex]\frac{3}{R_{x} }[/tex]

Rₙ = [tex]\frac{R_x}{3}[/tex] -------------------------(v)

But from equation (iv), Rₓ = [tex]\frac{R_{0} }{3}[/tex]

Substitute this value into equation (v) as follows;

Rₙ = [tex]\frac{R_{0} }{3*3}[/tex]

Rₙ = [tex]\frac{R_{0} }{9}[/tex] ----------------------(vi)

Now substitute the value of R₀ = 12.1Ω into equation (vi) as follows;

Rₙ = [tex]\frac{12.1}{9}[/tex]

Rₙ = 1.34Ω

Therefore, the resistance of this new wire is 1.34Ω