Question

A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one piece is 7.0 times the resistance of the other
Determine the length of the short piece.(% of length of the wire)?
Determine the resistance of the short piece?
Determine the resistance of the long piece?

Answer 1

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R=$13\Omega$

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper  wire=Sum of resistance of two piece of wires

$r+7r=13$

$8r=13$

$r=\frac{13}{8}$ohm

Resistance of long piece of wire=$7\times\frac{13}{8}=\frac{91}{8}\Omega$

Resistance of short piece of wire =$\frac{13}{8}\Omega$

Resistivity of wire and cross section area of wire remains same .

Let L be the total  length  of wire and L' be the length of short  piece of wire.

We know that

$R=\frac{\rho L}{A}$=$\frac{\rho}{A}L=KL$

$\frac{R}{L}=K$

Where K=$\frac{\rho}{A}$=Constant

Using the formula

$\frac{13}{L}=\frac{\frac{13}{8}}{L'}$

$\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}$

$L'=\frac{L}{8}$

Length of short piece of wire=L'=$\frac{L}{8}$

Length of long piece of  wire=$L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L$

% of length of short piece of   wire=$\frac{\frac{L}{8}}{L}\times 100=12.5%$%

The resistance of the short piece=$\frac{13}{8}\Omega$

The resistance of the long piece=$\frac{91}{8}\Omega$