Question

An overhead transmission cable for electrical power is 2000 m long and consists of two parallel copper wires, each encased in insulating material. A short circuit has developed somewhere along the length of the cable where the insulation has worn thin and the two wires are in contact. As a power company employee, you must locate the short so that repair crews can be sent to that location. Both ends of the cable have been disconnected from the power grid. At one end of the cable (point A), you connect the ends of the two wires to a$\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{V}$ battery that has negligible internal resistance and measure that 2.86 A of current flows through the battery. At the other end of the cable (point B ), you attach those two wires to the battery and measure that 1.65 A of current flows through the battery. How far is the short from point A ?

Short answer

The distance between the short circuit from the point A is 732.5 m.

Step-by-step solution

Define the ohm’s law and resistance ( R ).

According to Ohm’s law, the current flowing through the conductor is directly proportional to the voltage across the two points.

$V=IR$

Where, Iis current in ampere $\left(A\right)$, R is resistance in ohms$\left(\mathrm{\Omega }\right)$ and V is the potential difference volt$\left(V\right)$ .

Also,$V=EL$

Where, $E$is the electric field across conductor and $L$is length of the conductor.

The ratio of V to I for a particular conductor is called its resistance R :

$R=\frac{V}{I}$or$\frac{\rho L}{A}$

Where,$\rho$ is resistivity role="math" localid="1664170807826" $\Omega ·\mathrm{m}$, $L$is length in m and A is area in $m^2$.

Determine the distance between the point .

Given that, the length $L$of two wires is 2000 m, the current $I_A$at point is A, current $I_B$at point B is 1.65 A and the voltage V at both points is 9.0 V . Consider short circuit at point C , length between the points A and C is $L_{CA}$and distance between the point B and C is $L_{CB}$.

So,$L_{CB}=2000-L_{CA}$

And${\rho }_{CA}={\rho }_{CB}$

$$\begin{gathered} \frac{R_{CA}A}{L_{CA}}=\frac{R_{CB}A}{L_{CB}} \\ \frac{R_{CB}}{L_{CA}}=\frac{R_{CB}}{2000-L_{CA}} \\ {R}_{CA}\left(2000-L_{CA}\right)=L_{CA}{R}_{CB} \\ {L}_{CA}=\frac{2000R_{CA}}{R_{CA}+R_{CB}}.......\left(1\right) \end{gathered}$$

Substitute the values in equation

$$\begin{gathered} L_{CA}=\frac{2000R_{CA}}{R_{CA}+R_{CB}} \\ =\frac{2000\left({\displaystyle \frac{V}{I_A}}\right)}{\left({\displaystyle \frac{V}{I_A}}\right)+\left({\displaystyle \frac{V}{I_B}}\right)} \\ =\frac{2000\left(9.0/2.86\right)}{\left(9.0/2.86\right)+\left(9.0/1.65\right)} \\ =732.5\mathrm{m} \end{gathered}$$

Hence, the distance between the short circuit from the point A is $732.5\mathrm{m}$.