A 10-km-long underground cable extends east to west and consists of two parallel wires, each of which has resistance 13 Ω/km. An electrical short develops at distance xfrom the west end when a conducting path of resistance Rconnects the wires (Figure). The resistance of the wires and the short is then 100 Ω when measured from the east end and 200 Ω when measured from the west end. What are

(a) xand

(b) R?

$$\begin{gathered} L=\text{10km} \\ \alpha =13\frac{\text{Ω}}{\text{km}} \\ R\text{'}+R=100,\text{whenmeasuredfromeastend.} \\ R\text{'}+R=200\text{,whenmeasuredfromwestend.} \end{gathered}$$

Write two equations forthe total resistance when measured from east end and west end. Solving them will give the values of x and R.

Formulae are as follow:

$R=R\text{'}+R$

Where, R is resistance.

The total resistance when measured from east end is,

$$\begin{gathered} R_e=R\text{'}+R \\ {R}_e=2\alpha (L-x)+R \\ 100=2\left(13\right)(10-x)+R \\ 26x-R=160\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots 1) \end{gathered}$$

The total resistance when measured from west end is,

$$\begin{gathered} R_w=R\text{'}+R \\ {R}_w=2\alpha x+R \\ 200=2(13\times x)+R \\ 26x+R=200\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots 2) \end{gathered}$$

Adding 1) and 2), we get

$$\begin{gathered} 52x=360 \\ \therefore x=\text{6.9km} \end{gathered}$$

Hence, the value of x is $\text{6.9km}$

Inserting the value of x in equation 1) gives,

$$\begin{gathered} R=26\left(6.92\right)-160 \\ R=19.92~20\text{Ω} \end{gathered}$$

Hence, the value of R is $20\Omega .$

Therefore, find the resistance of the short from the total resistance of the wire from both the ends.