Question

An underground cable 8.3 km long extends east to west and consists of two parallel wires, each of which has resistance 14 2/km. An electrical short develops at distance x from the west end when a conducting path of resistance R connects the wires (see the figure). The resistance of the wires and the short is 170 2 when measured from the east end and 110 when measured from the west end. What are (a) x and (b) R? Conducting path Fast West Units (a) Number (b) Number Units Click if you would like to Show Work for this question: Open Show Work

Answer 1

Given that

Total length of cable L = 8.3 km

resistance of each wire of cable Rc = 14 ohm/km

resistance of wire and short when measure from east end R1 = 170 ohm

resistance of wire and short when measure from west end R2 = 110 ohm

Let suppose there are two points P and Q on both wires where conducting path exists

R1 -110 ohm R2 = 170 ohm

So total resistance of left side part (APQB) of cable = 110 ohm

Rap + Rpq + Rqb = 110 ohm

Rap=Rqb = 14 ohm/ km = 14x

Rpq = R

so

14x +R +14x = 110

28x + R = 110 --------------------(1)

Now right side of cable

Rpc +Rcd +Rdq = 170 ohm

Rpc = Rdq = 14(8.3-x)

Rcd = R

so

14(8.3-x) + R + 14(8.3-x) = 170

R -28x + 232.4 = 170

28x - R = 62.4 ---------------------(2)

adding equation (1) and (2)

56x = 172.4

x = 3.07 km

putting this value in equation (1)

28(3.07) +R = 110

R = 24.04 ohm

Good Luck !