A Varley Bridge is connected to a faulty three-core copper cable by two identical copper leads of resistance Rl.
(a) Show that for the initial reading (connection to earth);
2Rx = 2Rc – Ri .............................. 1
where Rc is the resistance of the cable core Ri is the initial reading of the bridge
Rx is the cable resistance to the fault from the bridge
and for the final reading:
2Rc = R f – 2Rl .............................. 2
where Rl is a lead resistance
Rf is the final reading resistance.
Then by substituting (2) in (1) and rearranging the equation, show:
Rx + Rl = R f R–f Ri Rc + Rl
(b) By multiplying the rhs brackets and collecting terms, show the effect of the leads is given by:
R = Rf–Ri R – Rl Ri
x c Rf
i.e. = effect with no leads – ratio of initial and final readings
× lead resistance
(c) Determine the distance to the fault by modifying the expression in (b) and using Rx = x
where x is the cable distance to the fault and L is the length of a cable core.
Derive an expression for x, the distance to the fault.
(d) Using R=ρL/A
where ρ is the resistivity,
L is the length
and A is the cross-sectional area of a cable core
The Attempt at a Solution
I can work out all the way to section d, but then have no idea what formula to start with