Question : A Varley Bridge is connected to a faulty three-core copper cable by two identical copper leads of resistance R_l.

Q1) Show that for the initial reading (connection to earth);

2R x = 2R c - Ri ............................. (1)

Where Rc is the resistance of the cable core

Riis the initial reading of the bridge

Rx is the cable resistance to the fault from the bridge

And for the final reading:

2R c = 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 = (Rf - Ri) / Rf x (Rc + Rl)

Q2) By multiplying the rhs brackets and collecting terms, show the effect of the leads is given by:

Rx = (Rf - Ri) / Rf x Rc - (Rl x Ri) / Rf

i.e. = effect with no leads - ratio of initial and final readings × lead resistance

Q3) Determine the distance to the fault by modifying the expression in (b) and using

Rx / Rc = x / L

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.

Q4) Using:

R = ρL / A

Where ρ is the resistivity,

L is the length

And A is the cross-sectional area of a cable core

show that R_l = R_cA_cL_l/A_lL_c, (where L_c =L)

and by substituting for R_l, show that the distance to the fault is given by:

x = [(R_f - R_i)/R_f - R_i/R_f A_cL_l/A_l,L]L

Q5) Determine the distance to a fault on a 200 m. 120 mm² copper cable if copper 10 mm² test leads of length 10 m are used and R_i/R_f = 0.2.