\(V _{ L }= I _{ L } R _{ L }\)

\(8=10 \times 10^{-3} \times R _{ L _{\max }}\)

\(\frac{4}{5} \times 10^{3}= R _{ L _{\max }}\)

\(800= R _{ L _{\max }}\)

\(I = I _{ Z }+ I _{ L }\)

\(I _{ L }=10 mA\)

If \(I _{ Z }=0\)

\(I _{ L _{\max }}=20 mA\)

\(V _{ L }= I _{ L _{\max }} \times R _{ L _{\min }}\)

\(\frac{8}{20} \times 10^{3}= R _{ L _{\min }}\)

\(400= R _{ L _{\min }}\)

\(\frac{ R _{ L _{\max }}}{ R _{ L _{\min }}}=\frac{800}{400}=2\)