Correct Answer - C
`I_(L)=(V)/(R_(L))=(50)/(10xx10^(3))=5xx10^(-3)A=5 mA`
for `80V V_(s)=30V`
`I_(S)=(V_(S))/(R_(S))=(30)/(5xx10^(3))=6 mA`
`:. I_(z)=1mA` (minimum current)
for `120V V_(S)=70V`
`I_(S)=(V_(S))/(R_(S))=(70)/(5xx10^(3))=14mA`
`:. I_(z)=14-5=9mA` (maximum current)