Correct Answer - A
An n-p-n transistor in a common-emitter mode with connections as given is shown in figure. Collector emitter voltage,
`V_("CE")=V_("CC")-I_C R_L`
`therefore R_L=(V_("CC")-V_("CE"))/I_C`
`=(8V-4V)/(4xx10^(-3))=10^3 Omega=1 kOmega`
As `beta_(dc)=I_C/I_B`
`therefore I_B=I_C/beta_(dc)=(4xx10^(-3))/100A=4xx10^(-5)A`
Base-emitter voltage,
`V_(BE)=V_("CC")-I_B R_B`
`therefore R_B=(V_("CC")-V_(BE))/I_B=(8V-0.6V)/(4xx10^(-5)A)`
`=1.85xx10^5 Omega= 185 kOmega`