The BJT is in active region since VCE
= VC = 0.9V > 0.3V (therefore not in SAT), and a current is
flowing in the BJT (therefore not in cutoff).
IC = 10µA therefore, the
base current is IB = IC/b = 10/60 = 0.167µA.
BJT in active: VBE = VB
= 0.7 (since VE = 0 = ground)
Therefore, the voltage across RB is
VC – VB = 0.9 – 0.7 = 0.2V.
Now, RB = VRB/IB
= 0.2/0.167µ = 1.2 MOhms.
The voltage across RC is VRC
= 1.5 – VC = 1.5 – 0.9 = 0.6V
Using KCL, the current flowing
through RC is IRC = IB + IC = 0.167
+ 10 = 10.167 µA.
This gives RB = VRC/IRC
= 0.6/10.167µ = 3.6 MOhms.