The expected answers were:

• For a bipolar junction transistor

β  =      α   
         1  - α

α  =       β   
            1+ β
Given that    VBE = 0.6V
                    α= 0.98
VCC = IBRB + VBE +IERE
VCC = VCE + IcRC + IERE

β  =      α   
         1  - α
    =   
    =   49

    = 25.9µA
    =  26µA

VCE = VCC - ICRC -ICRE
      = VCC –IC(RC+RE)
      = 12 – 1.27mA(5.6K + 1K)
VCE = 3.62V
IC = βIB = 49 x 25.9µA = 1.27µA

The question was on establishing the d.c. (operating point) parameters of a BJT connected in the common-emitter mode. It was quite popular with the candidates. However, the performance was about average.