The expected answers were:
(a)(i) V_B = 2.2kΩ x lOv / 2.2kΩ + 6.8kΩ
= 22 /9
= 2.44v

(ii)V_E = V_B - V_BE
= 2.44v - O.7v
= 1.74v

(iii)I_E =V_E / 1kΩ
= 1.74mA
(iv)V_CE = 10 - (I_E x 2.7k) - V_E
= 10 - 1.74 x 2.7 - 1.74
= 10 - 1.74 - 4.7
= 10 - 6.44
= 3.56 V
(b) (i) Switch
(ii) Voltage regulators
(iii) Amplifiers
(iv) Oscillators
This question was very similar to Question 1 of November/ December 2009 examination which was also on establishing the d.c. (operating point) parameters of a BJT connected in the common emitter mode which are
readily got on using readily obtained circuit theory laws.
The question was on Bipolar Transistor and other Semiconductor Devices. It was quite unpopular with the candidates. The performance was well below average.