The output ac voltage is 2.0 V. So, the ac collector current iC = 2.0/2000 = 1.0 mA. The signal current through the base is, therefore given by iB = iC/β = 1.0 mA/100 = 0.010 mA. The dc base current has to be 10 × 0.010 = 0.10 mA. RB = (VBB−VBE) / IB Assuming VBE = 0.6 V , RB = (2.0 . 0.6) / 0.10 = 14 kΩ