Since, emitter junction is F.B. So BJT is not in cut-offmode. It maybe either active or saturation region. Let us assume transistor in active region. Then, applying KVL at base-emitter loop. 2.7 - 40 K IB -0.7 = 0 40 K IB = 2 IB = 0.05 mA
Applying KVL at emitter collector loop.
Since, VCE < 0.5, so transistor is not in active region it is saturation region.
In a bipolar transistor, the increase in the depletion region of the reverse biased junction, tends to decrease the width of the base with increased reverse voltage. This in turn causes the current gain to increase with increased voltage resulting in a nearly linear increase in current even with the device in the saturation.