POTENTIAL (Logic0) than, there is no voltage DROP across Base Emitter junction and hence no current will flow through the Transistor (Cut Off state). therefore output is at high potential (Logic1).In a Common Emitter transistor, if input A is at high potential (Logic1) than, there will be a voltage drop across Base Emitter junction and hence current will flow through the Transistor (Saturation state). therefore output is at low potential (Logic0).Application:⇒ It can be NOT or NAND (because the output is inversion of input in these TWO CASES)⇒ NAND Gate (output is inversion of product of both the inputs)ABA⋅B\(\overline {A \cdot B}\)0001010110011110 Now, ∴ Output, X = A⋅BNote:Most of the STUDENTS in solving quickly can answer this to be NAND.As the first part of the circuit given a NAND output but this NAND output is again fed to a circuit whose output is again a NAND operation of its input. That is, NAND operation of a NAND output is AND output.