The collector current in a common base amplifier using n - p -n transistor is 24 ~mA . If 80 % of the electrons released by the emitter is accepted by the collector, then the base current is numerically:

Step-by-step solution

As per question, we have, I _ c =24 ~mA , and =0.8 Now, I _ E = I _ c = 24 0.8 =30 ~mA As, I _ E = I _ B + I _ C Hence, I _ B = I _ E - I _ C =6 ~mA (where, the symbols have their usual meanings)