An n-p-n transistor has three leads A, B and C. Connecting B and C...

Now, according to the question, A lead of the transistor is connected to the positive lead of the ammeter and the ammeter gives a deflection. We know that in an n-p-n transistor, the emitter and the collector are negatively doped, while the base is positively doped. So, in this type of transistor the electric field is from the base to the other terminals. Thus, the lead A must be the base of the transistor.
Now, the lead C is connected to the negative lead of the ammeter. Also, the leads B and C are connected by moist fingers. So, some conduction is there in between these two terminals. Therefore, both the emitter and the collector are connected to the negative end of the terminal. But the moist fingers are less conductive than the lead. So the lead B is not directly connected to the negative end.
We are having a large deflection in the ammeter. This means that there is a huge amount of current from the base to the other terminal. This is possible only when the terminal adjacent to the base terminal has a high number of charge carriers. As the emitter is most highly doped, so the lead C must correspond to the emitter.
Now the only terminal left for the third lead B is the collector. So, the lead B must be the collector.
Thus, A is the Base, B is the collector, and C is the emitter.