In an $N-P-N$ transistor amplifier,the collector current is $9 \ mA$. If $90\%$ of the electrons emitted from the emitter reach the collector,then:

An $N-P-N$ transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of $4 \ mA$. The terminal of an $8 \ V$ battery is connected to the collector through a load resistance $R_L$ and to the base through a resistance $R_B$. The collector-emitter voltage $V_{CE} = 4 \ V$,base-emitter voltage $V_{BE} = 0.6 \ V$ and base current amplification factor $\beta_{d.c.} = 100$. Calculate the values of $R_L$ and $R_B$.

In the case of $NPN$ transistors,the collector current is always less than the emitter current because:

In which of the configurations of a transistor, the power gain is highest?

Consider an $NPN$ transistor amplifier in common-emitter configuration. The current gain of the transistor is $100$. If the collector current changes by $1\, mA$,what will be the change in emitter current in $mA$?

In a $NPN$ transistor,$10^8$ electrons enter the emitter in $10^{-8} \ s$. If $1\%$ of electrons are lost in the base,the fraction of current that enters the collector and the current amplification factor $\beta$ are respectively: