When $NPN$ transistor is used as an amplifier,

For a common emitter amplifier,the current gain is $50$. If the emitter current is $6.6 \ mA$,then the value of the current gain when it acts as a common base amplifier is:

In a common-emitter transistor,the current gain $\beta$ is $49$. If there is a change of $5.0 \ \mu A$ in the base current,what will be the change in the collector current and the emitter current?

In an $n-p-n$ transistor,$200$ electrons enter the emitter in $10^{-8} \ s$. If $1 \%$ of electrons are lost in the base,then the current that enters the emitter and the current amplification factor are respectively $\left[e=1.6 \times 10^{-19} \ C\right]$

In a $NPN$ transistor,the collector current is $24 \, mA$. If $80\%$ of electrons reach the collector,what is the base current in $mA$?

$A$ $n-p-n$ transistor is connected in common-emitter configuration in which the collector supply is $8\, V$ and the voltage drop across the load resistance of $800\,\Omega$ connected in the collector circuit is $0.8\, V$. If the current amplification factor $\alpha$ is $25/26$,the collector-emitter voltage $V_{CE}$ and the base current $I_B$ will be: