In a common base amplifier circuit,calculate the change in base current if the change in emitter current is $2\, mA$ and $\alpha = 0.98$.

In an $NPN$ transistor,$10^{10}$ electrons enter the emitter region in $10^{-6} \, s$. If $2 \%$ of electrons are lost in the base region,then the collector current and the current amplification factor $(\beta)$ are,respectively:

When $n-p-n$ transistor is used as an amplifier :

In a transistor,when the emitter current changes by $9.85 \,mA$,the collector current changes by $9.5 \,mA$. Then the base current is (in $\,mA$)

$A$ transistor is connected in common emitter configuration. The collector supply is $8 V$ and the voltage drop across a resistor of $800 \Omega$ in the collector circuit is $0.5 V$. If the current gain factor $\alpha$ is $0.96$,the base current is

For a $CE$ transistor amplifier, the current amplification factor is $59$ and the emitter current is $6.6 \, mA$. Then the base current is