In an $NPN$ 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 are respectively:

In a $PNP$ transistor working as a common-base amplifier,current gain is $0.96$ and emitter current is $7.2 \, mA$. The base current is.....$mA$

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

The current gain $\beta$ of a transistor is $50$. The input resistance of the transistor when used in the common emitter mode is $1\,k\Omega$. The peak value of the collector current for an input peak voltage of $0.01\,V$ is ...... $\mu A$.

For a $CE$ amplifier,current gain is $69$. If the emitter current is $7\, mA$,then the base current and collector current will be:

For a constant collector-emitter voltage of $8\,V$,the collector current of a transistor changed from $4\,mA$ to $6\,mA$,whereas the base current changed from $20\,\mu A$ to $25\,\mu A$. If the transistor is in the active state,the small signal current gain (current amplification factor) will be: