An $n-p-n$ transistor power amplifier in $C-E$ configuration gives

$A$ $npn$ transistor is connected in common emitter configuration in a given amplifier. $A$ load resistance of $800 \,\,\Omega$ is connected in the collector circuit and the voltage drop across it is $0.8 \,\, V$. If the current amplification factor is $0.96$ and the input resistance of the circuit is $192 \,\,\Omega$,the voltage gain and the power gain of the amplifier will respectively be

The part of a transistor which is heavily doped to produce a large number of majority carriers is

$10^{10}$ electrons enter the emitter of a junction transistor in a time of $0.4 \mu s$. If $5 \%$ of the electrons are lost in the base, then the collector current is (in $\text{ mA}$)

In an $N-P-N$ transistor circuit,the common emitter configuration is used. $A$ load resistor $R_L$ is connected to the collector and a base resistor $R_B$ is connected to the base. Given $V_{CC} = 8V$,$V_{CE} = 4V$,$V_{BE} = 0.6V$,collector current $I_C = 4mA$,and current gain $\beta = 100$,find the values of $R_L$ and $R_B$ respectively.

In an $NPN$ transistor,the collector current is $10\, mA$. If $90\%$ of the electrons emitted reach the collector,then: