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:

The input resistance of a common emitter transistor amplifier,if the output resistance is $500\,k\Omega$,the current gain $\alpha = 0.98$ and power gain is $6.0625 \times 10^6$,is.......$\Omega$.

In a transistor,the base is very lightly doped as compared to the emitter because by doing so:

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 $R_B$.

The input resistance of a silicon transistor is $100\,\Omega$. Base current is changed by $40\,\mu A$,which results in a change in collector current by $2\,mA$. This transistor is used as a common emitter amplifier with a load resistance of $4\,k\Omega$. The voltage gain of the amplifier is:

$A$ transistor is used in an amplifier circuit in common emitter mode. If the base current changes by $100 \,\mu A$, it brings a change of $10 \,mA$ in collector current. If the load resistance is $2 \,k \Omega$ and input resistance is $1 \,k \Omega$, the value of power gain is $x \times 10^{4}$. The value of $x$ is . . . . . . .