The region in the output voltage versus input voltage graph where a transistor can be used as an amplifier is

$A$ common emitter amplifier has a voltage gain of $50$, an input resistance of $100 \ \Omega$, and an output resistance of $200 \ \Omega$. What is the power gain of the amplifier?

$A$ transistor has $3$ impurity regions of different doping levels. In the order of increasing doping level,the regions are

An $npn$ transistor operates as a common emitter amplifier with a power gain of $10^{6}$. The input circuit resistance is $100\, \Omega$ and the output load resistance is $10\, k\Omega$. The common emitter current gain $\beta$ will be (Round off to the nearest integer).

$A$ $n-p-n$ 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:

For a transistor,$\alpha_{dc}$ and $\beta_{dc}$ are the current ratios,then the value of $\frac{\beta_{dc}-\alpha_{dc}}{\alpha_{dc} \times \beta_{dc}}$ is: