If $\alpha$ and $\beta$ are the current gains in the $CB$ and $CE$ configurations respectively of the transistor circuit,then $\frac{\beta - \alpha}{\alpha \beta} = $

In a common emitter configuration with suitable bias,it is given that $R_L$ is the load resistance and $R_{BE}$ is the small signal dynamic resistance (input side). Then,the voltage gain,current gain,and power gain are given,respectively,by: [$\beta$ is the current gain,$I_B$,$I_C$,$I_E$ are respectively base,collector,and emitter currents]

What is the power gain in a $CE$ amplifier,where input resistance is $3 \, k\Omega$ and load resistance is $24 \, k\Omega$,given $\beta = 6$?

Assertion: $A$ transistor amplifier in common emitter configuration has a low input impedance.
Reason: The base-to-emitter region is forward-biased.

In the study of a transistor as an amplifier,if $\alpha = \frac{I_{C}}{I_{E}}$ and $\beta = \frac{I_{C}}{I_{B}}$,where $I_{C}$,$I_{B}$,and $I_{E}$ are the collector,base,and emitter currents respectively,then:

An $n-p-n$ transistor with current gain $\beta=100$ in common emitter configuration is shown in the figure. The output voltage of the amplifier will be $.....V$.