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

For a transistor,$\alpha$ and $\beta$ are given as $\alpha = \frac{I_{C}}{I_{E}}$ and $\beta = \frac{I_{C}}{I_{B}}$. Then the correct relation between $\alpha$ and $\beta$ is:

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).

In case of a bipolar transistor, $\beta = 45$. The potential drop across the collector resistance of $1 \ k\Omega$ is $5 \ V$. The base current is approximately: (in $\mu A$)

For a common emitter amplifier,the current gain is $50$. If the emitter current is $6.6 \ mA$,then the value of the current gain when it acts as a common base amplifier is:

In the given circuit,a transistor with a current gain $\beta = 100$ is used. To have $V_{CE} = 5 \text{ V}$ and $V_{BE} = 0$,what should be the value of the base resistance $R_b$?