What is the voltage gain in a common emitter amplifier,where input resistance is $3 \, \Omega$ and load resistance is $24 \, \Omega$,and the current gain $\beta = 0.6$?

For a common-emitter amplifier, the voltage gain is $40$. Its input and output impedances are $100 \ \Omega$ and $400 \ \Omega$, respectively. The power gain of the $CE$ amplifier will be

$10^{10}$ 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 $\text{ mA}$)

The current amplification factor of a transistor is $50$. The input resistance when used in common emitter mode is $1 \text{ k}\Omega$. The peak value of the collector current for an a.c. input voltage of $0.01 \text{ V}$ peak is:

In an amplifier circuit,the value of the load resistance $R_L$ is $2 \text{ k}\Omega$. $A$ change of $10 \text{ }\mu\text{A}$ in the base current of the transistor results in a change of $1 \text{ mA}$ in the collector current. The change in the collector-emitter voltage $(V_{CE})$ is ....... $V$.

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 $\mu A$)