The phase difference between the input voltage and the output voltage in a common emitter amplifier is (in $^{\circ}$)

In the transistor circuit shown,assume that the voltage drop between the base and the emitter is $0.5\ V$. What will be the ratio of the voltage across resistances $R_2$ and $R_1$ in order to make this circuit function as a source of constant current,$I = 1\ mA$?

In a common emitter transistor amplifier,the output voltage and input voltage have a phase difference of

The current amplification factor of a transistor in common emitter configuration is $80$. If the emitter current is $2.43 \text{ mA}$, then the base current is (in $\mu \text{A}$)

$A$ transistor having $\alpha = 0.99$ is used in a common base amplifier. If the load resistance is $4.5 \, k\Omega$ and the dynamic resistance of the emitter junction is $50 \, \Omega$,the voltage gain of the amplifier will be:

Consider an $NPN$ transistor amplifier in common-emitter configuration. The current gain of the transistor is $100$. If the collector current changes by $1\, mA$,what will be the change in emitter current in $mA$?