In the case of a common emitter transistor amplifier,the ratio of collector current to emitter current is $0.96$. Then,the current gain $(\beta)$ of the amplifier is:

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:

The graph between the input voltage $(V_{i})$ and the output voltage $(V_{o})$ of a transistor connected in common emitter configuration is shown in the figure. The active, saturation and cutoff regions of the transistor are respectively

Consider an amplifier circuit wherein a transistor is used in common emitter mode. The change in collector current and base current respectively are $4 mA$ and $20 \mu A$ when a signal of $40 mV$ is added to the base-emitter voltage. If the load resistance is $10 k\Omega$, then the power gain in the circuit is:

In the circuit shown, the value of $\beta$ of the transistor is $48$. If the base current supplied is $200 \mu A$, what is the voltage at the terminal $Y$ (in $V$)?

$A$ $npn$ transistor is connected in common emitter configuration in a given amplifier. $A$ load resistance of $800 \,\,\Omega$ is connected in the collector circuit and the voltage drop across it is $0.8 \,\, V$. If the current amplification factor is $0.96$ and the input resistance of the circuit is $192 \,\,\Omega$,the voltage gain and the power gain of the amplifier will respectively be