$A$ transistor has $3$ impurity regions of different doping levels. In the order of increasing doping level,the regions are

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

In the circuit shown in the figure,the input voltage $V_i$ is $20\, V$,$V_{BE} = 0$,and $V_{CE} = 0$. The values of $I_B$,$I_C$,and $\beta$ are given by:

In common emitter mode of a transistor, the d.c. current gain $(\beta)$ is $20$, and the emitter current $(I_E)$ is $7 \text{ mA}$. The collector current $(I_C)$ is:

$A$ transistor is connected in common emitter circuit configuration. The collector supply voltage is $10 \, V$ and the voltage drop across a resistor of $1000 \, \Omega$ in the collector circuit is $0.6 \, V$. If the current gain factor $(\beta)$ is $24$,then the base current is $.... \, \mu A$. (Round off to the Nearest Integer)

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 $mA$?