In a $CE$ transistor,when the base current increases by $60 \mu A$,the change in base-emitter voltage is $1.2 \ V$. Then the input resistance is: (in $\Omega$)

For a common emitter amplifier,the voltage gain is $50$,the input resistance is $200 \ \Omega$,and the output resistance is $400 \ \Omega$. The power gain of the amplifier will be:

In the circuit shown in the figure,the current gain $\beta = 100$ for an $npn$ transistor. What should be the base resistor $R_B$ so that $V_{CE} = 5 \text{ V}$,given $V_{BE} = 0 \text{ V}$?

In a common emitter transistor amplifier circuit,the current gain $\alpha = 0.8$. If the base current changes by $3 \mu A$,then the collector current changes by: (in $\mu A$)

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

The correct relation between $\alpha$ (ratio of collector current to emitter current) and $\beta$ (ratio of collector current to base current) of a transistor is :