$A$ common emitter amplifier is designed with an $NPN$ transistor $(\alpha = 0.99)$. The input impedance is $1 \ k\Omega$ and the load resistance is $10 \ k\Omega$. The voltage gain 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 $V_{BE} = 0 \text{ V}$?

When a signal is applied to the input of a transistor,it is found that the output signal is phase-shifted by $180^{\circ}$. The transistor configuration is:

In $n-p-n$ transistor, in $CE$ configuration:

In the common-base configuration,a transistor has a current amplification factor of $0.95$. If the transistor is used in a common-emitter configuration and the base current changes by $2 \mu A$,then the change in the collector current is: (in $\mu A$)

In an $n-p-n$ transistor,$200$ electrons enter the emitter in $10^{-8} \ s$. If $1 \%$ of electrons are lost in the base,then the current that enters the emitter and the current amplification factor are respectively $\left[e=1.6 \times 10^{-19} \ C\right]$