The symbol given below represents:

In the figure,given that $V_{BB}$ supply can vary from $0$ to $5.0 \,V$,$V_{CC} = 5\,V$,$\beta_{dc} = 200$,$R_B = 100\,k\Omega$,$R_C = 1\,k\Omega$ and $V_{BE} = 1.0\,V$. The minimum base current and the input voltage at which the transistor will go to saturation,will be respectively:

An $n-p-n$ transistor power amplifier in $C-E$ configuration gives

An $n-p-n$ transistor is connected in common-emitter configuration as shown in the figure. If the collector current is $5 \text{ mA}$,$V_{BE} = 0.6 \text{ V}$,$V_{CE} = 3 \text{ V}$,and the common-emitter current amplification factor $\beta = 50$,then the values of $R_1$ and $R_2$ are respectively.

In a common emitter amplifier,a.c. current gain is $40$ and input resistance is $2 \ k\Omega$. The load resistance is given as $10 \ k\Omega$. Then the voltage gain is

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 $\left[e=1.6 \times 10^{-19} \ C\right]$