In a common emitter transistor circuit,the base current is $40 \ \mu A$. Find the value of $V_{BE}$. (in $V$)

In the transistor circuit shown,the base current is $35 \mu A$. The value of the resistor $R_b$ is......$ k \Omega$.

The current gain of a transistor in common emitter configuration is $80$. The resistances in collector and base sides of the circuit are $5 \text{ k}\Omega$ and $1 \text{ k}\Omega$ respectively. If the input voltage is $2 \text{ mV}$, the output voltage is: (in $\text{ V}$)

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

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.

$A$ transistor having $\alpha=0.8$ is connected in a common-emitter configuration. When the base current changes by $6 \text{ mA}$, the change in collector current is: (in $\text{ mA}$)