In a transistor amplifier,$\beta = 62$,$R_L = 5000\, \Omega$,and the internal resistance of the transistor is $R_i = 500\, \Omega$. The voltage amplification of the amplifier will be:

In the given common-emitter circuit, an $npn$ transistor with $\beta = 100$ is connected. What is the output voltage of the amplifier?

For a common base configuration of a $PNP$ transistor,$\frac{I_C}{I_E} = 0.96$. Then,the maximum current gain in common emitter configuration will be:

$A$ $n-p-n$ transistor is connected in common-emitter configuration in which the collector supply is $8\, V$ and the voltage drop across the load resistance of $800\,\Omega$ connected in the collector circuit is $0.8\, V$. If the current amplification factor $\alpha$ is $25/26$,the collector-emitter voltage $V_{CE}$ and the base current $I_B$ will be:

In a common emitter amplifier,a change of $0.2 \text{ mA}$ in the base current causes a change of $5 \text{ mA}$ in the collector current. If the input resistance is $2 \text{ k}\Omega$ and the voltage gain is $75$,what is the load resistance used in the circuit?

If the base current is $100 \mu A$ and the collector current is $3 \ mA$,then the values of $\beta$,$\alpha$,and $I_e$ are respectively: