In the case of $NPN$ transistors,the collector current is always less than the emitter current because:

$A$ transistor is operated in common-emitter configuration at $V_{CE} = 2\, V$ such that a change in the base current from $100\,\mu A$ to $200\,\mu A$ produces a change in the collector current from $5\, mA$ to $10\, mA$. The current gain is

What is the voltage gain of a common-emitter amplifier with an input resistance of $3 \ \Omega$ and a load resistance of $24 \ \Omega$? (Given $\beta = 6$)

In an $n-p-n$ transistor, $95 \%$ of emitted electrons reach the collector. If the base current is $2 \text{ mA}$, then the collector current is (in $\text{ mA}$)

For a given transistor,if $\alpha = 0.95$,find the common emitter current gain $\beta$.

Identify the terminals of this $BJT$ and the type of $BJT$ $(NPN$ or $PNP)$. The multimeter is set to diode test mode (indicated by the symbol),and the black probe is connected to the $COM$ port while the red probe is connected to the $V\Omega$ port. Based on the connections shown in the image,determine the terminal configuration (from left to right) and the transistor type.