An $npn$ transistor operates as a common emitter amplifier with a power gain of $10^{6}$. The input circuit resistance is $100\, \Omega$ and the output load resistance is $10\, k\Omega$. The common emitter current gain $\beta$ will be (Round off to the nearest integer).

Consider an $NPN$ transistor amplifier in common-emitter configuration. The current gain of the transistor is $100$. If the collector current changes by $1\, mA$,what will be the change in emitter current in $mA$?

For a constant collector-emitter voltage of $8\,V$,the collector current of a transistor changed from $4\,mA$ to $6\,mA$,whereas the base current changed from $20\,\mu A$ to $25\,\mu A$. If the transistor is in the active state,the small signal current gain (current amplification factor) will be:

In an $n-p-n$ transistor circuit,the collector current is $10 \text{ mA}$. If $95\%$ of the electrons emitted reach the collector,then the base current is nearly: (in $mA$)

In the following common emitter circuit,$\beta=100$ and $V_{CE}=7 \text{ V}$. If $V_{BE}$ is negligible,then the base current is (in $\text{ mA}$)

Assertion: In a transistor,the base is made thin.
Reason: $A$ thin base makes the transistor stable.