In an $n-p-n$ common emitter $(CE)$ transistor,the collector current changes from $5\,mA$ to $16\,mA$ for a change in base current from $100\,\mu A$ to $200\,\mu A$,respectively. The current gain of the transistor is:

In a common emitter transistor amplifier, an input signal of $10 \ mV$ is applied. Due to this signal, the change in base current is $50 \ \mu A$ and the corresponding change in collector current is $5 \ mA$. If the load resistance in the collector-emitter circuit is $5 \ k\Omega$, the change in output voltage will be ..... $V$.

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

In the shown common emitter amplifier circuit,$\beta = 80$ and $V_{BE} = 0.7\,V$. The value of $R_C$ is:

In a transistor amplifier,the base-emitter junction is forward-biased and the collector-emitter junction is reverse-biased. The current gain is defined as:

$A$ transistor is connected in common emitter configuration. The collector supply is $8 V$ and the voltage drop across a resistor of $800 \Omega$ in the collector circuit is $0.5 V$. If the current gain factor $\alpha$ is $0.96$,the base current is