In the study of a transistor as an amplifier,if $\alpha = \frac{I_C}{I_E} = 0.98$ and $\beta = \frac{I_C}{I_B} = 49$,where $I_C, I_B,$ and $I_E$ are collector,base,and emitter currents respectively,then $\left(\frac{1}{\alpha} - \frac{1}{\beta}\right)$ is equal to:

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 a common base transistor circuit,the current gain is $0.98$. On changing emitter current by $5.00 \,mA$,the change in collector current is ......... $mA$.

For a transistor,current gain $(\beta) = 50$. To change the collector current by $350 \mu A$,the base current should be changed by:

In the transistor circuit shown,assume that the voltage drop between the base and the emitter is $0.5\ V$. What will be the ratio of the voltage across resistances $R_2$ and $R_1$ in order to make this circuit function as a source of constant current,$I = 1\ mA$?