The input voltage applied to a $CE$ amplifier is $v_i = 0.2 \sin(1000t) \text{ V}$. If the voltage gain of the amplifier is $10$,the equation for the output voltage is ....... $\text{V}$.

In a common base circuit of a transistor,the amplification factor is $0.95.$ The base current when the emitter current is $2\,mA,$ is ....$mA$

For a transistor to act as a switch,it must be operated in

The input signal given to a $CE$ amplifier having a voltage gain of $150$ is $V_{in} = 2 \cos(15t + \frac{\pi}{3}) \text{ V}$. The corresponding output signal will be:

In an $NPN$ transistor,$10^{10}$ electrons enter the emitter in $10^{-6} \ s$ and $2\%$ of the electrons recombine with holes in the base. The current gains $\alpha$ and $\beta$ are respectively:

$A$ transistor is operating in the active mode. $v_1$ is the potential barrier across the base-emitter junction and $v_2$ is the potential barrier across the collector-base junction. $b_1$ is the width of the depletion layer of the base-emitter junction and $b_2$ is the width of the collector-base junction.