If $\alpha$ and $\beta$ are the current gains in the $CB$ and $CE$ configurations respectively of the transistor circuit,then $\frac{\beta - \alpha}{\alpha \beta} = $

In a common base amplifier,the phase difference between the input signal voltage and the output voltage is

In a transistor, if $\frac{I_C}{I_E} = \alpha$ and $\frac{I_C}{I_B} = \beta$. If $\alpha$ varies between $\frac{20}{21}$ and $\frac{100}{101}$, then the value of $\beta$ lies between:

In a common emitter $(CE)$ amplifier having a voltage gain $G$,the transistor used has a transconductance of $0.03\,mho$ and a current gain of $25$. If the above transistor is replaced with another one with a transconductance of $0.02\,mho$ and a current gain of $20$,the new voltage gain will be:

$A$ $n-p-n$ transistor is connected in common-emitter configuration in which the collector supply is $8\, V$ and the voltage drop across the load resistance of $800\,\Omega$ connected in the collector circuit is $0.8\, V$. If the current amplification factor $\alpha$ is $25/26$,the collector-emitter voltage $V_{CE}$ and the base current $I_B$ will be:

The voltage gain of a transistor in common emitter configuration is $160$. The resistances in base and collector sides of the circuit are $1 \text{ k}\Omega$ and $4 \text{ k}\Omega$ respectively. If the change in base current is $100 \mu A$, then the change in output current is: