In a common emitter configuration,an $N-P-N$ transistor is connected. The current gain of the transistor is $100$. If the collector current changes by $1 \ mA$,by how much will the emitter current change in $mA$?

An $N-P-N$ transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of $4 \ mA$. The terminal of an $8 \ V$ battery is connected to the collector through a load resistance $R_L$ and to the base through a resistance $R_B$. The collector-emitter voltage $V_{CE} = 4 \ V$,base-emitter voltage $V_{BE} = 0.6 \ V$ and base current amplification factor $\beta_{d.c.} = 100$. Calculate the values of $R_L$ and $R_B$.

For the given Common Emitter $(CE)$ transistor circuit,the base current is $I_B = 0.04 \text{ mA}$. If the current gain $\beta = 100$ and the collector-emitter voltage $V_{CE} = 12 \text{ V}$,find the value of the collector resistance $R_C$ in $\text{k}\Omega$.

For a transistor,$\alpha_{dc}$ and $\beta_{dc}$ are the current ratios. Then the value of $\frac{\beta_{dc}-\alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}}$ is

The region in the output voltage versus input voltage graph where a transistor can be used as an amplifier is

In a transistor,the value of $\alpha$ varies between $\frac{20}{21}$ and $\frac{100}{101}$. Then the value of $\beta$ varies between