In an $NPN$ transistor,the collector current is $20 \, mA$. If $90\%$ of the electrons emitted reach the collector,then:

In a common emitter transistor amplifier,the output voltage and input voltage have a phase difference of

In an amplifier circuit,the value of the load resistance $R_L$ is $2 \text{ k}\Omega$. $A$ change of $10 \text{ }\mu\text{A}$ in the base current of the transistor results in a change of $1 \text{ mA}$ in the collector current. The change in the collector-emitter voltage $(V_{CE})$ is ....... $V$.

In a $CE$ transistor amplifier,the signal voltage across the collector resistance is $2.5 \ V$. The input signal voltage is $0.02 \ V$. If the base and collector resistances are $1.5 \ k\Omega$ and $2.5 \ k\Omega$,then the current amplification factor is

In an $NPN$ transistor,$10^{10}$ electrons enter the emitter region in $10^{-6} \, s$. If $2 \%$ of electrons are lost in the base region,then the collector current and the current amplification factor $(\beta)$ are,respectively:

For a transistor,$\frac{1}{\alpha_{DC}} - \frac{1}{\beta_{DC}}$ is equal to [where $\alpha_{DC}$ and $\beta_{DC}$ are current amplification factors].