In the following common emitter circuit, if $\beta=100$, $V_{CE}=7 \, V$, $V_{BE}$ is negligible, and $R_C=2 \, k\Omega$, then $I_B=$ (in $\, mA$)?

When a signal is applied to the input of a transistor,it is found that the output signal is phase-shifted by $180^{\circ}$. The transistor configuration is:

In a transistor (common emitter configuration) the ratio of power gain to voltage gain is ($\alpha$ and $\beta$ are current ratios)

Consider an amplifier circuit wherein a transistor is used in common emitter mode. The change in collector current and base current respectively are $4 mA$ and $20 \mu A$ when a signal of $40 mV$ is added to the base-emitter voltage. If the load resistance is $10 k\Omega$, then the power gain in the circuit is:

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:

The output characteristics of a transistor in common emitter $(CE)$ mode is the graph obtained by plotting