The components in the circuit given below are ideal. If $R=2 \mathrm{k} \Omega$ and $C=1 \mu \mathrm{~F}$, the -3 dB cut-off frequency of the circuit in Hz is

Consider the common emitter amplifier shown below with the following circuit parameters:

$$\beta = 100,\,{g_m} = 0.3861\,{\rm A}/V,\,{r_0} = \infty ,\,{r_\pi } = 259\,\Omega, $$
$${R_s} = 1\,K\Omega ,{R_B} = 93\,K\Omega ,\,{R_C} = 250\,\Omega, $$
$${R_L} = 1\,K\Omega ,\,{C_1} = \infty \,\,and\,\,{C_2} = 4.7\,\mu F.$$

The lower cut-off frequency due to C₂ is

Consider the common emitter amplifier shown below with the following circuit parameters:

$$\beta = 100,\,{g_m} = 0.3861\,{\rm A}/V,\,{r_0} = \infty ,\,{r_\pi } = 259\,\Omega, $$
$${R_s} = 1\,K\Omega ,{R_B} = 93\,K\Omega ,\,{R_C} = 250\,\Omega, $$
$${R_L} = 1\,K\Omega ,\,{C_1} = \infty \,\,and\,\,{C_2} = 4.7\,\mu F.$$

The Resistance seen by the source Vs is

An npn BJT has g_m = 38 mA/V, $${C_\mu }\, = {10^{ - 14}}$$ F, $${C_\pi }\, = 4\, \times {10^{ - 13}}\,F$$ and DC current gain $$\beta \, = \,90$$. For this transistor f_T and $${f_\beta }$$ are