Q.Calculate the frequency and feedback ratio of the circuit shown below.

This is a Colpitts tank: inductor $L = 10\,\mu\text{H}$ with series capacitors $C_1 = 0.22\,\mu\text{F}$ and $C_2 = 0.47\,\mu\text{F}$, so $f = \dfrac{1}{2\pi\sqrt{L\,C_{eq}}}$ with $C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2}$ and feedback ratio $\beta = \dfrac{C_1} …

[!TLDR]

The series capacitors give $C_{eq} \approx 0.1499\,\mu\text{F}$, so with $L = 10\,\mu\text{H}$ the frequency is $f \approx 0.13\,\text{MHz}$; the feedback ratio is $\beta = C_1/C_2 \approx 0.46$.

The shown network is a Colpitts feedback tank: an inductor across two series capacitors whose junction is grounded. The feedback voltage appears across $C_2$ and the output across $C_1$, so $\beta = \dfrac{C_1}{C_2}$; the frequency is the tank resonance with $C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2}$.

Step 1 — equivalent capacitance: $C_{eq} = \dfrac{(0.22\times10^{-6})(0.47\times10^{-6})}{(0.22 + 0.47)\times10^{-6}} = \dfrac{0.1034\times10^{-6}}{0.69} \approx 1.499\times10^{-7}\,\text{F}$. …