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

This is a Hartley tank: capacitor $C = 1000\,\text{pF}$ with a tapped inductor $L_1 = 6.8\,\text{mH}$ (output) and $L_2 = 1.2\,\text{mH}$ (feedback), so $f = \dfrac{1}{2\pi\sqrt{LC}}$ with $L = L_1 + L_2$ and $\beta = \dfrac{L_2}{L_1}$. …

[!TLDR]

With $L = L_1 + L_2 = 8\,\text{mH}$ and $C = 1000\,\text{pF}$, $f \approx 56.298\,\text{kHz}$; the feedback ratio is $\beta = L_2/L_1 \approx 0.176$.

The shown network is a Hartley feedback tank: a single capacitor across a centre-tapped inductor. The feedback voltage is developed across $L_2$ and the output across $L_1$, so $\beta = \dfrac{L_2}{L_1}$; the frequency is the tank resonance with $L = L_1 + L_2$.

Step 1 — total inductance: $L = 6.8\,\text{mH} + 1.2\,\text{mH} = 8\times10^{-3}\,\text{H}$.

Step 2 — frequency: $f = \dfrac{1}{2\pi\sqrt{(8\times10^{-3})(1000\times10^{-12})}} = \dfrac{1}{2\pi\sqrt{8\times10^{-12}}} \approx 5.63\times10^{4}\,\text{Hz} = 56.298\,\text{kHz}$.

Step 3 — feedback ratio: $\beta = \dfrac{L_2}{L_1} = \dfrac{1.2}{6.8} \approx 0.176$. …