Question

A radio set is retuned from the 108MHz to 90.5MHz station using the variable capacitor.By what factor is the capacitance changed

Answer 1

Let $f_1 = 108\times 10^6Hz$, $f_2 = 90.5\times10^6Hz$. The formulas for resonant frequencies are:

$$f_1 = \dfrac{1}{2\pi\sqrt{C_1L}}\\ f_2 = \dfrac{1}{2\pi\sqrt{C_2L}}$$

where $C_1, C_2$ are corresponding capacitances. Dividing the first expression by the second and expressing ratio $C_2/C_1$, obtain:

$$\dfrac{C_2}{C_1} = \left( \dfrac{f_1}{f_2} \right)^2\\ \dfrac{C_2}{C_1} = \left( \dfrac{108\times 10^6}{90.5\times10^6} \right)^2 \approx 1.42$$

Answer. The capacitance changed by a factor of 1.42.