Question

You want to double the resonance angular frequency of an L-R-C series circuit by changing only the pertinent circuit elements all by the same factor. (a) Which ones should you change? (b) By what factor should you change them?

Short answer

a) To double the angular frequency, we change the values of inductor and capacitor.

b) The inductance L and the capacitance C must be halved to their initial values.

Step-by-step solution

Step-1: Formulas used  

Resonance angular frequency is given by

$\mathit{\omega }\mathbf{=}\frac{\mathbf{1}}{\mathbf{L}\mathbf{C}}$ . where L is the inductance and C is the capacitance.

Step-2: Parameters required and calculations for resonance angular frequency

The angular frequency is independent of the resistance R, so no need to change the resistance. The angular frequency is related to the Inductance L and the Capacitance C only.

${\omega }_1=\frac{1}{\sqrt{\left(L\right)\left(C\right)}}$

Now, the new resonance frequency

role="math" localid="1663916144195" $$\begin{gathered} {\omega }_2=2\left({\omega }_1\right) \\ {\omega }_2=\frac{2}{\sqrt{\left(L\right)\left(C\right)}} \\ \frac{1}{\sqrt{\left({\mathrm{L}}_{\mathrm{f}}\right)\left({\mathrm{C}}_{\mathrm{f}}\right)}}=\frac{2}{\sqrt{\left(\mathrm{L}\right)\left(\mathrm{C}\right)}} \\ \left({\mathrm{L}}_{\mathrm{f}}\right)\left({\mathrm{C}}_{\mathrm{f}}\right)=\frac{\left(\mathrm{L}\right)\left(\mathrm{C}\right)}{4} \\ \left({\mathrm{L}}_{\mathrm{f}}\right)\left({\mathrm{C}}_{\mathrm{f}}\right)=\left(\frac{\mathrm{L}}{2}\right)\left(\frac{\mathrm{C}}{2}\right) \end{gathered}$$

Therefore, a) To double the angular frequency, we change the values of inductor and capacitor. b) The inductance L and the capacitance C must be halved to their initial values.