Solved Problem on RLC Circuits
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a) A LC circuit has an inductance of L0 and a capacitance of C0. Another LC circuit has inductance L = nL0 and capacitance L = nL0. What is the ratio of the frequencies of oscillations between the latter and the frequency of oscillations of the first circuit?
b) An LC circuit oscillates with frequency f0, for an inductance L0 and a capacitance C0. Keeping the same value of C0, replace the coil with another one with inductance L = nL0. Determine the new resonant frequency.
Solution
a) Data of the problem:
| Circuit 1 | Circuit 2 | |
|---|---|---|
| Inductance | L0 | nL0 |
| Capacitance | C0 | nC0 |
Table 1
The natural frequency of oscillations in a circuit is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\omega_{0}=\frac{1}{\sqrt{LC\;}}} \tag{I}
\end{gather}
\]
writing the expression (I) for the two circuits
\[
\begin{gather}
\omega_{01}=\frac{1}{\sqrt{L_{0}C_{0}\;}} \tag{II}
\end{gather}
\]
\[
\begin{gather}
\omega_{02}=\frac{1}{\sqrt{nL_{0}nC_{0}\;}}\\[5pt]
\omega_{02}=\frac{1}{\sqrt{n^{2}L_{0}C_{0}\;}}\\[5pt]
\omega_{02}=\frac{1}{n\sqrt{L_{0}C_{0}\;}} \tag{III}
\end{gather}
\]
In an LC circuit, the resonant angular frequency is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\omega =2\pi f} \tag{IV}
\end{gather}
\]
writing the expression (IV) for the two circuits
\[
\begin{gather}
\omega_{1}=2\pi f_{0} \tag{V}
\end{gather}
\]
\[
\begin{gather}
\omega_{2}=2\pi f \tag{VI}
\end{gather}
\]
dividing expression (VI) by expression (V)
\[
\begin{gather}
\frac{\omega_{2}}{\omega_{1}}=\frac{\cancel{2\pi} f}{\cancel{2\pi} f_{0}}\\[5pt]
\frac{\omega_{2}}{\omega_{1}}=\frac{f}{f_{0}} \tag{VII}
\end{gather}
\]
substituting expressions (II) and (III) into expression (VII)
\[
\begin{gather}
\frac{\frac{1}{n\sqrt{L_{0}C_{0}\;}}}{\frac{1}{\sqrt{L_{0}C_{0}\;}}}=\frac{f}{f_{0}}\\[5pt]
\frac{1}{n\cancel{\sqrt{L_{0}C_{0}\;}}}\frac{\cancel{\sqrt{L_{0}C_{0}\;}}}{1}=\frac{f}{f_{0}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\frac{f}{f_{0}}=\frac{1}{n}}
\end{gather}
\]
b) Data of the problem:
| Circuit 1 | Circuit 2 | |
|---|---|---|
| Inductance | L0 | nL0 |
| Capacitance | C0 | C0 |
| Frequency | f0 | x = f |
Table 1
Writing the (I) expression for the two circuits
\[
\begin{gather}
\omega_{01}=\frac{1}{\sqrt{L_{0}C_{0}\;}} \tag{VIII}
\end{gather}
\]
\[
\begin{gather}
\omega_{02}=\frac{1}{\sqrt{nL_{0}C_{0}\;}} \tag{IX}
\end{gather}
\]
writing the expression (IV) for the two circuits
\[
\begin{gather}
\omega_{1}=2\pi f_{0} \tag{X}
\end{gather}
\]
\[
\begin{gather}
\omega_{2}=2\pi f \tag{XI}
\end{gather}
\]
dividing expression (XI) by expression (X)
\[
\begin{gather}
\frac{\omega_{2}}{\omega_{1}}=\frac{\cancel{2\pi} f}{\cancel{2\pi} f_{0}}\\[5pt]
\frac{\omega_{2}}{\omega_{1}}=\frac{f}{f_{0}} \tag{XII}
\end{gather}
\]
substituting expressions (VIII) and (IX) into expression (XII)
\[
\begin{gather}
\frac{\frac{1}{\sqrt{nL_{0}C_{0}\;}}}{\frac{1}{\sqrt{L_{0}C_{0}\;}}}=\frac{f}{f_{0}}\\[5pt]
\frac{1}{\sqrt{n\;}\cancel{\sqrt{L_{0}C_{0}\;}}}\frac{\cancel{\sqrt{L_{0}C_{0}\;}}}{1}=\frac{f}{f_{0}}\\[5pt]
\frac{1}{\sqrt{n\;}}=\frac{f}{f_{0}}\\[5pt]
f=f_{0}\frac{1}{\sqrt{n\;}}\\[5pt]
f=f_{0}\frac{1}{\sqrt{n\;}}\frac{\sqrt{n\;}}{\sqrt{n\;}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{f=f_{0}\frac{\sqrt{n\;}}{n}}
\end{gather}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .