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A circuit containing resistance $\mathrm{R}_1$, inductance $\mathrm{L}_1$ and capacitance $\mathrm{C}_1$ connected in series resonates at the same frequency ' $\mathrm{f}_{\mathrm{r}}$ ' as another circuit containing $\mathrm{R}_2, \mathrm{~L}_2$ and $\mathrm{C}_2$ in series. If two circuits are connected in series, then the new frequency at resonance is
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Verified Answer
The correct answer is:
$\mathrm{f}_{\mathrm{r}}$
When $\mathrm{R}_1, \mathrm{~L}_1$ and $\mathrm{C}_1$ are connected in series the resonant (angular) frequency is given by
$$
\omega_{\mathrm{r}}=\frac{1}{\sqrt{\mathrm{L}_1 \cdot \mathrm{C}_1}}
$$
Also, when $\mathrm{R}_2, \mathrm{~L}_2, \mathrm{C}_2$ are connected in series
Then $\omega_{\mathrm{r}}=\frac{1}{\sqrt{\mathrm{L}_2 \cdot \mathrm{C}_2}}$
By (1) and (2): $\mathrm{L}_1 \mathrm{C}_1=\mathrm{L}_2 \mathrm{C}_2$
If the two circuits are connected in series then the equivalent inductance is given by $\mathrm{L}=\mathrm{L}_1+\mathrm{L}_2$ and the equivalent capacitance is given by
$$
\begin{aligned}
& \mathrm{C}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2} \\
& \therefore \mathrm{LC}=\left(\mathrm{L}_1+\mathrm{L}_2\right) \cdot \frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2} \\
& =\frac{\mathrm{L}_1 \mathrm{C}_1 \mathrm{C}_2+\mathrm{L}_2 \mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}\left[\because \mathrm{L}_1 \mathrm{C}_1=\mathrm{L}_2 \mathrm{C}_2\right] \\
& =\frac{\mathrm{L}_2 \mathrm{C}_2\left(\mathrm{C}_2+\mathrm{C}_1\right)}{\mathrm{C}_1+\mathrm{C}_2}=\mathrm{L}_2 \mathrm{C}_2 \\
& \therefore \omega=\frac{1}{\sqrt{\mathrm{L}_2 \mathrm{C}_2}}=\omega_{\mathrm{r}}
\end{aligned}
$$
$$
\omega_{\mathrm{r}}=\frac{1}{\sqrt{\mathrm{L}_1 \cdot \mathrm{C}_1}}
$$
Also, when $\mathrm{R}_2, \mathrm{~L}_2, \mathrm{C}_2$ are connected in series
Then $\omega_{\mathrm{r}}=\frac{1}{\sqrt{\mathrm{L}_2 \cdot \mathrm{C}_2}}$
By (1) and (2): $\mathrm{L}_1 \mathrm{C}_1=\mathrm{L}_2 \mathrm{C}_2$
If the two circuits are connected in series then the equivalent inductance is given by $\mathrm{L}=\mathrm{L}_1+\mathrm{L}_2$ and the equivalent capacitance is given by
$$
\begin{aligned}
& \mathrm{C}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2} \\
& \therefore \mathrm{LC}=\left(\mathrm{L}_1+\mathrm{L}_2\right) \cdot \frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2} \\
& =\frac{\mathrm{L}_1 \mathrm{C}_1 \mathrm{C}_2+\mathrm{L}_2 \mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}\left[\because \mathrm{L}_1 \mathrm{C}_1=\mathrm{L}_2 \mathrm{C}_2\right] \\
& =\frac{\mathrm{L}_2 \mathrm{C}_2\left(\mathrm{C}_2+\mathrm{C}_1\right)}{\mathrm{C}_1+\mathrm{C}_2}=\mathrm{L}_2 \mathrm{C}_2 \\
& \therefore \omega=\frac{1}{\sqrt{\mathrm{L}_2 \mathrm{C}_2}}=\omega_{\mathrm{r}}
\end{aligned}
$$
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