MHT CET · Physics · Alternating Current
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
- A \(\frac{3}{4} f_r\)
- B \(\frac{3}{2} f_r\)
- C \(2 \mathrm{f}_{\mathrm{r}}\)
- D \(\mathrm{f}_{\mathrm{r}}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{f}_{\mathrm{r}}\)
Step-by-step Solution
Detailed explanation
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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