MHT CET · Physics · Alternating Current
The resonant frequency of a series LCR circuit is ' \(\mathrm{f}_R\) '. The circuit is connected to a sinusoidally alternating e.m.f. of frequency ' \(2 \mathrm{f}_{\mathrm{R}}\) '. The inductive reactance becomes. ' \(\mathrm{X}_{\mathrm{L}_1}\) ' and capacitive reactance becomes ' \(\mathrm{X}_{\mathrm{C}_1}\) ' after changing the frequency. ' \(\mathrm{X}_{C_1}\) ' is equal to
- A \(2 \mathrm{X}_{\mathrm{L}_1}\)
- B \(\frac{1}{4} \mathrm{X}_{\mathrm{L}_1}\)
- C \(\frac{1}{2} \mathrm{X}_{\mathrm{L}_1}\)
- D \(\mathrm{x}_{\mathrm{L}_1}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{4} \mathrm{X}_{\mathrm{L}_1}\)
Step-by-step Solution
Detailed explanation
\(X_C=\frac{1}{\omega^2}\) and \(X_L=(\omega L)\)
We know, \(\omega=2 \pi f \Rightarrow f \propto \omega\) if \(f\) is doubled
\(\therefore \mathrm{X}_{\mathrm{C}_1}=\frac{\mathrm{X}_{\mathrm{C}}}{2}\) and \(\mathrm{X}_{\mathrm{L}_1}=2 \mathrm{X}_{\mathrm{L}}\)
We know, \(\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}\)
\(\Rightarrow 2 \mathrm{X}_{\mathrm{C}_1}=\frac{\mathrm{XL}_1}{2} \Rightarrow\left(\mathrm{X}_{\mathrm{C}_1}=\frac{\mathrm{X}_{\mathrm{L}_1}}{4}\right)\)
We know, \(\omega=2 \pi f \Rightarrow f \propto \omega\) if \(f\) is doubled
\(\therefore \mathrm{X}_{\mathrm{C}_1}=\frac{\mathrm{X}_{\mathrm{C}}}{2}\) and \(\mathrm{X}_{\mathrm{L}_1}=2 \mathrm{X}_{\mathrm{L}}\)
We know, \(\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}\)
\(\Rightarrow 2 \mathrm{X}_{\mathrm{C}_1}=\frac{\mathrm{XL}_1}{2} \Rightarrow\left(\mathrm{X}_{\mathrm{C}_1}=\frac{\mathrm{X}_{\mathrm{L}_1}}{4}\right)\)
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