MHT CET · Physics · Alternating Current
In a series LCR circuit, \(\mathrm{C}=2 \mu \mathrm{F}, \mathrm{L}=1 \mathrm{mH}\) and \(\mathrm{R}=10 \Omega\). The ratio of the energies stored in the inductor and the capacitor, when the maximum current flows in the circuit, is
- A \(5: 1\)
- B \(3: 2\)
- C \(1: 2\)
- D \(1: 5\)
Answer & Solution
Correct Answer
(A) \(5: 1\)
Step-by-step Solution
Detailed explanation
In resonance condition (current is maximum),
\(\therefore \quad \mathrm{X}_{\mathrm{c}}=\mathrm{X}_{\mathrm{L}}\)
\(\therefore \quad\) The ratio of energies in the inductor and capacitor is:
\(\begin{aligned}
& \frac{\mathrm{U}_{\mathrm{L}}}{\mathrm{U}_{\mathrm{C}}}=\frac{\mathrm{LI}^2}{\mathrm{CV}^2}=\frac{\mathrm{L}}{\mathrm{CR}^2} \quad \ldots\left(\because \frac{\mathrm{I}}{\mathrm{V}}=\frac{1}{\mathrm{R}}\right) \\
& \frac{\mathrm{U}_{\mathrm{L}}}{\mathrm{U}_{\mathrm{C}}}=\frac{10^{-3}}{2 \times 10^{-6} \times 10^2} \\
& \frac{\mathrm{U}_{\mathrm{L}}}{\mathrm{U}_{\mathrm{C}}}=\frac{5}{1}
\end{aligned}\)
\(\therefore \quad \mathrm{X}_{\mathrm{c}}=\mathrm{X}_{\mathrm{L}}\)
\(\therefore \quad\) The ratio of energies in the inductor and capacitor is:
\(\begin{aligned}
& \frac{\mathrm{U}_{\mathrm{L}}}{\mathrm{U}_{\mathrm{C}}}=\frac{\mathrm{LI}^2}{\mathrm{CV}^2}=\frac{\mathrm{L}}{\mathrm{CR}^2} \quad \ldots\left(\because \frac{\mathrm{I}}{\mathrm{V}}=\frac{1}{\mathrm{R}}\right) \\
& \frac{\mathrm{U}_{\mathrm{L}}}{\mathrm{U}_{\mathrm{C}}}=\frac{10^{-3}}{2 \times 10^{-6} \times 10^2} \\
& \frac{\mathrm{U}_{\mathrm{L}}}{\mathrm{U}_{\mathrm{C}}}=\frac{5}{1}
\end{aligned}\)
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