In a series LCR circuit, $\mathrm{C}=2 \mu \mathrm{F}, \mathrm{L}=1 \mathrm{mH}$ and $\mathrm{R}=10 \Omega$.…
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
$5: 1$
$3: 2$
$1: 2$
$1: 5$
Solution
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:
$\begin{aligned}
& \frac{\mathrm{U}_{\mathrm{L}}}{\mathrm{U}_{\mathrm{C}}}=\frac{\mathrm{LI}^2}{\mathrm{CV}^2}=\frac{\mathrm{L}}{\mathrm{CR}^2} \quad \ldots\left(\because \frac{\mathrm{I}}{\mathrm{V}}=\frac{1}{\mathrm{R}}\right) \\
& \frac{\mathrm{U}_{\mathrm{L}}}{\mathrm{U}_{\mathrm{C}}}=\frac{10^{-3}}{2 \times 10^{-6} \times 10^2} \\
& \frac{\mathrm{U}_{\mathrm{L}}}{\mathrm{U}_{\mathrm{C}}}=\frac{5}{1}
\end{aligned}$
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