Resistor of $2 \Omega$, inductor of $100 \mu \mathrm{H}$ and capacitor of $400 \mathrm{pF}$ are connected in…
Resistor of $2 \Omega$, inductor of $100 \mu \mathrm{H}$ and capacitor of $400 \mathrm{pF}$ are connected in series across a source of $\mathrm{e}_{\mathrm{rms}}=0.1$ Volt. At resonance, voltage drop across inductor is
$25 \mathrm{~V}$
$2.5 \mathrm{~V}$
$250 \mathrm{~V}$
$20 \mathrm{~V}$
Solution
At resonance condition,
$\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}$
The impedance is given as:
$\begin{aligned}
& \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2} \\
& \mathrm{Z}=\mathrm{R}=2 \Omega \\
& \mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{e}_{\mathrm{mms}}}{\mathrm{R}} \\
& \mathrm{I}_{\mathrm{rms}}=\frac{0.1}{2}=0.05 \mathrm{~A} \\
& \omega=\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{10^{-4} \times 4 \times 10^{-10}}} \\
& \omega=5 \times 10^6
\end{aligned}$
$\therefore \quad$ The voltage-drop across the inductor is,
$\begin{aligned}
& \mathrm{V}=\mathrm{I}_{\mathrm{rms}} \times \mathrm{X}_{\mathrm{L}}=\mathrm{I}_{\mathrm{rms}} \times \mathrm{L} \omega \\
& \mathrm{V}=0.05 \times 10^{-4} \times 5 \times 10^6 \\
& \mathrm{~V}=25 \mathrm{~V}
\end{aligned}$