Question 370: AC fundamentals

Question

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

Accepted Answer

At resonance condition, $$\mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}$$
The impedance is given as:
$$\begin{aligned}
\mathrm{Z} & =\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}ight)^2} \
	herefore \quad \mathrm{Z} & =\mathrm{R}=2 \Omega \
\qquad\mathrm{I}_{\mathrm{rms}} & =\frac{\mathrm{e}_{\mathrm{ms}}}{\mathrm{R}} \
\mathrm{I}_{\mathrm{rms}} & =\frac{0.1}{2}=0.05 \mathrm{~A} \
\omega & =\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{10^{-4} 	imes 4 	imes 10^{-10}}} \
\omega & =5 	imes 10^6
\end{aligned}$$
$$	herefore \quad$$ The voltage-drop across the inductor is,
$$\begin{aligned}
& V=I_{\mathrm{rms}} 	imes X_L=I_{\mathrm{rms}} 	imes L \omega \
& V=0.05 	imes 10^{-4} 	imes 5 	imes 10^6 \
& V=25 \mathrm{~V}
\end{aligned}$$

Answer

25 V

Answer

2.5 V

Answer

250 V

Answer

20 V

AC fundamentals

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

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