Question 433: AC fundamentals

Question

An alternating emf of $$0.2 \mathrm{~V}$$ is applied across an L-C-R series circuit having $$R=4 \Omega, C=80 \mu \mathrm{F}$$ and $$L=200 \mathrm{~mH}$$. At resonance the voltage drop across the inductor is

Accepted Answer

Given, $$V_{\mathrm{rms}}=0.2 \mathrm{~V}, R=4 \Omega$$,
$$C=80 \mu \mathrm{F}=80 	imes 10^{-6} \mathrm{~F}$$
and $$L=200 \mathrm{~mH}=200 	imes 10^{-3} \mathrm{H}$$
The impedance of the series L-C-R circuit,
$$Z=\sqrt{R^2+\left(X_L-X_Cight)^2}$$
At resonance, $$X_L=X_C$$
$$\Rightarrow \quad Z=R=4 \Omega$$
$$	herefore$$ Voltage drop across inductor,
$$\begin{aligned}
\left(V_{\mathrm{rms}}ight)_L & =I_{\mathrm{rms}} 	imes X_L \
& =\frac{V_{\mathrm{rms}}}{Z} 	imes \omega L \
& \begin{aligned}
& =\frac{V_{\mathrm{rms}}}{R} 	imes \frac{L}{\sqrt{L C}} \quad\left(\because \omega=\frac{1}{\sqrt{L C}}ight) \
& =\frac{V_{\mathrm{rms}}}{R} 	imes \sqrt{\frac{L}{C}} \
& =\frac{0.2}{4} 	imes \sqrt{\frac{200 	imes 10^{-3}}{80 	imes 10^{-6}}} \
& =0.05 	imes \sqrt{2500} \
& =0.05 	imes 50=2.5 \mathrm{~V}
\end{aligned}
\end{aligned}$$

Answer

10V

Answer

2.5V

Answer

1V

Answer

5V

AC fundamentals

An alternating emf of $0.2 \mathrm{~V} is applied across an L-C-R series circuit having R=4 \Omega, C=80 \mu \mathrm{F} and L=200 \mathrm{~mH}$. At resonance the voltage drop across the inductor is

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