Question 151: AC fundamentals

Question

A capacitor of capacitance $$100 \mu \mathrm{F}$$ is charged to a potential of $$12 \mathrm{~V}$$ and connected to a $$6.4 \mathrm{~mH}$$ inductor to produce oscillations. The maximum current in the circuit would be :

Accepted Answer

By energy conservation
$$\begin{aligned}
& \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_{	ext {max }}^2 \
& \mathrm{I}_{\max }=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \mathrm{V} \
& =\sqrt{\frac{100 	imes 10^{-6}}{6.4 	imes 10^{-3}}} 	imes 12 \
& =\frac{12}{8}=\frac{3}{2}=1.5 \mathrm{~A}
\end{aligned}$$

Answer

2.0 A

Answer

3.2 A

Answer

1.5 A

Answer

1.2 A

AC fundamentals

A capacitor of capacitance $100 \mu \mathrm{F} is charged to a potential of 12 \mathrm{~V} and connected to a 6.4 \mathrm{~mH}$ inductor to produce oscillations. The maximum current in the circuit would be :

Choose the correct answer: