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$$
\begin{array}{l}{\text { A } 7.50 \text { -nF capacitor is charged up to } 12.0 \mathrm{V} \text { , then discon- }} \\ {\text { nected from the power supply and connected in series through a }} \\ {\text { coil. The period of oscillation of the circuit is then measured to be }} \\ {8.60 \times 10^{-5} \text { s. Calculate: (a) the inductance of the coil; (b) the }} \\ {\text { maximum charge on the capacitor; (c) the total energy of the cir- }} \\ {\text { cuit; (d) the maximum current in the circuit. }}\end{array}
$$

(a) $$
L=\frac{T^{2}}{4 \pi^{2} c}
$$                $$
T=\frac{1}{f}-\frac{1}{\omega / 2 \pi}=\frac{2 \pi}{w}
$$

$$
L=\frac{T^{2}}{4 \pi^{2} c}
$$$$
=\frac{8.6 * 10^{-3}}{4 \pi 2\left(7.5*10^{-9}\right)}
$$$$
=2.5 * 10^{-2} \mathrm{H}
$$

$$
= 25 \mathrm{mH}
$$

(b) $$
Q \cdot c v
=$$$$
7.5*10^{-9} * 12=9*10^{-8} \mathrm{C}
$$

(c) $$
U=\frac{Q^{2}}{2 C}=
$$$$
\frac{\left(4*10^{-8}\right)^{2}}{2\left(7 \cdot 5 * 10^{-9}\right)}=5.4 * 10^{-7} J
$$

(d) $$
U_{l}+U_{c}
$$$$
=U_{total}
$$$$
\rightarrow U_{L}=U_{total=}
$$$$
\frac{1}{2}LI^{2}=5 \cdot 4 * 10^{-7}
$$

$$
U_{L}=U_{total} 
$$

$$
\frac{1}{2} L I^{2}=5.4 * 10^{-7}
$$

$$
\rightarrow I=\sqrt{\frac{2 U_{total}}{L}}
$$$$
=\sqrt{\frac{2 * 5 \cdot 4 * 10^{-7}}{2.5 * 10^{-2}}}
$$$$
=6.58 * 10^{-3} \mathrm{A}
$$

$$
=6.58 \mathrm{mA}
$$

$$
\begin{array}{l}{\text { In an } L-C \text { circuit, } L=85.0 \mathrm{mH} \text { and } C=3.20 \mu \mathrm{F} . \text { Dur- }} \\ {\text { ing the oscillations the maximum current in the inductor is } 0.850 \mathrm{mA} \text { . }} \\ {\text { (a) What is the maximum charge on the capacitor? (b) What is the }} \\ {\text { magnitude of the charge on the capacitor at an instant when the }} \\ {\text { current in the inductor has magnitude } 0.500 \mathrm{mA} \text { ? }}\end{array}
$$

(a) $$
\frac{1}{2} l{ i}^{2}
$$$$
\frac{a^{2}}{\beta c}=\frac{Q_{2}}{2 c}
\leftarrow$$

$$
q=Q \rightarrow i=0
$$

$$
i=i_{\max } \longrightarrow
$$$$
q=0
$$

$$
\frac{1}{2} Li_{\max }^{2}=
$$$$
\frac{Q^{2}}{2 C} \rightarrow Q=i_{max} \sqrt{L C}
$$

\(∴ Q=0.85 * 10^{-3}* \sqrt{(0.085)\left(3.2*10^{-6}\right)} =4.43*10^{-7} \mathrm{C} \)

(b) $$
q=\sqrt{Q^{2}-L C i^{2}}
$$$$
=\sqrt{\left(4 \cdot 43*10^{-7}\right)^{2}-(0.085)(3.2*10^{-6})*(0.5*10^{-3})^2} 
$$

\(∴ q=3.58*10^{-7} \mathrm{C} \)

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