Chapter 32, Problem 53P

(a)

To determine

The frequency of oscillation of the $LC$ circuit.

Answer to Problem 53P

The frequency of oscillation of the $LC$ circuit is $503\text{Hz}$.

Explanation of Solution

Write the expression for the frequency of oscillation of the $LC$ circuit.

  $f=\frac{1}{2\pi \sqrt{LC}}$                                                                                                   (I)

Here, $f$ is the frequency of oscillation, $L$ is the inductance of inductor and $C$ is the capacitance of the capacitor.

Conclusion:

Substitute $0.100\text{H}$ for $L$ and $1.00\mu \text{F}$ for $C$ in equation (I) to find $f$.

  $\begin{array}{c}f=\frac{1}{2\pi \sqrt{\left(0.100\text{H}\right)\left(1.00\mu \text{F}\right)}}\\ =\frac{1}{2\pi \sqrt{\left(0.100\text{H}\right)\left(1.00\times {10}^{-6}\text{F}\right)}}\\ =503\text{Hz}\end{array}$

Therefore, the frequency of oscillation of the $LC$ circuit is $503\text{Hz}$.

(b)

To determine

The maximum charge at the capacitor.

Answer to Problem 53P

The maximum charge at the capacitor is $12.0\mu \text{C}$.

Explanation of Solution

Write the expression for the maximum charge at the capacitor.

  $Q_{\max }=C\Delta V_{\max }$                                                                                   (II)

Here, $Q_{\max }$ is the maximum charge on the capacitor and $\Delta V_{\max }$ is the maximum potential difference.

Conclusion:

Substitute $1.00\mu \text{F}$ for $C$ and $12.0\text{V}$ for $\Delta V_{\max }$ in equation (II) to find $Q_{\max }$.

  $\begin{array}{c}{Q}_{\max }=\left(1.00\mu \text{F}\right)\left(12.0\text{V}\right)\\ =\left(1.00\times {10}^{-6}\text{F}\right)\left(12.0\text{V}\right)\\ =12.0\times {10}^{-6}\text{C}\\ \text{=12}\text{.0}\mu \text{C}\end{array}$

Therefore, the maximum charge at the capacitor is $12.0\mu \text{C}$.

(c)

To determine

The maximum current in the inductor.

Answer to Problem 53P

The maximum current in the inductor is $37.9\text{mA}$.

Explanation of Solution

Write the expression for the maximum energy stored in capacitor.

  $U_{\text{C,}\max }=\frac{1}{2}C{\left(\Delta V_{\max }\right)}^2$                                                                                                   (III)

Here, $U_{\text{C,max}}$ is the maximum energy stored in the capacitor.

Write the expression for the maximum energy stored in inductor.

  $U_{\text{L,}\max }=\frac{1}{2}L{I}_{\max }^2$                                                                                                             (IV)

Here, $U_{\text{L,}\max }$ is the maximum energy stored in the inductor and $I_{\max }$ is the maximum current in the inductor.

Equate equation (III) and equation (IV) to find $I_{\max }$.

  $\begin{array}{c}\frac{1}{2}C{\left(\Delta V_{\max }\right)}^2=\frac{1}{2}L{I}_{\max }^2\\ C{\left(\Delta V_{\max }\right)}^2=L{I}_{\max }^2\\ I_{\max }^2=\frac{C{\left(\Delta V_{\max }\right)}^2}{L}\\ I_{\max }=\sqrt{\frac{C}{L}}\Delta V_{\max }\end{array}$                                                                                            (V)

Conclusion:

Substitute $1.00\mu \text{F}$ for $C$, $0.100\text{H}$ for $L$ and $12.0\text{V}$ for $\Delta V_{\max }$ in equation (V) to find $I_{\max }$.

  $\begin{array}{c}{I}_{\max }=\sqrt{\frac{1.00\mu \text{F}}{0.100\text{H}}}\left(12.0\text{V}\right)\\ =\sqrt{\frac{1.00\times {10}^{-6}\text{F}}{0.100\text{H}}}\left(12.0\text{V}\right)\\ =0.0379\text{A}\\ \text{=37}\text{.9}\text{mA}\end{array}$

Therefore, the maximum current in the inductor is $37.9\text{mA}$.

(d)

To determine

The total energy in the circuit at $t=3.00\text{s}$.

Answer to Problem 53P

The total energy in the circuit at $t=3.00\text{s}$ is $72.0\mu \text{J}$.

Explanation of Solution

Write the expression for the total energy stored in $LC$ circuit.

  $U=\frac{Q_{\max }^2}{2C}$                                                                                                   (VI)

Here, $U$ is the total energy stored, $Q_{\max }$ is the maximum charge on the capacitor and $C$ is the capacitance of the capacitor.

Conclusion:

Substitute $12.0\mu \text{C}$ for $Q_{\max }$ and $1.00\mu \text{F}$ for $C$ in equation (VI) to find $U$.

  $\begin{array}{c}U=\frac{{\left(12.0\mu \text{C}\right)}^2}{2\left(1.00\mu \text{F}\right)}\\ =72.0\mu \text{J}\end{array}$

Therefore, the total energy in the circuit at $t=3.00\text{s}$ is $72.0\mu \text{J}$.