Chapter 33, Problem 23PQ

In the LC circuit in Figure 33.11, the inductance is L = 19.8 mH and the capacitance is C = 19.6 mF. At some moment, U_B = U_E= 17.5 mJ. a. What is the maximum charge stored by the capacitor? b. What is the maximum current in the circuit? c. At t = 0, the capacitor is fully charged. Write an expression for the charge stored by the capacitor as a function of lime. d. Write an expression for the current as a function of time.

To determine

The maximum charge stored by the capacitor.

Answer to Problem 23PQ

The maximum charge stored by the capacitor is $3.7\times {10}^{-2}\text{C}$.

Explanation of Solution

Write the expression to calculate the total; energy.

    $E_{\text{T}}=U_{\text{B}}+U_{\text{E}}$                                         (I)

Here, $E_{\text{T}}$ is total energy, $U_{\text{B}}$ is energies stored in magnetic field, and $U_{\text{E}}$ is energy stored in electric field.

Write the expression to calculate the maximum charge.

    $E_{\text{T}}=\frac{Q^2{}_{\max }}{2C}$                                                (II)

Here, $Q_{\max }$ is maximum charge, and $C$ is capacitance.

Conclusion:

Substitute $17.5\text{mJ}$ for $U_{\text{B}}$ and $U_{\text{E}}$ in equation (I) to calculate $E_{\text{T}}$.

    $\begin{array}{c}{E}_{\text{T}}=17.5\text{mJ}\times \left(\frac{1\text{J}}{{10}^3\text{mJ}}\right)+17.5\text{mJ}\times \left(\frac{1\text{J}}{{10}^3\text{mJ}}\right)\\ =17.5\times {10}^{-3}\text{J}+17.5\times {10}^{-3}\text{J}\\ =\text{35}\times {\text{10}}^{-3}\text{J}\end{array}$

Substitute $19.6\text{mF}$ for $C$, and $\text{35}\times {\text{10}}^{-3}\text{J}$ for $E_{\text{T}}$ in equation (II) to calculate $Q_{\max }$.

    $\begin{array}{c}\text{35}\times {\text{10}}^{-3}\text{J}=\frac{Q^2{}_{\max }}{2\times 19.6\text{mF}\times \left(\frac{1\text{F}}{{10}^{-3}\text{mF}}\right)}\\ Q^2{}_{\max }=1372\times {10}^{-6}{\text{C}}^2\\ Q_{\max }=\sqrt{1372\times {10}^{-6}{\text{C}}^2}\\ Q_{\max }=3.7\times {10}^{-2}\text{C}\end{array}$

Therefore, the maximum charge stored by the capacitor is $3.7\times {10}^{-2}\text{C}$.

To determine

The maximum current in the circuit.

Answer to Problem 23PQ

The maximum current in the capacitor is $1.88\text{A}$.

Explanation of Solution

Write the expression to calculate the maximum current.

    $E_T=\frac{1}{2}L{I}^2{}_{\max }$                                                    (III)

Here, $I_{\max }$ is maximum current, and $L$ is inductance.

Conclusion:

Substitute $19.8\times {10}^{-3}\text{H}$ for $L$, and $\text{35}\times {\text{10}}^{-3}\text{J}$ for $E_T$ in Equation (III) to calculate $I_{\max }$.

    $\begin{array}{c}\text{35}\times {\text{10}}^{-3}\text{J}=\frac{1}{2}\times 19.8\times {10}^{-3}\text{H}\times I^2{}_{\max }\\ I^2{}_{\max }=3.53{\text{A}}^{\text{2}}\\ I_{\max }=\sqrt{3.53{\text{A}}^{\text{2}}}\\ I_{\max }=1.88\text{A}\end{array}$

Therefore, the maximum current in the capacitor is $1.88\text{A}$.

To determine

The expression for the charge stored by the capacitor as function of time.

Answer to Problem 23PQ

The expression for the charge stored by the capacitor as function of time is $Q\left(t\right)=3.7\times {10}^{-2}\cos \left(50.8t\right)$.

Explanation of Solution

Write the expression to calculate the angular frequency.

    $\omega =\frac{1}{\sqrt{LC}}$                                                             (IV)

Here, $\omega$ is angular frequency.

Write the expression for the charge stored by the capacitor as function of time $Q\left(t\right)$.

    $Q\left(t\right)=Q_{\max }\cos \left(\omega t+\varphi \right)$                                       (V)

Here, $Q_{\max }$ is maximum charge.

Conclusion:

Substitute $19.8\times {10}^{-3}\text{H}$ for $L$, and $19.6\times {10}^{-3}\text{F}$ for $C$ in Equation (IV) to calculate $\omega$.

    $\begin{array}{c}\omega =\frac{1}{\sqrt{19.8\times {10}^{-3}\text{H}\times 19.6\times {10}^{-3}\text{F}}}\\ =\frac{1}{\sqrt{388.08\times {10}^{-6}}}\text{radian}/\text{s}\\ =50.8\text{radian}/\text{s}\end{array}$

Substitute $0$ for $\varphi$, $50.8\text{radian}/\text{s}$ for $\omega$, and $3.7\times {10}^{-2}\text{C}$ for $Q_{\max }$ in Equation (V) to calculate $Q\left(t\right)$.

    $\begin{array}{c}Q\left(t\right)=3.7\times {10}^{-2}\text{C}\cos \left(50.8\text{radian}/\text{s}\times t+0\right)\\ =3.7\times {10}^{-2}\cos \left(50.8t\right)\end{array}$

Therefore, the expression for the charge stored by the capacitor as function of time is $Q\left(t\right)=3.7\times {10}^{-2}\cos \left(50.8t\right)$.

To determine

The expression for the current as function of time.

Answer to Problem 23PQ

The expression for the current as function of time is $I\left(t\right)=-1.88\cos \left(50.8t\right)$.

Explanation of Solution

Write the expression for current as function of time $I\left(t\right)$.

    $I\left(t\right)=-I_{\max }\cos \left(\omega t+\varphi \right)$                                       (VI)

Here, $I_{\max }$ is maximum current.

Conclusion:

Substitute $0$ for $\varphi$, $50.8\text{radian}/\text{s}$ for $\omega$, and $1.88\text{A}$ for $I_{\max }$ in equation (VI) to calculate $I\left(t\right)$

    $\begin{array}{c}I\left(t\right)=-1.88\text{A}\cos \left(50.8\text{radian}/\text{s}\times t+0\right)\\ =-1.88\cos \left(50.8t\right)\end{array}$

Therefore, the expression for the current as function of time is $I\left(t\right)=-1.88\cos \left(50.8t\right)$.