Chapter 32, Problem 32.54P

(a)

To determine

The energy stored in the capacitor at any time $t$.

Answer to Problem 32.54P

The energy stored in the capacitor at any time $t=2.00\text{ms}$ is $4.29\text{J}$.

Explanation of Solution

Given info: The inductance of the inductor is $3.30\text{H}$, the capacitance of the capacitor is $840\text{pF}$, the initial charge of the capacitor is $105\text{μC}$ and the time is $2.00\text{ms}$.

Formula to calculate the angular frequency of the LC circuit is,

$\omega =\frac{1}{\sqrt{LC}}$

Substitute $3.30\text{H}$ for $L$ and $840\text{pF}$ of $C$ to find the $\omega$.

$\begin{array}{c}\omega =\frac{1}{\sqrt{\left(3.30\text{H}\right)840\text{pF}\left(\frac{{10}^{-12}\text{F}}{\text{1}\text{pF}}\right)}}\\ =1.8\times {10}^4\text{s}\end{array}$

Formula to calculate the energy stored in the capacitor at any time $t$ is,

$E_{\text{c}}=\frac{Q{\left(t\right)}^2}{2C}$ (1)

Formula to calculate the instantaneous value of the charge is,

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

Here,

$Q\left(t\right)$ is the instantaneous value of the charge.

$Q_{\max }$ is the maximum charge of the capacitor.

$\omega$ is the angular frequency.

$t$ is the time.

Substitute $Q_{\max }\cos \omega t$ for $Q\left(t\right)$ in equation (1).

$E_{\text{c}}=\frac{{\left(Q_{\max }\cos \omega t\right)}^2}{2C}$ (2)

Here,

$E_{\text{c}}$ is the energy stored in the capacitor at any time $t$.

Substitute $105\text{μC}$ for $Q_{\max }$, $1.8\times {10}^4\text{s}$ for $\omega$, $840\text{pF}$ of $C$ and $2.00\text{ms}$ for $t$ in the  equation (2) to find the $E_{\text{c}}$.

$\begin{array}{c}{E}_{\text{c}}=\frac{{\left(105\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\cos \left(\left(1.8\times {10}^4\text{s}\right)2.00\text{ms}\left(\frac{{10}^{-3}\text{s}}{1\text{ms}}\right)\right)\right)}^2}{2\times 840\text{pF}\left(\frac{{10}^{-12}\text{F}}{\text{1}\text{pF}}\right)}\\ =4.29\text{J}\end{array}$

Thus, the energy stored in the capacitor at any time $t=2.00\text{ms}$ is $4.29\text{J}$.

Conclusion:

Therefore, the energy stored in the capacitor at $t=2.00\text{ms}$ is $4.29\text{J}$.

(b)

To determine

The energy stored in the inductor at any time $t$.

Answer to Problem 32.54P

The energy stored in the capacitor at any time $t=2.00\text{ms}$ is $\mathrm{2.03.}\text{J}$.

Explanation of Solution

Given info: The inductance of the inductor is $3.30\text{H}$, the capacitance of the capacitor is $840-\text{pF}$ and the initial charge of the capacitor is $105-\text{μC}$ and the time is $2.00\text{ms}$.

Formula to calculate the current in the circuit is,

$i\left(t\right)=\frac{dQ\left(t\right)}{dt}$

Substitute $Q_{\max }\cos \omega t$ for $Q\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\frac{d\left(Q_{\max }\cos \omega t\right)}{dt}\\ =-\omega Q_{\max }\sin \omega t\end{array}$

Formula to calculate the energy stored in the inductor at ant time $t$ is,

$E_{\text{I}}=\frac{1}{2}Li(t)$ (3)

Here,

$E_{\text{I}}$ is the energy stored in the inductor.

$i(t)$ is the current at time $t$.

Substitute $-\omega Q_{\max }\sin \omega t$ for $i(t)$ in equation (3).

$E_{\text{I}}=\frac{1}{2}L{\left(-\omega Q_{\max }\sin \omega t\right)}^2$ (4)

Substitute $105\text{μC}$ for $Q_{\max }$, $1.8\times {10}^4\text{rad}/\text{s}$ for $\omega$, $3.30\text{H}$ for $L$ and $2.00\text{ms}$ for $t$ in the  equation (4) to find the $U$.

$\begin{array}{c}{E}_{\text{I}}=\frac{1}{2}3.30\text{H}{\left(-\left(1.8\times {10}^4\text{rad}/\text{s}\right)105\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\sin \left(1.8\times {10}^4\text{s}\right)2.00\text{ms}\left(\frac{{10}^{-3}\text{s}}{1\text{ms}}\right)\right)}^2\\ =\mathrm{2.03.}\text{J}\end{array}$

Thus, the energy stored in the capacitor at any time $t=2.00\text{ms}$ is $\mathrm{2.03.}\text{J}$.

Conclusion:

Therefore, the energy stored in the capacitor at any time $t=2.00\text{ms}$ is $\mathrm{2.03.}\text{J}$.

(c)

To determine

The total energy in the circuit.

Answer to Problem 32.54P

The total energy in the circuit is $6.56\text{J}$.

Explanation of Solution

Given info: The inductance of the inductor is $3.30\text{H}$, the capacitance of the capacitor is $840\text{pF}$ and the initial charge of the capacitor is $105\text{μC}$ and the time is $2.00\text{ms}$.

Formula to calculate the total energy in the circuit is,

$E=\frac{{\left(Q_{\max }\right)}^2}{2C}$

Here,

$E$ is the total energy in the circuit.

Substitute $105\text{μC}$ for $Q_{\max }$ and $840\text{pF}$ for $C$ to find the $E_{\text{C}}$.

$\begin{array}{c}E=\frac{{\left(105\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)}^2}{2\times 840\text{pF}\left(\frac{{10}^{-12}\text{F}}{\text{1}\text{pF}}\right)}\\ =6.56\text{J}\end{array}$

Thus, the total energy in the circuit is $6.56\text{J}$.

Conclusion:

Therefore, the total energy in the circuit is $6.56\text{J}$.