Chapter 7, Problem 69E

(a)

To determine

Find the energy stored in the element at time $t=0$ for the circuit shown in Figure 7.83.

Answer to Problem 69E

The energy stored in the element at time $t=0$ for the circuit shown in Figure 7.83 is $405\mu \text{J}$.

Explanation of Solution

Formula used:

Write a general expression to calculate the energy stored in an capacitor.

$w=\frac{1}{2}C{v}^2\left(t\right)$        (1)

Here,

$C$ is the value of capacitance, and

$v\left(t\right)$ is the voltage across the capacitor.

Given data:

Refer to Figure 7.83 in the textbook.

The value of initial voltage across the capacitor $v_C\left(0\right)=v\left(0\right)$ is $9\text{V}$.

Calculation:

Substitute $0$ for $t$ in equation (1) to find $w$ at time $t=0$.

$w=\frac{1}{2}C{\left(v\left(0\right)\right)}^2$

Substitute $10\mu \text{F}$ for $C$ and $9\text{V}$ for $v\left(0\right)$ in equation (1) to find $w$.

$\begin{array}{c}w=\frac{1}{2}\left(10\mu \text{F}\right){\left(9\text{V}\right)}^2\\ =\frac{1}{2}\left(10\times {10}^{-6}\text{F}\right)\left(81{\text{V}}^2\right)\left\{\because 1\mu ={10}^{-6}\right\}\\ =405\times {10}^{-6}\text{F}\cdot \text{V}\\ =405\times {10}^{-6}\left(\frac{\text{A}\cdot \text{s}}{\text{V}}\cdot {\text{V}}^2\right)\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\end{array}$

Simplify the above equation to find $w$.

$\begin{array}{c}w=405\times {10}^{-6}\left(\text{A}\cdot \text{s}\cdot \text{V}\right)\\ =405\times {10}^{-6}\left(\frac{\text{C}}{\text{s}}\cdot \text{s}\cdot \text{V}\right)\left\{\because 1\text{A}=\frac{1\text{C}}{1\text{s}}\right\}\\ =405\times {10}^{-6}\text{C}\cdot \text{V}\\ =405\mu \text{J}\left\{\begin{array}{l}\because 1\text{J}=1\text{C}\cdot 1\text{V}\\ \text{1}\mu {\text{=10}}^{-6}\end{array}\right\}\end{array}$

Conclusion:

Thus, the energy stored in the element at time $t=0$ for the circuit shown in Figure 7.83 is $405\mu \text{J}$.

(b)

To determine

Explain whether the energy stored in the capacitor remains same for time $t>0$.

Answer to Problem 69E

No, the energy is not same in the capacitor for time $t>0$.

Explanation of Solution

Refer to Figure 7.83, it shows an $RC$ circuit. When an DC voltage is applied to the capacitor the charges up with an increasing voltage. Since the energy of capacitor is directly proportional to the voltage, for an increasing voltage energy also increases.

Due to the presence of resistor in a circuit, there is a voltage drop across the resistor which reduces the voltage across the capacitor. Therefore, the resistor slowly dissipate the energy stored in a capacitor for time $t>0$.

Conclusion:

Thus, the energy is not same in the capacitor for time $t>0$.

(c)

To determine

Determine the value of $v\left(t\right)$ at time $t=460\text{ms}$, $t=920\text{ms}$,  and $t=2.3\text{s}$ using transient simulation.

Answer to Problem 69E

The value of $v\left(t\right)$ at time $t=460\text{ms}$ is, $t=920\text{ms}$ is, and $t=2.3\text{s}$ is

Explanation of Solution

Calculation:

Create the new schematic in LTspice with series connected resistor and inductor of given circuit as shown in Figure 1.

Using SPICE Directive mention the command .ic V(Cap_voltage)=9 as shown in Figure 2.

Enter the stop time as 2.5s, time to start saving data as 0, and maximum Timestep as 10ms in Edit simulation Cmd as shown in Figure 3. Use label net option and mention Cap_voltage.

After adding the Spice directives the circuit shows as in Figure 4.

Now run the simulation and place the probe at the node of capacitor, the plot of the voltage across the capacitor with respect to time is shown as shown in Figure 5.

By placing the cursor on the graph, we obtain the current values for different time as shown in below.

For time $t=0\text{s}v\left(t\right)=9\text{V}$

For time $t=460\text{ms}v\left(t\right)=3.31\text{V}$

For time $t=920\text{ms},v\left(t\right)=1.22\text{V}$

For time $t=2.3\text{s},v\left(t\right)=60.4\text{mV}$

Conclusion:

Thus, the value of $v\left(t\right)$ at time $t=460\text{ms}$ is $3.31\text{V}$, $t=920\text{ms}$ is $1.22\text{V}$,  and $t=2.3\text{s}$ is $60.4\text{mV}$.

(d)

To determine

Find the fraction of initial energy remains in the capacitor at time $t=460\text{ms}$ and $t=2.3\text{s}$.

Answer to Problem 69E

The fraction of initial energy remains in the capacitor at time $t=460\text{ms}$ is $13.6\%$ and $t=2.3\text{s}$ is $0.0045\%$.

Explanation of Solution

Calculation:

Refer to part (b), the value of voltage at time $t=460\text{ms}$ is $v\left(460\text{ms}\right)=3.31\text{V}$ and the value of voltage at time $t=2.3\text{s}$ is $v\left(2.3\text{s}\right)=60.4\text{mV}$

Substitute $460\text{ms}$ for $t$ in equation (1) to find $w$ at time $t=460\text{ms}$.

$w\left(460\text{ms}\right)=\frac{1}{2}C{\left(v\left(460\text{ms}\right)\right)}^2$

Substitute $3.31\text{V}$ for $v\left(460\text{ms}\right)$, and $10\mu \text{F}$ for $C$ in above equation to find $w\left(460\text{ms}\right)$.

$\begin{array}{c}w\left(460\text{ms}\right)=\frac{1}{2}\left(10\mu \text{F}\right){\left(3.31\text{V}\right)}^2\\ =\frac{1}{2}\left(10\times {10}^{-6}\text{F}\right)\left(11{\text{V}}^2\right)\left\{\because 1\mu ={10}^{-6}\right\}\\ =55\times {10}^{-6}\text{F}\cdot {\text{V}}^2\left\{\because 1\mu ={10}^{-6}\right\}\\ =55\times {10}^{-6}\left(\frac{\text{A}\cdot \text{s}}{\text{V}}\cdot {\text{V}}^2\right)\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\end{array}$

Simplify the above equation to find $w\left(460\text{ms}\right)$.

$\begin{array}{c}w\left(460\text{ms}\right)=55\times {10}^{-6}\left(\text{A}\cdot \text{s}\cdot \text{V}\right)\\ =55\times {10}^{-6}\left(\frac{\text{C}}{\text{s}}\cdot \text{s}\cdot \text{V}\right)\left\{\because 1\text{A}=\frac{1\text{C}}{1\text{s}}\right\}\\ =55\times {10}^{-6}\text{C}\cdot \text{V}\\ =55\mu \text{J}\left\{\begin{array}{l}\because 1\text{J}=1\text{C}\cdot 1\text{V}\\ \text{1}\mu {\text{=10}}^{-6}\end{array}\right\}\end{array}$

Substitute $2.3\text{s}$ for $t$ in equation (1) to find $w$ at time $t=2.3\text{s}$.

$w\left(2.3\text{s}\right)=\frac{1}{2}C{\left(v\left(2.3\text{s}\right)\right)}^2$

Substitute $60.4\text{mV}$ for $v\left(2.3\text{s}\right)$, and $10\mu \text{F}$ for $C$ in above equation to find $w\left(2.3\text{s}\right)$.

$\begin{array}{c}w\left(2.3\text{s}\right)=\frac{1}{2}\left(10\mu \text{F}\right){\left(60.4\text{mV}\right)}^2\\ =\frac{1}{2}\left(10\times {10}^{-6}\text{F}\right){\left(60.4\times {10}^{-3}\text{V}\right)}^2\left\{\begin{array}{l}\because 1\mu ={10}^{-6}\\ 1\text{m}={10}^{-3}\end{array}\right\}\\ =\frac{1}{2}\left(10\times {10}^{-6}\text{F}\right)\left(3.648\times {10}^{-3}{\text{V}}^2\right)\\ =18.2\times {10}^{-9}\text{F}\cdot {\text{V}}^2\end{array}$

Simplify the above equation to find $w\left(2.3\text{s}\right)$.

$\begin{array}{c}w\left(2.3\text{s}\right)=18.2\times {10}^{-9}\left(\frac{\text{A}\cdot \text{s}}{\text{V}}\cdot {\text{V}}^2\right)\left\{\because 1\text{F}=\frac{\text{1}\text{A}\cdot 1\text{s}}{\text{1}\text{V}}\right\}\\ =18.2\times {10}^{-9}\left(\text{A}\cdot \text{s}\cdot \text{V}\right)\\ =18.2\times {10}^{-9}\left(\frac{\text{C}}{\text{s}}\cdot \text{s}\cdot \text{V}\right)\left\{\because 1\text{A}=\frac{1\text{C}}{1\text{s}}\right\}\\ =18.2\times {10}^{-9}\text{C}\cdot \text{V}\end{array}$

$w\left(2.3\text{s}\right)=18.2\text{nJ}\left\{\begin{array}{l}\because 1\text{J}=1\text{C}\cdot 1\text{V}\\ \text{1}{\text{n=10}}^{-9}\end{array}\right\}$

The fraction of initial energy remains stored in the capacitor at time $t=460\text{ms}$ is calculated as follows:

$\text{Fraction of energy at }t=460\text{ms}=\frac{w\left(460\text{ms}\right)}{w\left(0\right)}\times 100\%$

Substitute $405\mu \text{J}$ for $w\left(0\right)$, and $55\mu \text{J}$ for $w\left(460\text{ms}\right)$ in above equation to find $\text{Fraction of energy at }t=460\text{ms}$.

$\begin{array}{c}\text{Fraction of energy at }\left(t=460\text{ms}\right)=\frac{55\mu \text{J}}{405\mu \text{J}}\times 100\%\\ =13.6\%\end{array}$

The fraction of initial energy remains stored in the capacitor at time $t=2.3\text{s}$ is calculated as follows:

$\text{Fraction of energy at }t=2.3\text{s}=\frac{w\left(2.3\text{s}\right)}{w\left(0\right)}\times 100\%$

Substitute $405\mu \text{J}$ for $w\left(0\right)$, and $18.2\text{nJ}$ for $w\left(2.3\text{s}\right)$ in above equation to find $\text{Fraction of energy at }t=2.3\text{s}$.

$\begin{array}{c}\text{Fraction of energy at}\left(t=2.3\text{s}\right)=\frac{18.2\text{nJ}}{405\mu \text{J}}\times 100\%\\ =\frac{18.2\times {10}^{-9}\text{J}}{405\times {10}^{-6}\text{J}}\times 100\%\\ =0.0045\%\end{array}$

Conclusion:

Thus, the fraction of initial energy remains in the capacitor at time $t=460\text{ms}$ is $13.6\%$ and $t=2.3\text{s}$ is $0.0045\%$.