Chapter 7, Problem 8P

For the circuit in Fig. 7.88, if

$v=10e^{-4t}\text{V}\text{and}i=\mathrm{0..2}{e}^{-4t}\text{A},t>0$

(a)    Find R and C.

(b)    Determine the time constant.

(c)    Calculate the initial energy in the capacitor.

(d)    Obtain the time it takes to dissipate 50 percent of the initial energy.

Figure 7.88

For Prob. 7.8.

To determine

Find the value of resistance R and capacitance C in the circuit.

Answer to Problem 8P

The value of resistance R in the circuit is $50\text{Ω}$ and the value of capacitance C in the circuit is $5\text{mF}$.

Explanation of Solution

Given data:

The voltage across the capacitor $\left(v\right)$ is $10e^{-4t}\text{V}$.

The current flows through the circuit $\left(i\right)$ is $0.2e^{-4t}\text{A}$.

Formula used:

Write the expression to find the voltage across the capacitor for a source-free RC circuit.

$v=V_0{e}^{-\frac{t}{\tau }}$ (1)

Here,

$V_0$ is the initial voltage at $t=0$, and

$\tau$ is the time constant for RC circuit.

Write the expression to find the time constant for RC circuit.

$\tau =RC$ (2)

Here,

R is the resistance of the resistor, and

C is the capacitance of the capacitor.

Write the expression to find the current through the capacitor.

$i=C\frac{dv}{dt}$ (3)

Here,

$v$ is the voltage across the capacitor.

Calculation:

Refer to Figure 7.88 in the textbook. In the circuit, the direction of current is given as leaving form the positive terminal the capacitor, therefore in equation (3), the current direction is taken as negative. The equation (3) becomes,

$-i=C\frac{dv}{dt}$ (4)

Substitute $10e^{-4t}\text{V}$ for $v$ and $0.2e^{-4t}\text{A}$ for $i$ in equation (4).

$\begin{array}{c}-0.2e^{-4t}\text{A}=C\frac{d}{dt}\left(10e^{-4t}\text{V}\right)\\ =C\left(10\left(-4e^{-4t}\right)\frac{\text{V}}{\text{s}}\right)\\ =C\left(-40e^{-4t}\frac{\text{V}}{\text{s}}\right)\end{array}$

Rearrange the equation to find capacitance C.

$C=\frac{-0.2e^{-4t}\text{A}}{-40e^{-4t}\frac{\text{V}}{\text{s}}}$

$C=5\times {10}^{-3}\frac{\text{A}\cdot \text{s}}{\text{V}}$ (5)

Substitute the unit F for $\frac{\text{A}\cdot \text{s}}{\text{V}}$ in equation (5) to find the capacitance C in farads.

$\begin{array}{c}C=5\times {10}^{-3}\text{F}\\ =5\text{mF}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Compare the given voltage across the capacitor $v=10e^{-4t}\text{V}$ and equation (1) for the time constant $\tau$.

$\tau =\frac{1}{4}\text{s}$

Substitute $\frac{1}{4}\text{s}$ for $\tau$ in equation (2).

$RC=\frac{1}{4}\text{s}$ (6)

Substitute $5\text{mF}$ for C in equation (6).

$R\left(5\text{mF}\right)=\frac{1}{4}\text{s}$

Rearrange the equation to find resistance R.

$\begin{array}{c}R=\frac{\left(\frac{1}{4}\text{s}\right)}{5\text{mF}}\\ =\frac{\left(\frac{1}{4}\text{s}\right)}{5\times {10}^{-3}\text{F}}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

$R=50\frac{\text{s}}{\text{F}}$ (7)

Substitute the unit $\Omega$ for $\frac{\text{s}}{\text{F}}$ in equation (7) to find the resistance R in ohms.

$R=50\Omega$

Conclusion:

Thus, the value of resistance R in the circuit is $50\text{Ω}$ and the value of capacitance C in the circuit is $5\text{mF}$.

To determine

Calculate the value of time constant for the RC circuit.

Answer to Problem 8P

The value of time constant for the RC circuit is $0.25\text{s}$.

Explanation of Solution

Given data:

Refer part (a),

The value of resistance is $\left(R\right)$ is $50\text{Ω}$.

The value of capacitance $\left(C\right)$ is $5\text{mF}$.

Calculation:

Substitute $50\text{Ω}$ for $R$ and $5\text{mF}$ for $C$ in equation (2) to find the time constant $\tau$.

$\tau =\left(50\text{Ω}\right)\left(5\text{mF}\right)$ (8)

Substitute the units $\frac{\text{V}}{\text{A}}$ for $\Omega$ and $\frac{\text{A}\cdot \text{s}}{\text{V}}$ for F in equation (8) to find the time constant $\tau$ in seconds.

$\begin{array}{c}\tau =\left(50\frac{\text{V}}{\text{A}}\right)\left(5\text{m}\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\\ =\left(50\frac{\text{V}}{\text{A}}\right)\left(5\times {10}^{-3}\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\left\{\because 1\text{k}={10}^3,1\text{m}={10}^{-3}\right\}\\ =0.25\text{s}\end{array}$

Conclusion:

Thus, the value of time constant for the RC circuit is $0.25\text{s}$.

To determine

Find the initial energy in the capacitor $W_{\text{C}}\left(0\right)$.

Answer to Problem 8P

The initial energy in the capacitor $W_{\text{C}}\left(0\right)$ is $250\text{mJ}$.

Explanation of Solution

Given data:

Refer part (a),

The value of capacitance $\left(C\right)$ is $5\text{mF}$.

The initial voltage $\left(V_0\right)$ at $t=0$ is .

Formula used:

Write the expression to find the initial energy in the capacitor.

$W_{\text{C}}\left(0\right)=\frac{1}{2}C{V}_0^2$ (9)

Calculation:

The voltage across the capacitor,

$v=10e^{-4t}\text{V}$ (10)

Compare the equation (10) and (1) for the initial voltage $V_0$.

$V_0=10\text{V}$

Substitute $10\text{V}$ for $V_0$ and $5\text{mF}$ for C in equation (9) to find the initial energy $W_{\text{C}}\left(0\right)$.

$W_{\text{C}}\left(0\right)=\frac{1}{2}\left(5\text{mF}\right){\left(10\text{V}\right)}^2$ (11)

Substitute the unit $\frac{\text{J}}{{\text{V}}^{\text{2}}}$ for F in equation (11) to find the initial energy $W_{\text{C}}\left(0\right)$ in joules.

$\begin{array}{c}{W}_{\text{C}}\left(0\right)=\frac{1}{2}\left(5\text{m}\frac{\text{J}}{{\text{V}}^{\text{2}}}\right){\left(10\text{V}\right)}^2\\ =0.25\text{J}\end{array}$

Convert the unit J to mJ.

$\begin{array}{c}{W}_{\text{C}}\left(0\right)=250\times {10}^{-3}\text{J}\\ =250\text{mJ}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Conclusion:

Thus, the initial energy in the capacitor $W_{\text{C}}\left(0\right)$ is $250\text{mJ}$.

To determine

Find the time taken to dissipate 50 percent of the initial energy.

Answer to Problem 8P

The time taken to dissipate 50 percent of the initial energy is $86.6\text{ms}$.

Explanation of Solution

The energy in the capacitor to dissipate 50 percent of the initial energy is,

$W=\frac{1}{2}{W}_{\text{C}}\left(0\right)=\frac{1}{2}C{v}^2$ (12)

Substitute $250\text{mJ}$ for $W_{\text{C}}\left(0\right)$, $5\text{mF}$ for C and $10e^{-4t}\text{V}$ for v in equation (12) to find t.

$\frac{1}{2}\left(250\text{mJ}\right)=\frac{1}{2}\left(5\text{mF}\right){\left(10e^{-4t}\text{V}\right)}^2$

$250\times {10}^{-3}\text{J}=\left(5\times {10}^{-3}\text{F}\right){\left(10e^{-4t}\text{V}\right)}^2\left\{\because 1\text{m}={10}^{-3}\right\}$ (13)

Substitute the unit $\frac{\text{J}}{{\text{V}}^{\text{2}}}$ for F in equation (13).

$\begin{array}{l}250\times {10}^{-3}\text{J}=\left(5\times {10}^{-3}\frac{\text{J}}{{\text{V}}^{\text{2}}}\right){\left(10e^{-4t}\text{V}\right)}^2\\ 50={\left(10e^{-4t}\right)}^2\\ 50=100e^{-8t}\\ 1=2e^{-8t}\end{array}$

Rearrange the equation as follows.

$\frac{1}{2}=e^{-8t}$

$0.5=e^{-8t}$ (14)

Take ln on both sides of the equation (14) to find the time t in seconds.

$\begin{array}{l}\ln \left(0.5\right)=\ln \left(e^{-8t}\right)\\ -0.69314=-8t\\ 0.69314=8t\end{array}$

Rearrange the equation as follows.

$\begin{array}{c}t=\frac{0.69314}{8}\text{s}\\ =0.0866\text{s}\end{array}$

Convert the unit s to ms.

$\begin{array}{c}t=86.6\times {10}^{-3}\text{s}\\ =86.6\text{ms}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Conclusion:

Thus, the time taken to dissipate 50 percent of the initial energy is $86.6\text{ms}$.