Chapter 7, Problem 20P

For the circuit in Fig. 7.100,

v = 90e^−50t V

and

i = 30e^−50t A, t > 0

1. (a) Find L and R.
2. (b) Determine the time constant.
3. (c) Calculate the initial energy in the inductor.
4. (d) What fraction of the initial energy is dissipated in 10 ms?

Figure 7.100

To determine

Find the value of resistance R and inductance L in the Figure 7.100.

Answer to Problem 20P

The value of resistance R in the circuit is $3\text{Ω}$ and the value of inductance L in the circuit is $60\text{mH}$.

Explanation of Solution

Given data:

The voltage $v$ is $90e^{-50t}\text{V}$.

The current $i$ is $30e^{-50t}\text{A}$.

Formula used:

Write the expression to find the voltage across the inductor for the given circuit.

$v=-L\frac{di}{dt}$ (1)

Here,

$\frac{di}{dt}$ is the change in current with respect to time, and

L is the inductance of the inductor.

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

$\tau =\frac{L}{R}$ (2)

Here,

R is the resistance of the resistor.

Write the general expression to find the voltage response in the RL Circuit.

$v\left(t\right)=V_0{e}^{-\frac{t}{\tau }}$ (3)

Here,

$\tau$ is the time constant for the RL circuit, and

$V_0$ is the initial voltage at the inductor terminal.

Calculation:

Substitute $90e^{-50t}\text{V}$ for $v$ and $30e^{-50t}\text{A}$ for $i$ in equation (1) to find the inductance L.

$\begin{array}{l}90e^{-50t}\text{V}=-L\frac{d\left(30e^{-50t}\text{A}\right)}{dt}\\ 90e^{-50t}=-30L\left(-50e^{-50t}\right)\end{array}$

Rearrange the equation as follows,

$\begin{array}{c}L=\frac{90e^{-50t}}{-30\left(-50e^{-50t}\right)}\\ =\frac{90}{1500}\text{H}\\ =0.06\text{H}\end{array}$

Convert the unit H to mH.

$\begin{array}{c}L=0.06\times {10}^3\times {10}^{-3}\text{H}\\ =60\times {10}^{-3}\text{H}\\ =60\text{mH}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Compare the given voltage $v=90e^{-50t}\text{V}$ and equation (3) to find the time constant $\tau$.

$\frac{1}{\tau }=50$ (4)

Rearrange the equation (4) to find the time constant $\tau$.

$\tau =\frac{1}{50}\text{s}$

Substitute $\frac{1}{50}\text{s}$ for $\tau$ and $0.06\text{H}$ for L in equation (2).

$\frac{1}{50}\text{s}=\frac{0.06\text{H}}{R}$ (5)

Rearrange the equation (5) to find the resistance R in ohms.

$\begin{array}{c}R=\frac{\left(0.06\text{H}\right)\left(50\right)}{1\text{s}}\\ =3\Omega \left\{\because 1\Omega =\frac{1\text{H}}{1\text{s}}\right\}\end{array}$

Conclusion:

Thus, the value of resistance R in the circuit is $3\text{Ω}$ and the value of inductance L in the circuit is $60\text{mH}$.

To determine

Find the time constant $\tau$ for the given RL circuit.

Answer to Problem 20P

The time constant $\tau$ for the given RL circuit is $20\text{ms}$.

Explanation of Solution

Given data:

Refer to part (a).

The value of resistance R is $3\text{Ω}$.

The value of inductance L is $60\text{mH}$.

Calculation:

Substitute $3\text{Ω}$ for R and $0.06\text{H}$ for L in equation (2) to find the time constant $\tau$.

$\tau =\frac{0.06\text{H}}{3\text{Ω}}$ (6)

Substitute the units $\frac{\text{V}}{\text{A}}$ for $\Omega$ and $\frac{\text{V}\cdot \text{s}}{\text{A}}$ for H in equation (6) to find the time constant $\tau$ in seconds.

$\begin{array}{c}\tau =\frac{\left(0.06\frac{\text{V}\cdot \text{s}}{\text{A}}\right)}{\left(3\frac{\text{V}}{\text{A}}\right)}\\ =0.02\text{s}\end{array}$

Convert the unit s to ms.

$\begin{array}{c}\tau =0.02\times {10}^3\times {10}^{-3}\text{s}\\ =20\times {10}^{-3}\text{s}\\ =20\text{ms}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Conclusion:

Thus, the time constant $\tau$ for the given RL circuit is $20\text{ms}$.

To determine

Find the value of initial energy stored in the inductor.

Answer to Problem 20P

The value of initial energy stored in the inductor is $27\text{J}$.

Explanation of Solution

Given data:

Refer to part (a).

The value of inductance L is $60\text{mH}$.

Formula used:

Write the expression to find the energy stored in the inductor.

$w=\frac{1}{2}L{i}^2$ (7)

Here,

$i$ is the current in the RL circuit.

Calculation:

The given current is,

$i=30e^{-50t}\text{A}$

The initial current in the RL circuit at $t=0$ is calculated as follows,

$\begin{array}{c}i=30e^{-50\left(0\right)}\text{A}\\ =30\text{A}\end{array}$

Substitute $60\text{mH}$ for L and $30\text{A}$ for i in equation (7) to find the initial energy W in the inductor.

$\begin{array}{c}w=\frac{1}{2}\left(60\text{mH}\right){\left(30\text{A}\right)}^2\\ =\frac{1}{2}\left(60\times {10}^{-3}\text{H}\right){\left(30\text{A}\right)}^2\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{1}{2}\left(60\times {10}^{-3}\frac{\text{J}}{{\text{A}}^2}\right){\left(30\text{A}\right)}^2\left\{\because 1\text{H}=\frac{\text{1}\text{J}}{\text{1}{\text{A}}^2}\right\}\\ =27\text{J}\end{array}$

Conclusion:

Thus, the value of initial energy stored in the inductor is $27\text{J}$.

To determine

Find the fraction of the energy dissipated in the first 10 ms.

Answer to Problem 20P

The fraction of the energy dissipated in the first 10 ms is $63.21\%$.

Explanation of Solution

Given data:

Refer to part (c).

The value of initial energy w stored in the inductor is $27\text{J}$.

Formula used:

Write the expression to find the energy stored in the inductor.

$w_{10}=\frac{1}{2}L{\left(i\left(t\right)\right)}^2$ (8)

Calculation:

The given current is,

$i=i\left(t\right)=30e^{-50t}\text{A}$

The current in the RL circuit at $t=10\text{ms}$ is calculated as follows,

$\begin{array}{c}i\left(t\right)=30e^{-50\left(10\text{ms}\right)}\text{A}\\ =18.196\text{A}\end{array}$

Substitute $60\text{mH}$ for L and $18.196\text{A}$ for i(t) in equation (8) to find the energy $w_{10}$ in the inductor.

$\begin{array}{c}{w}_{10}=\frac{1}{2}\left(60\text{mH}\right){\left(18.196\text{A}\right)}^2\\ =\frac{1}{2}\left(60\times {10}^{-3}\text{H}\right){\left(18.196\text{A}\right)}^2\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\frac{1}{2}\left(60\times {10}^{-3}\frac{\text{J}}{{\text{A}}^2}\right){\left(18.196\text{A}\right)}^2\left\{\because 1\text{H}=\frac{\text{1}\text{J}}{\text{1}{\text{A}}^2}\right\}\\ =9.933\text{J}\end{array}$

The fraction of the energy dissipated in the first 10 ms is calculated as follows.

$\text{Fraction of the energy dissipated in the first 10 ms}=\frac{w-w_{10}}{w}\times 100$ (9)

Substitute $27\text{J}$ for w and $9.933\text{J}$ for $w_{10}$ in equation (9).

$\begin{array}{c}\text{Fraction of the energy dissipated in the first 10 ms}=\frac{27\text{J}-9.933\text{J}}{27\text{J}}\times 100\\ =0.6321\times 100\\ =63.21\%\end{array}$

Conclusion:

Thus, the fraction of the energy dissipated in the first 10 ms is $63.21\%$.