Chapter 5, Problem 5.60HP

To determine

The value of the voltage $v_0\left(t\right)$ across the inductor.

To sketch:

A graph of voltage for time $0<t<2\text{s}$ .

Answer to Problem 5.60HP

The value of the voltage $v_0\left(t\right)$ across the inductor is

  $v_0\left(t\right)=\{\begin{array}{l}5000e^{-10t}\text{V}\text{for}\text{0}\le t\le t_0\\ -6000e^{-10\left(t-t_0\right)}\text{V}\text{for}{t}_0\le t\le t_1\\ 2000e^{-10\left(t-t_1\right)}\text{V}\text{for}{t}_1\le t\le t_2\\ -1000e^{-10\left(t-t_2\right)}\text{V}\text{for}{t}_2\le t\le t_3\\ 5000e^{-10\left(t-t_3\right)}\text{V}\text{for}t\ge t_3\end{array}$

The waveform for the inductor voltage $v_0\left(t\right)$ is shown in Figure 3.

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

Mark the time interval on the source current waveform and redraw the circuit.

The required diagram is shown in Figure 2

  

The expression for the initial current through the inductor is given by,

  $i_L\left(0^{-}\right)=0\text{A}$

The expression for the initial current through the inductor for time $t=0^{+}$ is given by,

  $i_L\left(0^{+}\right)=0\text{A}$

For the circuit in DC steady state, inductor acts as the short circuit and the current through it is given by,

  $i_L\left(t_0\right)=i_S$

Substitute $10\text{A}$ for $i_S$ in the above equation.

  $i_L\left(t_0\right)=10\text{A}$

The expression for the time constant of the circuit is given by,

  $\tau =\frac{L}{R}$

Substitute $1\text{H}$ for $L$ and $10\Omega$ for $R$ in the above equation.

  $\begin{array}{c}\tau =\frac{1\text{H}}{10\Omega }\\ =0.1\text{sec}\end{array}$

The expression for the complete solution for current is given by,

  $i_L\left(t\right)=i_L\left(t_0\right)+\left[i_L\left(0^{-}\right)-i_L\left(t_0\right)\right]e^{\frac{-t}{\tau }}$

Substitute $10\text{A}$ for $i_L\left(t_0\right)$, $0\text{A}$ for $i_L\left(0^{-}\right)$ and $0.1\text{sec}$ for $\tau$ in the above equation.

  $\begin{array}{c}{i}_L\left(t\right)=10\text{A}+\left[0-10\text{A}\right]e^{\frac{-t}{0.1}}\\ =10-10e^{\frac{-t}{0.1}}\text{A}\end{array}$

The expression for the voltage across the inductor for the interval $0\le t\le t_0$ is given by,

  $v_0\left(t\right)=L\frac{d{i}_L\left(t\right)}{dt}$

Substitute $50\text{H}$ for $L$ and $10-10e^{-10t}\text{A}$ for $i_L\left(t\right)$ in the above equation.

  $\begin{array}{c}{v}_0\left(t\right)=\left(50\text{H}\right)\frac{d\left[10-10e^{-10t}\text{A}\right]}{dt}\\ =5000e^{-10t}\text{V}\end{array}$

The expression for the current through the inductor for the time interval $t_0\le t\le t_1$ is given by,

  $i_L\left(t_0\right)=10\text{A}$

For DC state the inductor is short circuited and the current through it is given by,

  $i_L\left(t_1\right)=i_S$

Substitute $-2\text{A}$ for $i_S$ in the above equation.

  $i_L\left(t_1\right)=-2\text{A}$

The expression for the complete solution for current is given by,

  $i_L\left(t\right)=i_L\left(t_1\right)+\left[i_L\left(t_0\right)-i_L\left(t_1\right)\right]e^{\frac{-\left(t-t_0\right)}{\tau }}$

Substitute $-2\text{A}$ for $i_L\left(t_1\right)$, $10\text{A}$ for $i_L\left(t_0\right)$ and $0.1\text{sec}$ for $\tau$ in the above equation.

  $\begin{array}{c}{i}_L\left(t\right)=-2\text{A}+\left[10\text{A}-\left(-2\text{A}\right)\right]e^{\frac{-\left(t-t_0\right)}{0.1}}\\ =-2+12e^{-10\left(t-t_0\right)}\text{A}\end{array}$

The expression for the voltage across the inductor for the interval $t_0\le t\le t_1$ is given by,

  $v_0\left(t\right)=L\frac{d{i}_L\left(t\right)}{dt}$

Substitute $50\text{H}$ for $L$ and $-2+12e^{-10\left(t-t_0\right)}\text{A}$ for $i_L\left(t\right)$ in the above equation.

  $\begin{array}{c}{v}_0\left(t\right)=\left(50\text{H}\right)\frac{d\left[-2+12e^{-10\left(t-t_0\right)}\text{A}\right]}{dt}\\ =-6000e^{-10\left(t-t_0\right)}\text{V}\end{array}$

The expression for the current through the inductor for the time interval $t_1\le t\le t_2$

  $i_L\left(t_1\right)=-2\text{A}$

For DC state, the inductor is short circuited and the current through it is given by,

  $i_L\left(t_2\right)=i_S$

Substitute $2\text{A}$ for $i_S$ in the above equation.

  $i_L\left(t_2\right)=2\text{A}$

The expression for the complete solution for current is given by,

  $i_L\left(t\right)=i_L\left(t_2\right)+\left[i_L\left(t_1\right)-i_L\left(t_2\right)\right]e^{\frac{-\left(t-t_1\right)}{\tau }}$

Substitute $-2\text{A}$ for $i_L\left(t_1\right)$, $2\text{A}$ for $i_L\left(t_2\right)$ and $0.1\text{sec}$ for $\tau$ in the above equation.

  $\begin{array}{c}{i}_L\left(t\right)=-2\text{A}+\left[-2\text{A}-2\text{A}\right]e^{\frac{-\left(t-t_1\right)}{0.1}}\\ =2-4e^{-10\left(t-t_1\right)}\text{A}\end{array}$

The expression for the voltage across the inductor for the interval $t_1\le t\le t_2$ is given by,

  $v_0\left(t\right)=L\frac{d{i}_L\left(t\right)}{dt}$

Substitute $50\text{H}$ for $L$ and $2-4e^{-10\left(t-t_1\right)}\text{A}$ for $i_L\left(t\right)$ in the above equation.

  $\begin{array}{c}{v}_0\left(t\right)=\left(50\text{H}\right)\frac{d\left[2-4e^{-10\left(t-t_1\right)}\text{A}\right]}{dt}\\ =2000e^{-10\left(t-t_1\right)}\text{V}\end{array}$

The expression for the current through the inductor for the time interval $t_2\le t\le t_3$ .

  $i_L\left(t_2\right)=2\text{A}$

For DC state the inductor is short circuited and the current through it is given by,

  $i_L\left(t_3\right)=i_S$

Substitute $0\text{A}$ for $i_S$ in the above equation.

  $i_L\left(t_3\right)=0\text{A}$

The expression for the complete solution for current is given by,

  $i_L\left(t\right)=i_L\left(t_2\right)+\left[i_L\left(t_1\right)-i_L\left(t_2\right)\right]e^{\frac{-\left(t-t_1\right)}{\tau }}$

Substitute $2\text{A}$ for $i_L\left(t_2\right)$, $0\text{A}$ for $i_L\left(t_3\right)$ and $0.1\text{sec}$ for $\tau$ in the above equation.

  $\begin{array}{c}{i}_L\left(t\right)=0\text{A}+\left[2\text{A}-0\text{A}\right]e^{\frac{-\left(t-t_2\right)}{0.1}}\\ =2e^{-10\left(t-t_2\right)}\text{A}\end{array}$

The expression for the voltage across the inductor for the interval $t_2\le t\le t_3$ is given by,

  $v_0\left(t\right)=L\frac{d{i}_L\left(t\right)}{dt}$

Substitute $50\text{H}$ for $L$ and $2e^{-10\left(t-t_2\right)}\text{A}$ for $i_L\left(t\right)$ in the above equation.

  $\begin{array}{c}{v}_0\left(t\right)=\left(50\text{H}\right)\frac{d\left[2e^{-10\left(t-t_2\right)}\text{A}\right]}{dt}\\ =-1000e^{-10\left(t-t_2\right)}\text{V}\end{array}$

The expression for the current through the inducer for the interval $t\ge t_3$ is given by,

  $i_L=0\text{A}$

The current through the inductor for the steady state is given by,

  $i_L\left(\infty \right)=i_S$

Substitute $10\text{A}$ for $i_S$ in the above equation.

  $i_L\left(\infty \right)=10\text{A}$

The expression for the complete solution for current is given by,

  $i_L\left(t\right)=i_L\left(\infty \right)+\left[i_L\left(t_3\right)-i_L\left(\infty \right)\right]e^{\frac{-\left(t-t_3\right)}{\tau }}$

Substitute $0\text{A}$ for $i_L\left(t_3\right)$, $10\text{A}$ for $10\text{A}$ and $0.1\text{sec}$ for $\tau$ in the above equation.

  $\begin{array}{c}{i}_L\left(t\right)=10\text{A}+\left[0\text{A}-10\text{A}\right]e^{\frac{-\left(t-t_3\right)}{\tau }}\\ =10\left[1-e^{-10\left(t-t_3\right)}\right]\text{A}\end{array}$

The expression for the voltage across the inductor for the interval $t\ge t_3$ is given by,

  $v_0\left(t\right)=L\frac{d{i}_L\left(t\right)}{dt}$

Substitute $50\text{H}$ for $L$ and $10\left[1-e^{-10\left(t-t_3\right)}\right]\text{A}$ for $i_L\left(t\right)$ in the above equation.

  $\begin{array}{c}{v}_0\left(t\right)=\left(50\text{H}\right)\frac{d\left[10\left[1-e^{-10\left(t-t_3\right)}\right]\text{A}\right]}{dt}\\ =5000e^{-10\left(t-t_3\right)}\text{V}\end{array}$

Thus, the expression for the voltage across the interval is given by,

  $v_0\left(t\right)=\{\begin{array}{l}5000e^{-10t}\text{V}\text{for}\text{0}\le t\le t_0\\ -6000e^{-10\left(t-t_0\right)}\text{V}\text{for}{t}_0\le t\le t_1\\ 2000e^{-10\left(t-t_1\right)}\text{V}\text{for}{t}_1\le t\le t_2\\ -1000e^{-10\left(t-t_2\right)}\text{V}\text{for}{t}_2\le t\le t_3\\ 5000e^{-10\left(t-t_3\right)}\text{V}\text{for}t\ge t_3\end{array}$

From above expression, the waveform for the voltage across the circuit is shown below.

The required diagram is shown in Figure 3

  

Conclusion:

Therefore, the waveform for the inductor voltage $v_0\left(t\right)$ is shown in Figure 3.