Chapter 4, Problem 4.12HP

The voltage across a 0.5-mH inductor, Plotted as a function of time, is shown in Figure P4.12. Determine the current through the inductor at $t=6ms$ .

To determine

The current through the inductor at $t=6\text{ms}$ .

Answer to Problem 4.12HP

The current through the inductor at $t=6\text{ms}$ is $6\text{A}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The conversion from $\text{ms}$ to $\text{s}$ is given by,

  $1\text{ms}={10}^{-3}\text{s}$

The conversion from $\text{2}\text{ms}$ to $\text{s}$ is given by,

  $\text{2}\text{ms}=2\times {10}^{-3}\text{s}$

The conversion from $\text{4}\text{ms}$ to $\text{s}$ is given by,

  $4\text{ms}=4\times {10}^{-3}\text{s}$

The conversion from $\text{6}\text{ms}$ to $\text{s}$ is given by,

  $\text{6}\text{ms}=6\times {10}^{-3}\text{s}$

The conversion from $\text{mH}$ into $\text{H}$ is given by,

  $1\text{mH}={\text{10}}^{-3}\text{H}$

The conversion from $0.5\text{mH}$ into $\text{H}$ is given by,

  $0.5\text{mH}=0.5\times {\text{10}}^{-3}\text{H}$

From the graph the expression for the voltage between the points $\left(0,0\right)$ and $\left(2\text{ms},6\text{V}\right)$ is calculated as,

  $\begin{array}{c}v\left(t\right)=\frac{6-0}{2\times {10}^{-3}-0}\left(t-0\right)+0\\ =3000t\end{array}$

From the graph the expression for the voltage between the points $\left(2\text{ms},6\text{V}\right)$ and $\left(4\text{ms},-3\text{V}\right)$ is calculated as,

  $\begin{array}{c}v\left(t\right)=\left(\frac{-3-6}{4\times {10}^{-3}-2\times {10}^{-3}}\right)\left(t-2\times {10}^{-3}\right)+6\\ =-4500t+9+6\\ =15-4500t\end{array}$

From the graph the expression for the voltage at $t>4\text{ms}$ is given by,

  $v\left(t\right)=-3\text{V}$

The expression for the voltage across the inductor is given by,

  $v\left(t\right)=\{\begin{array}{l}3000t\text{V}\text{for}0\le t\le 2\text{ms}\\ 15-4500t\text{V}\text{for}2\le t\le 4\text{ms}\\ -3\text{V}\text{for}t>4\text{ms}\end{array}$

The initial current through the inductor is given by,

  $i\left(0\right)=0$

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

  $i_L\left(t\right)=\frac{1}{L}{\displaystyle {\int }_{t_0}^{t}v\left(t\right)dt}+i\left(t_0\right)$

Substitute $0.5\times {\text{10}}^{-3}\text{H}$ for $L$ and $3000t$ for $v\left(t\right)$ in the above equation.

  $i_L\left(t\right)=\frac{1}{0.5\times {\text{10}}^{-3}\text{H}}{\displaystyle {\int }_{t_0}^t\left(3000t\right)dt}+i\left(t_0\right)$ ............(1)

From equation (1) the current through the inductor for the time interval $0\le t\le 2\text{ms}$ is is given by,

  $i_L\left(t\right)=\frac{1}{0.5\times {\text{10}}^{-3}\text{H}}{\displaystyle {\int }_0^t\left(3000t\right)dt}+i\left(0\right)$

Substitute $0$ for $i\left(0\right)$ in the above equation.

  $\begin{array}{c}{i}_L\left(t\right)=\frac{1}{0.5\times {\text{10}}^{-3}\text{H}}{\displaystyle {\int }_0^t\left(3000t\right)dt}+i\left(0\right)\\ =\left(3\times {10}^6\right)t^2+0\end{array}$

Substitute $2\text{ms}$ for $t$ in the above equation.

  $\begin{array}{c}{i}_L\left(t\right)=\left(3\times {10}^6\right){\left(2\times {10}^{-3}\text{s}\right)}^2\\ =12\text{A}\end{array}$

From equation (1) the current through the inductor for the time interval $2\le t\le \text{4}\text{ms}$ is given by,

  $i_L\left(t\right)=\frac{1}{L}{\displaystyle {\int }_{2\text{ms}}^{t}v\left(t\right)dt}+i\left(2\text{ms}\right)$

Substitute $12\text{A}$ for $i\left(2\text{ms}\right)$, $0.5\times {\text{10}}^{-3}\text{H}$ for $L$ and $15-4500t\text{V}$ for $v\left(t\right)$ in the above equation.

  $\begin{array}{c}{i}_L\left(t\right)=\frac{1}{0.5\times {\text{10}}^{-3}\text{H}}{\displaystyle {\int }_{2\text{ms}}^t\left(15-4500t\text{V}\right)dt}+12\text{A}\\ =\left(2000\right){\left[15t-\frac{4500t^2}{2}\right]}_{2\text{ms}}^t+12\\ =\left(2000\right){\left[12t-225t^2\right]}_{2\text{ms}}^t+12\\ \text{=}\left(2000\right)\left[15t-2250t^2-0.03+0.009\right]+12\end{array}$

Solve further as,

  $\begin{array}{c}{i}_L\left(t\right)=30\times {10}^{3}t-4.5\times {10}^6{t}^2-42+12\\ =30\times {10}^{3}t-4.5\times {10}^6{t}^2-30\end{array}$

Substitute $4\text{ms}$ for $t$ in the above equation.

  $\begin{array}{c}{i}_L\left(t\right)=30\times {10}^3\left(4\times {10}^{-3}\text{s}\right)-4.5\times {10}^6{\left(4\times {10}^{-3}\text{s}\right)}^2-30\\ =120-72-30\\ =18\text{A}\end{array}$

The expression for the current through the inductor for interval $t\ge \text{4}\text{ms}$ is given by,

  $i_L\left(t\right)=\frac{1}{L}{\displaystyle {\int }_{4\text{ms}}^{t}v\left(t\right)dt}+i\left(\text{4}\text{ms}\right)$

Substitute $18\text{A}$ for $i\left(4\text{ms}\right)$, $0.5\times {\text{10}}^{-3}\text{H}$ for $L$ and $-3\text{V}$ for $v\left(t\right)$ in the above equation.

  $\begin{array}{c}{i}_L\left(t\right)=\frac{1}{0.5\times {\text{10}}^{-3}\text{H}}{\displaystyle {\int }_{4\text{ms}}^t\left(-3\right)dt}+18\text{A}\\ =\left(2000\right){\left[-3t\right]}_{4\text{ms}}^t+18\\ =\left(2000\right){\left[-3t+3\left(4\times {10}^{-3}\right)\right]}_{2\text{ms}}^t+18\\ \text{=}-6\times {10}^{-3}t+42\end{array}$

Substitute $6\text{ms}$ for $t$ in the above equation.

  $\begin{array}{c}6\text{ms}=-6\times {10}^{-3}\left(6\times {10}^{-3}\text{s}\right)+42\\ =-36+42\\ =6\text{A}\end{array}$

Conclusion:

Therefore, the current through the inductor at $t=6\text{ms}$ is $6\text{A}$ .