Chapter 6, Problem 2P

a.

To determine

Write the expressions of current $i\left(t\right)$ in the intervals $t<0,0\le t\le 25\text{ms},25\text{ms}\le t\le 50\text{ms,}\text{and}t>\text{50}\text{ms}$.

Answer to Problem 2P

The current $i\left(t\right)$ in the intervals $t<0,0\le t\le 25\text{ms},25\text{ms}\le t\le 50\text{ms,}\text{and}t>\text{50}\text{ms}$ are $\underset{\_}{0,-4t\text{A},-0.2+4t\text{A},\text{and}0}$ respectively.

Explanation of Solution

Given data:

Refer to the given figure in the respective question for the triangular current pulse.

Calculation:

PSPICE:

Design the circuit in PSPICE that contains the inductor and the given triangular pulse as shown in Figure 1.

Provide the simulation settings as shown in Figure 2.

Run the circuit and obtain the graph of current versus time as shown n Figure 3.

For $t<0\text{and}t>\text{50}\text{ms}$, the current $i\left(t\right)$ is zero since the input current pulse is zero.

From Figure 3, it is concluded that, for $0\le t\le 25\text{ms}\text{and}25\text{ms}\le t\le 50\text{ms}$, the current $i\left(t\right)$ are $-4t\text{A}\text{and}-0.2+4t\text{A}$ respectively.

Conclusion:

Thus, the current $i\left(t\right)$ in the intervals $t<0,0\le t\le 25\text{ms},25\text{ms}\le t\le 50\text{ms,}\text{and}t>\text{50}\text{ms}$ are $\underset{\_}{0,-4t\text{A},-0.2+4t\text{A},\text{and}0}$ respectively.

b.

To determine

Derive the expressions for the inductor voltage, power, and energy.

Answer to Problem 2P

The expressions for the inductor voltage, power, and energy are $\underset{\_}{v=\{\begin{array}{cc}0& t<0\\ -2\text{V}& 0\le t\le 25\text{ms}\\ 2\text{V}& 25\text{ms}\le t\le 50\text{ms}\\ 0& t>50\text{ms}\end{array}}$, $\underset{\_}{p=\{\begin{array}{cc}0& t<0\\ 8t\text{W}& 0\le t\le 25\text{ms}\\ 8t-0.4\text{W}& 25\text{ms}\le t\le 50\text{ms}\\ 0& t>50\text{ms}\end{array}}$, and $\underset{\_}{w=\{\begin{array}{cc}0& t<0\\ 4t^2\text{J}& 0\le t\le 25\text{ms}\\ 4t^2-0.4t+\left(10\times {10}^{-3}\right)\text{J}& 25\text{ms}\le t\le 50\text{ms}\\ 0& t>50\text{ms}\end{array}}$ respectively.

Explanation of Solution

Calculation:

Write the expression for the inductor voltage.

$v=L\frac{di}{dt}$        (1)

Substitute 50 mH for L and $-4t$ A for $i$ in Equation (1).

$\begin{array}{c}v=\left(50\text{mH}\right)\frac{d\left(-4t\right)}{dt}\left\{\because 0\le t\le 25\text{ms}\right\}\\ =\left(50\times {10}^{-3}\text{H}\right)\left(-4\right)\\ =-2\text{V}\end{array}$

Substitute 50 mH for L and $-0.2+4t\text{A}$ for $i$ in Equation (1).

$\begin{array}{c}v=\left(50\text{mH}\right)\frac{d}{dt}\left(-0.2+4t\right)\left\{\because 25\text{ms}\le t\le 50\text{ms}\right\}\\ =\left(50\times {10}^{-3}\text{H}\right)\left(4\right)\\ =2\text{V}\end{array}$

For $t<0\text{and}t>\text{50}\text{ms}$, the inductor voltage is zero since the input current is zero.

The inductor voltage expression is,

$v=\{\begin{array}{cc}0& t<0\\ -2\text{V}& 0\le t\le 25\text{ms}\\ 2\text{V}& 25\text{ms}\le t\le 50\text{ms}\\ 0& t>50\text{ms}\end{array}$

Consider the expression for the power in the inductor.

$p=vi$        (2)

Substitute $-2\text{V}$ for v and $-4t$ A for $i$ in Equation (2).

$\begin{array}{c}p=\left(-2\right)\left(-4t\right)\left\{\because 0\le t\le 25\text{ms}\right\}\\ =8t\text{W}\end{array}$

Substitute $2\text{V}$ for v and $\left(-0.2+4t\right)\text{A}$ for $i$ in Equation (2).

$\begin{array}{c}p=\left(2\right)\left(-0.2+4t\right)\text{A}\left\{\because 25\text{ms}\le t\le 50\text{ms}\right\}\\ =8t-0.4\text{W}\end{array}$

For $t<0\text{and}t>\text{50}\text{ms}$, the inductor power is zero since the input current is zero.

The power in the inductor is,

$p=\{\begin{array}{cc}0& t<0\\ 8t\text{W}& 0\le t\le 25\text{ms}\\ 8t-0.4\text{W}& 25\text{ms}\le t\le 50\text{ms}\\ 0& t>50\text{ms}\end{array}$

For $t<0\text{and}t>\text{50}\text{ms}$, the energy in the inductor is zero since the input current is zero.

Calculate the energy in the inductor.

$w=\{\begin{array}{cc}0& t<0\\ {\displaystyle {\int }_0^t\left(8x\right)dx}& 0\le t\le 25\text{ms}\\ {\displaystyle {\int }_{0.025}^t\left(8x-0.4\right)dx+2.5\times {10}^{-3}}& 25\text{ms}\le t\le 50\text{ms}\\ 0& t>50\text{ms}\end{array}$

Simplify the equation as follows.

$w=\{\begin{array}{cc}0& t<0\\ {\left[\frac{8x^2}{2}\right]}_0^t=4t^2\text{J}& 0\le t\le 25\text{ms}\\ {\left[4x^2-0.4x\right]}_{0.025}^t+\left(2.5\times {10}^{-3}\right)=4t^2-0.4t+\left(10\times {10}^{-3}\right)\text{J}& 25\text{ms}\le t\le 50\text{ms}\\ 0& t>50\text{ms}\end{array}$

Conclusion:

Thus, the expressions for the inductor voltage, power, and energy are $\underset{\_}{v=\{\begin{array}{cc}0& t<0\\ -2\text{V}& 0\le t\le 25\text{ms}\\ 2\text{V}& 25\text{ms}\le t\le 50\text{ms}\\ 0& t>50\text{ms}\end{array}}$, $\underset{\_}{p=\{\begin{array}{cc}0& t<0\\ 8t\text{W}& 0\le t\le 25\text{ms}\\ 8t-0.4\text{W}& 25\text{ms}\le t\le 50\text{ms}\\ 0& t>50\text{ms}\end{array}}$, and $\underset{\_}{w=\{\begin{array}{cc}0& t<0\\ 4t^2\text{J}& 0\le t\le 25\text{ms}\\ 4t^2-0.4t+\left(10\times {10}^{-3}\right)\text{J}& 25\text{ms}\le t\le 50\text{ms}\\ 0& t>50\text{ms}\end{array}}$ respectively.