Consider a 1H inductor. The current across the inductor is given by iL(t) = { t, 0 ≤ t ≤ 2 s 5/2 − t/4, 2 ≤ t ≤ 10 s 0, t ≥ 10 s (A). Find and plot the voltage across the inductor, as well as its power and energy for t 0.

Question

Consider a [latex]1 \mathrm{H}[/latex] inductor. The current across the inductor is given by

[latex]i_{L}(t)=\left\{\begin{aligned}t, \quad & 0 \leq t \leq 2 \mathrm{~s} \5 / 2-t / 4, \quad & 2 \leq t \leq 10 \mathrm{~s} \quad(\mathrm{~A}) .\0, \quad & t \geq 10 \mathrm{~s}\end{aligned} ight.[/latex]

Find and plot the voltage across the inductor, as well as its power and energy for [latex]t>0[/latex].

Accepted Answer

Using the expression for the voltage of an inductor, we have

[latex]v_{L}(t)=L \frac{d i_{L}(t)}{d t}=\left\{\begin{aligned}1, \quad & 010 \mathrm{~s}\end{aligned} ight.[/latex]

Then the current and voltage across the inductor can be plotted as follows.

Even when they are modeled ideally, inductors in resistive circuits cannot have discontinuous currents. Specifically, it takes time for an inductor (connected to some resistors) to store and release energy. Since the energy of an inductor is proportional to the square of its current, a jump in the current of an inductor requires a jump (instant change) in its energy that may not be physically possible. Such a discontinuous current at a particular time also needs an infinite voltage across the inductor.

Next, for the given voltage and current values above, the power of the inductor can be obtained as

[latex]p_{L}(t)=v_{L}(t) i_{L}(t)=\left\{\begin{aligned}t, \quad & 010 \mathrm{~s}\end{aligned} ight.[/latex]

While the energy can be found by integrating the power, it can also be obtained as

[latex]w_{L}(t)=\frac{1}{2} L\left[i_{L}(t) ight]^{2}=\left\{\begin{aligned}t^{2} / 2, \quad & 0 \leq t \leq 2 \mathrm{~s} \25 / 8-5 t / 8+t^{2} / 32, \quad & 2 \leq t \leq 10 \mathrm{~s}\quad(\mathrm{~J}) . \0, \quad & t \geq 10 \mathrm{~s}\end{aligned} ight.[/latex]

We note that, for [latex]0

Question 6.103: Consider a 1H inductor. The current across the inductor is g...

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