Answer
See the detailed answer below.
Work Step by Step
This problem is the reverse of the previous problem.
Recall the potential difference through an inductor is given by
$$\Delta V_L=-L\dfrac{dI}{dt}$$
$$\Delta V_L=-L\dfrac{\Delta I}{\Delta t}$$
Plug the known;
$$\Delta V_L=-(50\times 10^{-3})\dfrac{\Delta I}{\Delta t} $$
Hence,
$$\Delta I=\dfrac{\Delta V_L\Delta t}{-(50\times 10^{-3})}\tag 1$$
We have, from the given graph, 4 stages. We will use (1) to find the potential difference in each stage.
$\bullet$ For the first stage from $t=0$ s to $t=10$ ms, where $I_0=0.2$ A:
$$\Delta I=I_1-I_0=\dfrac{\Delta V_L\Delta t}{-(50\times 10^{-3})} $$
$$ I_1=\dfrac{\Delta V_L\Delta t_1}{-(50\times 10^{-3})} +I_0$$
Plug the known;
$$ I_1=\dfrac{(-1)(10-0)\times 10^{-3}}{-(50\times 10^{-3})} +0.20=\bf 0.40\;\rm A$$
$\bullet\bullet$ For the second stage from $t=10$ ms to $t=20$ ms, the potential difference is zero, so the current is constant through this stage.
Hence,
$$I_1=I_2=\bf 0.40\;\rm A$$
$\bullet\bullet\bullet$ For the third stage from $t=20$ ms to $t=30$ ms,
$$\Delta I=I_3-I_2=\dfrac{\Delta V_L\Delta t}{-(50\times 10^{-3})} $$
$$ I_3=\dfrac{\Delta V_L\Delta t_3}{-(50\times 10^{-3})} +I_2$$
Plug the known;
$$ I_3=\dfrac{(2)(30-20)\times 10^{-3}}{-(50\times 10^{-3})} +0.40=\bf 0\;\rm A$$
$\bullet\bullet\bullet\bullet $ For the fourth stage from $t=30$ ms to $t=40$ ms, the potential difference is zero, so the current is constant through this stage.
Hence,
$$I_3=I_4=\bf 0\;\rm A$$
$\Rightarrow$ Noting that $I_1,I_2,I_3$, and $I_4$ are the currents at the end of each stage. $I_1$ is the current at the end of the first stage, and so on.
Now we can draw the graph of the current through these four stages.
See the graph below.
