Hysteresis is a common phenomenon in the study of real physical systems. It refers to situations in which the current state of a system depends not only on the present conditions to which it is subjected, but also on its past history — i.e., on the “trajectory” by which the system arrived at its current state.
Although direct-current (DC) circuits typically do not exhibit hysteresis, in this problem we consider a circuit that does possess hysteretic behavior due to the presence of an component $X$, which models the behavior of a Shockley diode.
If the current $I$ through component $X$ is less than $I_{\mathrm{c}}$, its $I$–$V$ characteristic is described by
\[
U = 4U_0\,\frac{I}{I_0}\left(1 - \frac{I}{I_0}\right). \tag{1}
\]
If the current $I$ through component $X$ exceeds $I_{\mathrm{c}}$, its $I$–$V$ characteristic follows the hyperbolic relation
\[
U = \frac{A}{I}. \tag{2}
\]
The two operating regimes merge smoothly, so that the resulting $I(U)$ curve is differentiable.
Assume that the values $U_0$ and $I_0$ are known.
The voltage generated by the source at its terminals depends on the current $I$ flowing through it according to the relation
\begin{equation}
U = \mathcal{E} - I R_0. \tag{3}
\end{equation}
Here, the voltage $U$ is also the voltage across component $X$. Therefore, to determine the current that will establish in the circuit, one needs to plot line (3) on top of the $I$–$V$ characteristic of component $X$ and find the intersection point. Line (3) is called the load line and is used for the graphical analysis of the behavior of nonlinear elements.
The voltage $\mathcal{E}$ is gradually increasing up to the value $U_0$.
The voltage $\mathcal{E}$ continues to be gradually increased until the current in the circuit suddenly jumps when $\mathcal{E}$ reaches a certain value $\mathcal{E} = \mathcal{E}_\uparrow > U_0$.
After that, the voltage $\mathcal{E}$ is gradually decreased until the current in the circuit suddenly drops when $\mathcal{E}$ reaches a certain value $\mathcal{E} = \mathcal{E}_\downarrow < \mathcal{E}_\uparrow$.