(a)  ^2.00 Express $V_{1}$ and $V_{2}$ in terms of the other parameters defined above.

Critical voltages

The power heating the resistor is $P_{\text {el }}=V^{2} / R_{j}$. The thermal equilibrium is reached when $P_{\text {el }}=P=\alpha\left(T_{\text {eq }}-T_{0}\right)$. To avoid oscillations, the equilibrium temperature $T_{\text {eq }}$ must satisfy $T_{\text {eq }}< T_{c}$ if $R=R_{1}$ and $T_{\text {eq }}>T_{c}$ if $R=R_{2}$. Solving for $V$, we have
\begin{equation*}
V=\sqrt{R_{j} \alpha\left(T_{\mathrm{eq}}-T_{0}\right)} . \tag{1}
\end{equation*}The critical values therefore are

(b)  ^6.00 Assuming that $V_{1}<V<V_{2}$, sketch qualitatively how the temperature of the resistor $T$ depends on time $t$, and find the ratio $\left(T_{\text {max }}-T_{0}\right) /\left(T_{\text {min }}-T_{0}\right)$, where $T_{\text {max }}$ and $T_{\text {min }}$ denote the maximal and minimal values of $T$, respectively.

Temperature behaviour

In the oscillating regime, we have a time-dependent current $I(t)$. The power dissipated over the resistor is $P_{\mathrm{el}}(t)=$ $R(t) I(t)^{2}$. By assumption (ii), we may assume that the thermal equilibrium is reached very fast, i.e. $P_{\mathrm{el}}(t)=P(t)$. The temperature $T(t)$ is therefore determined by the current via
\begin{equation*}
T(t)=T_{0}+\frac{R(t) I(t)^{2}}{\alpha} \tag{3}
\end{equation*}If the resistance has value $R_{1}$, the current will increase, trying to reach $J_{1}=V / R_{1}$. The difference $I(t)-V / R_{1}$ will decay exponentially, with characteristic time $L / R_{1}$. The phase transition occurs once the critical current
$$I_{1}=\sqrt{\frac{\alpha\left(T_{c}-T_{0}\right)}{R_{1}}}
$$is reached. After the phase transition, the current will decrease, approaching the new equilibrium value $J_{2}=V / R_{2}$. Again, $I(t)-V / R_{2}$ will decay exponentially with characteristic time $L / R_{2}$, until the critical current
$$
I_{2}=\sqrt{\frac{\alpha\left(T_{c}-T_{0}\right)}{R_{2}}}
$$is reached. This behaviour is shown in Fig. 1.

Together with (3), we see that the temperature behaves like in Figure 2.

The maximum and minimum temperatures will be attained just after the phase transitions occur. We obtain that

(c)  ^2.00 Find the period of oscillations if $V=\sqrt{V_{1} V_{2}}$ and $R_{2}=16 R_{1}$.

Period of oscillations

If the phase transition occurs at $t=0$, with the resistance changing from $R_{j^{\prime}}$ to $R_{j}$, the current is given by
\begin{equation*}
I(t)=\frac{V}{R_{j}}+\left(I_{j^{\prime}}-\frac{V}{R_{j}}\right) \mathrm{e}^{-R_{j} t / L} \tag{5}
\end{equation*}until the next phase transition occurs when $I\left(t_{j}\right)=I_{j}$. Hence, the period is
\begin{equation*}
t_{1}+t_{2}=\frac{L}{R_{1}} \ln \left(\frac{I_{2}-V / R_{1}}{I_{1}-V / R_{1}}\right)+\frac{L}{R_{2}} \ln \left(\frac{I_{1}-V / R_{2}}{I_{2}-V / R_{2}}\right) \tag{6}
\end{equation*}Inserting the relations $R_{2}=\eta R_{1}$ and $V=\sqrt{V_{1} V_{2}}=$ $\eta^{1 / 4} \sqrt{R_{1} \alpha\left(T_{c}-T_{0}\right)}$, we obtain the period