A1  ^?? What is the minimum voltage $V_{\min}$ that must be applied to a discharge lamp for it to light up?

With a constant voltage across the lamp, the equation for the concentration of electrons turns to:\[\frac{\mathrm dn_\mathrm e}{\mathrm dt}=-\beta n_\mathrm e+\alpha\frac{V_{\min}^2}{F}n_\mathrm e=\left[\alpha\frac{V_{\min}^2}{F}-\beta\right]n_\mathrm e. \]The lamp will light up if the expression in parentheses is positive. Thus, the condition for $V_{\min}$ is:\[\alpha\frac{V_{\min}^2}{F}=\beta\implies\]

A2  ^?? Find the expressions for the equilibrium current through the lamp and the voltage across the lamp.

The equilibrium voltage across the lamp in this case must be $V_\min$. Indeed:

• if the voltage across the lamp is greater than $V_{\min}$, this would lead to an increase in electron concentration and an increase in current, but the current through the lamp should become smaller because the voltage drop across the internal resistance of the source has decreased;
• if the voltage across the lamp is less than $V_{\min}$, this would cause the concentration of electrons to drop and the current to decrease, but the current through the lamp must become larger because the voltage drop across the internal resistance of the source has increased.

The equilibrium current through the lamp will be equal to the current through the source, i.e.:

B1  ^?? Express the current and voltage on the lamp at some time through the concentration of electrons $n_\mathrm e$.

If the current through the lamp is $I$, then the voltage drop across the lamp is:\[V=\mathcal E-rI,\]and then from the equation for lamp resistance:\[R=\frac{F}{n_\mathrm e}=\frac{\mathcal E-rI}{I}\implies\]

Voltage drop across the lamp is:\[V=\mathcal E-rI=\mathcal E-r\frac{\mathcal E n_\mathrm e}{rn_\mathrm e+F}\implies\]

B2  ^?? Considering $n_\mathrm e$ to be nearly constant, average the equation for the concentration over time and find the equilibrium value of the electron concentration $n_{0}$.

Substituting the expressions obtained in the previous task into the equation for the electron concentration:\[0=-\beta n_{0}+\alpha\frac{V_0^2Fn_{0}}{(rn_{0}+F)^2}\cos^2\omega t.\]When averaging over time:\[\langle \cos^2\omega t\rangle_t=\dfrac{1}{2},\]so then:\[0=-\beta n_{0}+\alpha\frac{V_0^2Fn_{0}}{2(rn_{0}+F)^2}\\\beta=\alpha\frac{V_0^2F}{2(rn_{0}+F)^2}\\ rn_{0}+F=V_0\sqrt{\frac{\alpha F}{2\beta}}\implies\]

B3  ^?? Find $n_1$ and $\phi$.

Let's substitute\[n_{\mathrm e}(t)=\frac{1}{r}\left(V_0\sqrt{\frac{\alpha F}{2\beta}}-F\right)+n_1\cos(2\omega t+\phi)\]into the expression for electron concentration. Because we only need terms with $\cos(2\omega t)$, we will drop other ones:\[\begin{aligned}n_1\frac{\mathrm d\left[\cos(2\omega t+\phi)\right]}{\mathrm dt}&=-\beta n_1\cos(2\omega t+\phi)+\alpha\frac{V_0^2Fn_{0}}{2\left(F+rn_{0}\right)^2}\cos(2\omega t)=\\&=-\beta n_1\cos(2\omega t+\phi)+\frac{\beta}{r}\left(V_0\sqrt{\frac{\alpha F}{2\beta}}-F\right)\cos(2\omega t).\end{aligned}\]Since it's a linear trigonometric differential equation, it's convenient to rewrite it in complex form:\[2i\omega n_1e^{i \phi}=-\beta n_1e^{i\phi}+\frac{\beta}{r}\left(V_0\sqrt{\frac{\alpha F}{2\beta}}-F\right),\\n_1=\frac{\sqrt{2\beta F}\left(V_0\sqrt{\alpha}-\sqrt{2\beta F}\right)}{2r\sqrt{\beta^2+4\omega^2}},\quad -\phi=\arg\left(\beta+2i\omega\right)=\operatorname{arctg}\frac{2\omega}{\beta}.\]

B4  ^?? Determine $\delta$ for the lamp assuming $\delta\ll1$. What does this smallness condition correspond to in terms of the initial variables of the problem?

Since resistance is inversely proportional to electron concentration, we can rewrite $\delta$ in terms of concentration:\[\begin{aligned}\delta &=\frac{2n_1}{n_0}=\\&=2\cdot\frac{\sqrt{2\beta F}\left(V_0\sqrt{\alpha}-\sqrt{2\beta F}\right)}{2r\sqrt{\beta^2+4\omega^2}\cdot \dfrac{1}{r}\left(V_0\sqrt{\dfrac{\alpha F}{2\beta}}-F\right)}=\\&=\frac{2\beta}{\sqrt{\beta^2+4\omega^2}}\overset{\delta\ll1}\approx\frac{\beta}{\omega}.\end{aligned}\]

So, the smallness condition turns into: