A1  ^?? By averaging the equation for temperature, obtain the equation that relates $V_0$, $T_1$, and the other variables of the problem.

We can assume thermistor's resisitance to be approximately constant (because its relative change is far less than relative changes of current and voltage):\[R_1=R_0\exp\left[\frac{E_g}{k_BT_1}\right].\]Plugging this into the equation for temperature, we get:\[C\frac{\mathrm dT}{\mathrm dt}=-\kappa(T_1-T_0)+\frac{V_0^2}{R_0}\exp\left[-\frac{E_g}{k_BT_1}\right]\cos^2\omega t.\]Time-averaging this equation leaves us with:\[0=-\kappa(T_1-T_0)+\frac{V_0^2}{2R_0}\exp\left[-\frac{E_g}{k_BT_1}\right]\implies\]

A2  ^?? Calculate the activation energy $E_g$. Express your answer in electronvolts ($1~eV=1.60\cdot10^{-19}~J$). Boltzmann constatn is equal to $k_B=1.38\cdot10^{-23}~J/K$.

The simplest linearization of the dependence obtained in the previous task is:\[\ln\frac{t_1-t_0}{V_0^2}=\ln\frac{\kappa}{2R_0}+\frac{E_g}{k_B}\cdot\frac{1}{t_1+273~K}.\]Calculations of $\ln\dfrac{t_1-t_0}{V_0^2}$ and $\dfrac{1}{t_1+273~K}$ are shown in the table below:

$V_0,~V$	2.0	2.5	3.0	3.5	4.0	4.5	5.0
$t_1,~{}^\circ\mathrm C$	20.7	21.1	21.6	22.2	22.9	23.8	24.8
$\ln\dfrac{t_1-t_0}{V_0^2}$	0.175	0.176	0.178	0.180	0.181	0.188	0.192
$\dfrac{10^3}{t_1+273~K}$	3.405	3.400	3.394	3.388	3.380	3.369	3.358
$V_0,~V$	5.5	6.0	6.5	7.0	7.5	8.0	8.5
$t_1,~{}^\circ\mathrm C$	26.0	27.4	29.1	31.0	33.3	36.2	40.0
$\ln\dfrac{t_1-t_0}{V_0^2}$	0.198	0.206	0.215	0.224	0.236	0.253	0.277
$\dfrac{10^3}{t_1+273~K}$	3.344	3.329	3.310	3.289	3.265	3.234	3.195

Next, let's plot this dependency:

From the plot we find:\[\ln\frac{\kappa}{2R_0}=3.759,\quad \frac{E_g}{k_B}=2.074\cdot10^3.\]So:\[E_g=2074k_B=2074\cdot\frac{1.38\cdot10^{-23}}{1.60\cdot10^{-19}}~eV\approx0.18~eV.\]

A3  ^?? Calculate $R_0$ and $\kappa$.

Using room temperature $t_0$ ans resistance $R(t_0)$, we obtain $R_0$:\[\begin{aligned}&R=R_0\exp\left[\frac{E_g}{k_B(t_0+273~K)}\right]\implies\\&\implies R_0=R\exp\left[-\frac{E_g}{k_B(t_0+273~K)}\right]=8.60\cdot \exp\left[\frac{2074}{293}\right]=7.3\cdot10^{-3}~\Omega.\end{aligned}\]Finally, we find $\kappa$:\[\kappa=2R_0\exp[3.759]=0.62~\frac{W}{K}.\]

B1  ^?? Find expressions for $T_2$ and $\phi$.

Leaving only the terms containing $\cos(2\omega t)$ in the equation for temperature (the constant terms will be cancel out as a consequence of the previous task, and the terms with larger frequencies will be of higher order of smallness and irrelevant). So:\[CT_2\frac{\mathrm d\cos[2\omega t+\phi]}{\mathrm dt}=-\kappa T_2\cos[2\omega t+\phi]+\frac{V_0^2}{2R_0}\exp\left[-\frac{E_g}{k_BT}\right]\cos(2\omega t).\]Since the obtained equation turned out to be linear trigonometric equation, it is convenient to work with it in the complex notation:\[2i\omega CT_2e^{i\phi}=-\kappa T_2e^{i\phi}+\frac{V_0^2}{2R_0}\exp\left[-\frac{E_g}{k_BT}\right]=-\kappa T_2e^{i\phi}+\kappa(T_1-T_0)\\ T_2=\frac{\kappa(T_1-T_0)}{\sqrt{\kappa^2+4\omega^2C^2}}\overset{T_2\ll(T_1-T_0)}\approx \frac{\kappa(T_1-T_0)}{2\omega C},\quad \phi=-\operatorname{arctg}\frac{2\omega C}{\kappa}.\]

B2  ^?? Calculate thermistor's heat capacity $C$.

Due to small temperature fluctuations, the resistance of the thermistor fluctuates, which results in a hysteresis loop. Let's find how the current through the thermistor depends on time:\[2\omega C\gg \kappa\implies \phi\approx -\frac{\pi}{2}\implies T(t)=T_1+T_2\sin(2\omega t).\]Then:\[\begin{aligned}I(t)&=\frac{V(t)}{R(t)}=\\&=\frac{V_0}{R(T_1)}\cos(\omega t)\cdot\left[1-T_2\sin(2\omega t)\cdot\left.\frac{\mathrm dR(T)}{\mathrm dT}\right|_{T=T_1}\right]=\\&=\frac{V_0}{R(T_1)}\cos(\omega t)\cdot\left[1-T_2\sin(2\omega t)\cdot R(T_1)\cdot\left(-\frac{E_g}{k_BT_1^2}\right)\right]=\\&=\frac{V_0}{R(T_1)}\cos(\omega t)+\frac{V_0E_gT_2}{k_BT_1^2}\cos(\omega t)\sin(2\omega t).\end{aligned}\]The maximum value of the first-order correction to the current is achieved at:\[\cos(\omega t) \sin(2\omega t)\to\max\implies\\\begin{aligned}0&=\frac{\mathrm d}{\mathrm d(\omega t)}\big[\cos(\omega t) \sin(2\omega t)\big]=\\&=-\sin(\omega t)\sin(2\omega t)+2\cos(\omega t)\cos(2\omega t)=\\&=-2\sin^2(\omega t)\cos(\omega t)+2\cos(\omega t)\big[\cos^2(\omega t)-\sin^2(\omega t)\big]=\\&=2\cos(\omega t)\big[\cos^2(\omega t)-2\sin^2(\omega t)\big]=\\&=2\cos(\omega t)\big[1-3\sin^2(\omega t)\big]\implies\end{aligned}\\\sin(\omega t)=\sqrt\frac13,\quad \cos(\omega t) =\sqrt\frac23\]and is equal to:\[\delta I_{1\mathrm m}=\max_t|\delta I_1|=\frac{V_0E_gT_2}{k_BT_1^2}\cdot 2\cdot\sqrt\frac13\cdot \frac23=\frac{1}{3\sqrt3}\frac{V_0E_g}{k_BT_1^2}\frac{\kappa(T_1-T_0)}{\pi\nu C}.\]Since resistance is more sensitive to temperature than voltage (its relative change is larger), we will determine the temperature from resistance by measuring average hysteresis loop slope, $\langle R\rangle$:\[\langle R\rangle=R_0\exp\left[\frac{E_g}{k_BT_1}\right]\implies T_1=\frac{E_g}{k_B\ln\dfrac{\langle R\rangle}{R_0}}.\]We will also measure the maximum width of the hysteresis loop along the $I$ axis. This should be done at voltage:\[V'=V_0\cos(\omega t)_{\max}=V_0\sqrt\frac23\approx 0.82V_0.\]The loop's width will be equal to:\[2\delta I_{1\mathrm m}=\frac{1}{C}\frac{2\kappa V_0E_g(T_1-T_0)}{3\sqrt3\cdot\pi\nu k_BT_1^2}.\]Calculations for the linearized dependency $2\delta I_{1\mathrm m}\left(\frac{2\kappa V_0E_g(T_1-T_0)}{3\sqrt3\cdot\pi\nu k_BT_1^2}\right)$ are shown in a table below:

$№$	1	2	3	4	5	6
$V_0,~V$	4.5	6.7	8.0	8.8	9.3	9.6
$I_0,~A$	0.58	1.00	1.40	1.78	2.16	2.56
$\langle R\rangle,~\Omega$	7.76	6.70	5.71	4.94	4.31	3.75
$T_1,~K$	297.6	304.0	311.3	318.2	325.0	332.3
$\frac{2\kappa V_0E_g(T_1-T_0)}{3\sqrt3\cdot\pi\nu k_BT_1^2},~\frac{J\cdot A}{K}$	0.041	0.094	0.149	0.196	0.239	0.280
$V',~V$	3.7	5.5	6.5	7.2	7.6	7.8
$2\delta I_{1\mathrm m},~A$	0.07	0.12	0.16	0.20	0.23	0.26

Next, let's plot this dependency:

The plot's slope is equal to:\[\frac{1}{C}=1.27~\frac{K}{J}\implies C=0.79~\frac{J}{K}\]