$\vec{J}=-n e \vec{v}_d$

$\vec{v}_d=\frac{e \vec{E} \tau}{m}$

Mobility

$\mu=\frac{\left|v_d\right|}{|E|} =\frac{e \tau}{m}$

$\tau =10^{-14} \mathrm{~s}$

$e \sim 10^{-19}$

$40-45 \mathrm{~cm}^2 / \mathrm{v}-\mathrm{s} \text {. }$

$\sigma =\frac{n e^2 \tau}{m}$

$=n e \mu$

For semiconductors

$\sigma=e\left[\eta_e \mu_e + p_n \mu_n\right]$

For $S_{i}$:

$\mu_e \approx 1400 \mathrm{~cm}^2 / \mathrm{v}-\mathrm{s}$

$\mu_n \approx 460 \mathrm{~cm}^2 / \mathrm{v}-\mathrm{s}$

5. Ohm's Law

$V=I R$

$\vec{J}=\sigma \vec{E}$

6. $\rho(T)=\rho(T_0)\left[1+\alpha\left(T-T_0\right)\right]$ in the linear region.

7. Carbon Resistors.

$\alpha=.004 / ^0 K$

$\rho=2 \rho_0=\rho_0(1+\alpha \Delta T)$

$\Delta T=\frac{1}{\alpha}=\frac{1}{0.004}=250^{\circ} \mathrm{C}$

Resintance at $0^{\circ} \mathrm{C}$.

$R=R_0(1+\alpha \Delta T)$

$\beta=$ temperature coefficent of linear expansion

$L =L_0(1+\beta \Delta T)$

$V =V_0(1+\gamma \Delta T)$

$\approx L_0^3(1+3 \beta \Delta T)$

$A =L_0^2 \frac{(1+3 \beta \Delta T)}{(1+\beta \Delta T)} \approx A_0 \cdot(1+2 \beta \Delta T)$

$R =\rho \frac{L}{A}$

$\Delta R =\frac{L}{A} \cdot \Delta \rho + \frac{\rho}{A} \Delta L+\rho L\left(-\frac{\Delta A}{A^2}\right)$

$=\frac{L}{A} \Delta \rho + \frac{\rho}{A} \Delta L-\rho L \cdot \frac{\Delta A}{A^2}$

$\frac{\Delta R}{R}=\frac{\Delta \rho}{\rho}+\frac{\Delta L}{L}-\frac{\Delta A}{A}$

$\frac{\Delta \rho}{\rho}=\frac{\rho_0(1+\alpha \Delta T)-\rho_0}{\rho_0} \approx \alpha \Delta T$

$\beta \approx 2.5 \times 10^{-5}$

$\alpha = {4.3 \times 10^{-3}}$

$\frac{\Delta L}{L}=2.5 \times 10^{-5}$

$\frac{\Delta A}{A}=5 \times 10^{-5}$

$\alpha = .0009$

$\beta = 1.8 \times 10^{-4}$

$\frac{\Delta R}{R}=\frac{\Delta \rho}{\rho}+\frac{\Delta L}{L}$

$=\alpha \Delta T+\beta \Delta T$

$\varepsilon=$ work done per unit charge

$=\frac{d W}{d q} \quad \frac{\text { Joule. }}{\text { coulomb }}=\text { volt }$

Emf is not a Force

Higher potential

Lower potential

$E = \frac {V}{L}$

$\vec{J} =\sigma \vec{E}$

$\vec{E} =\rho \vec{J}$

$I =J A$

$E =\frac{V}{L}=\rho J = \rho \frac{I}{A}$

$V =({\rho} \frac{L}{A}) I$

Resistance $R =\rho \frac{L}{A}$

$\Delta V_{a b} =\Delta V_{a c}+\Delta V_{c b}$

$=I T+\Delta V_{c b}$

$=0+10$

If the circuit is open

Open circuit voltage $10 \mathrm{~V}$

$\frac{8.5}{R_L}=0.5$

$R_L=17 \Omega$

Start at C and go in a direction opposite to current direction.

$V_c+I R_2 =V_b$

$V_b+I R_1 =V_a$

$V_c+I R_2 =V_a-I R_1$

$V_a-V_c =I\left(R_1+R_2\right)$

$\therefore I =\frac{V_a-V_c}{R_1+R_2}=\frac{E}{R_1+R_2}$

$r_1=1 \Omega$ ,$r_2=1.5$

$R_L= 5.5 \Omega$

$V_A-\varepsilon_2+i r_2+i R_L+\varepsilon r_1+\varepsilon_1=V_A$

$i {\left[r_1+r_2+R_L\right]+\left[\varepsilon_1-\varepsilon_2\right]=0 }$

$i[8]=2 \quad i=\frac{1}{4} A$

$V_B= V_A-\epsilon_2+i \sqrt {2}$

$V_B-V_A=-3.625$