physics-class-12-unit-03-chapter-04-electromotive-force-and-ohms-law-current-and-electricity-l-4-11-qfza48rkwcc

### <Success style="font-size:25px"> Electromotive Force And Ohms Law Current And Electricity L-4 </Success>
### <primary style="font-size:24px"> Mobility </primary>
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* $ \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}$

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

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<footer style = "font-size: 18px"> $\rightarrow$ Electromotive Force and Ohm's Law $\rightarrow$ <span style = "color:blue"> Mobility </span> $\rightarrow$ Mobility $\rightarrow$ Mobility </footer>

### <Success style="font-size:25px"> Electromotive Force And Ohms Law Current And Electricity L-4 </Success>
### <primary style="font-size:24px"> Mobility </primary>
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- 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.

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<footer style = "font-size: 18px">Mobility $\rightarrow$ Mobility $\rightarrow$ <span style = "color:blue"> Mobility </span> $\rightarrow$ Linear Expansion $\rightarrow$ Thermal Expansion </footer>

### <Success style="font-size:25px"> Electromotive Force And Ohms Law Current And Electricity L-4 </Success>
### <primary style="font-size:24px"> Linear Expansion </primary>
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- $ \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)$

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<footer style = "font-size: 18px">Mobility $\rightarrow$ Mobility $\rightarrow$ <span style = "color:blue"> Linear Expansion </span> $\rightarrow$ Thermal Expansion $\rightarrow$ Thermal Expansion </footer>

### <Success style="font-size:25px"> Electromotive Force And Ohms Law Current And Electricity L-4 </Success>
### <primary style="font-size:24px"> Thermal Expansion </primary>
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- $R  =\rho \frac{L}{A}$

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

- $ =\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}$

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<footer style = "font-size: 18px">Mobility $\rightarrow$ Linear Expansion $\rightarrow$ <span style = "color:blue"> Thermal Expansion </span> $\rightarrow$ Thermal Expansion $\rightarrow$ Battery </footer>

### <Success style="font-size:25px"> Electromotive Force And Ohms Law Current And Electricity L-4 </Success>
### <primary style="font-size:24px"> Thermal Expansion </primary>
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- $\alpha = .0009$

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

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

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

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<footer style = "font-size: 18px">Linear Expansion $\rightarrow$ Thermal Expansion $\rightarrow$ <span style = "color:blue"> Thermal Expansion </span> $\rightarrow$ Battery $\rightarrow$ EMF </footer>

### <Success style="font-size:25px"> Electromotive Force And Ohms Law Current And Electricity L-4 </Success>
### <primary style="font-size:24px"> Battery </primary>
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- Electromotive force (EMF) moves charge from lower potential to higher potential

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<footer style = "font-size: 18px">Thermal Expansion $\rightarrow$ Thermal Expansion $\rightarrow$ <span style = "color:blue"> Battery </span> $\rightarrow$ EMF $\rightarrow$ Resistance </footer>

### <Success style="font-size:25px"> Electromotive Force And Ohms Law Current And Electricity L-4 </Success>
### <primary style="font-size:24px"> EMF </primary>
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- $\varepsilon=$ 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

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<footer style = "font-size: 18px">Thermal Expansion $\rightarrow$ Battery $\rightarrow$ <span style = "color:blue"> EMF </span> $\rightarrow$ Resistance $\rightarrow$ Open Circuit Voltage </footer>

### <Success style="font-size:25px"> Electromotive Force And Ohms Law Current And Electricity L-4 </Success>
### <primary style="font-size:24px"> Open Circuit Voltage </primary>
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- $\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 $10 \mathrm{~V}$

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<footer style = "font-size: 18px">EMF $\rightarrow$ Resistance $\rightarrow$ <span style = "color:blue"> Open Circuit Voltage </span> $\rightarrow$ Example $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Electromotive Force And Ohms Law Current And Electricity L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 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 $

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

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

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<footer style = "font-size: 18px">Open Circuit Voltage $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Example $\rightarrow$ Thank You </footer>

### <Success style="font-size:25px"> Electromotive Force And Ohms Law Current And Electricity L-4 </Success>
### <primary style="font-size:24px"> Example </primary>
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- $ 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$

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<footer style = "font-size: 18px">Example $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Thank You $\rightarrow$  </footer>