## 物理代写|几何光学代写Geometrical Optics代考|PHYSICS134A

2023年1月2日

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• Statistical Computing 统计计算
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## 物理代写|几何光学代写Geometrical Optics代考|Electric Energy and Power

In Fig. 5.8, we show a simple circuit consisting of a battery and a resistor. Consider a positive amount of charge $\Delta Q$ moves clockwise around the circuit. That is, the charge passes from point $a$ through the battery, then resistor and back to point $a$. Points $a$ and $d$ are grounded; that is, we take the electric potential at these two points to be zero. Electric potential energy $U$ increases as the charge moves from point $a$ to $b$ through the battery as
$$\Delta U=\Delta Q \Delta V=\Delta Q\left(\phi_b-\phi_a\right)$$

where $\phi_a$ and $\phi_b$ are electric potentials at the points $a$ and $b$, respectively.
In contrast, as the charge moves through the resistor, from $c$ to $d$, it loses the same amount of electric potential energy due to collisions with atoms of the resistor. That potential energy transfers into internal energy in the resistor. Note that we have neglected the resistance of the connecting wires; that is, electric potential energy faces no loss of energy in the paths $b c$ and $d a$. Therefore, when the charge arrives back at point $a$, it has lost all the electric potential energy, and hence its potential energy is zero.

The rate at which the charge $\Delta Q$ loses potential energy in going through the resistor is
$$\frac{\Delta U}{\Delta t}=\frac{\Delta Q}{\Delta t} \Delta V=I \Delta V$$
where $I$ is the current in circuit. The charge loses energy equals the electric power:
$$\mathcal{P}=I \Delta V$$
Using Ohm’s Law: $\Delta V=I R$, we get
$$\mathcal{P}=I^2 R=\frac{(\Delta V)^2}{R}$$
When $I$ expresses in amperes, $\Delta V$ in volts and $R$ in ohms, the SI unit of power is the watt.

## 物理代写|几何光学代写Geometrical Optics代考|Electromotive Force

A battery is a device that supplies electrical energy. Often, it is called either a source of electromotive force or emf source. In general, the internal resistance of the battery is neglected, and the potential difference between points $a$ and $b$, see Fig. 5.8, is equal to the emf $\epsilon$ of the battery:
$$\Delta V=\phi_b-\phi_a=\epsilon$$
Therefore, the current in the circuit, based on Ohm’s law, is
$$I=\frac{\Delta V}{R}=\frac{\epsilon}{R}$$
Because $\Delta V=\epsilon$, the power supplied by the emf source can be expressed as
$$\mathcal{P}=I \epsilon$$

Solution 5.1 The volume is
$$V=\frac{m}{\rho}=\frac{63.5 \mathrm{~g}}{8.95 \mathrm{~g} / \mathrm{cm}^3}=7.09 \mathrm{~cm}^3$$
Since in each mol of a substance there are $N_A=6.02 \times 10^{23}$ atoms, and since each atom contributes with one electron, the total number of free electrons is
$$n=\frac{6.02 \times 10^{23}}{7.09 \mathrm{~cm}^3}=8.49 \times 10^{28} \text { electrons } / \mathrm{m}^3$$
Using the equation of current $I=n q A v_d$, with $q=1.6 \times 10^{-19} \mathrm{C}$ being the absolute value of charge of electron, we get
$$v_d=\frac{I}{n q A}=2.22 \times 10^{-4} \mathrm{~m} / \mathrm{s}$$

# 几何光学代考

## 物理代写|几何光学代写Geometrical Optics代考|Electric Energy and Power

ΔU=ΔQΔV=ΔQ(ϕb−ϕa)

ΔUΔt=ΔQΔtΔV=IΔV

P=IΔV

P=I2R=(ΔV)2R

## 物理代写|几何光学代写Geometrical Optics代考|Electromotive Force

$$\Delta V=\phi_b-\phi_a=\epsilon$$

$$I=\frac{\Delta V}{R}=\frac{\epsilon}{R}$$

$$\mathcal{P}=I \epsilon$$

$$V=\frac{m}{\rho}=\frac{63.5 \mathrm{~g}}{8.95 \mathrm{~g} / \mathrm{cm}^3}=7.09 \mathrm{~cm}^3$$

$$n=\frac{6.02 \times 10^{23}}{7.09 \mathrm{~cm}^3}=8.49 \times 10^{28} \text { electrons } / \mathrm{m}^3$$


v_d=\frac{I}{n q A}=2.22 \times 10^{-4} \mathrm{~m} / \mathrm{s}

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## MATLAB代写

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