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# 电子代写|电路基础代写Circuit Fundamentals代考|EET1015C Voltage

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## 电子代写|电路基础代写Circuit Fundamentals代考|Voltage

As explained briefly in the previous section, to move the electron in a conductor in a particular direction requires some work or energy transfer. This work is performed by an external electromotive force (emf), typically represented by the battery in Fig. 1.3. This emf is also known as voltage or potential difference. The voltage $v_{a b}$ between two points $a$ and $b$ in an electric circuit is the energy (or work) needed to move a unit charge from $b$ to $a$; mathematically,
$$v_{a b} \triangleq \frac{d w}{d q}$$
where $w$ is energy in joules (J) and $q$ is charge in coulombs (C). The voltage $v_{a b}$ or simply $v$ is measured in volts (V), named in honor of the Italian physicist Alessandro Antonio Volta (1745-1827), who invented the first voltaic battery. From Eq. (1.3), it is evident that
1 volt $=1$ joule/coulomb $=1$ newton-meter/coulomb
Thus,
Voltage (or potential difference) is the energy required to move a unit charge from a reference point ( $-)$ to another point (+), measured in volts (V).
Figure $1.6$ shows the voltage across an element (represented by a rectangular block) connected to points $a$ and $b$. The plus (+) and minus ( $-)$ signs are used to define reference direction or voltage polarity. The $v_{a b}$ can be interpreted in two ways: (1) Point $a$ is at a potential of $v_{a b}$ volts higher than point $b$, or (2) the potential at point $a$ with respect to point $b$ is $v_{a b}$. It follows logically that in general
$$v_{a b}=-v_{b a}$$
For example, in Fig. 1.7, we have two representations of the same voltage. In Fig. 1.7(a), point $a$ is $+9 \mathrm{~V}$ above point $b$; in Fig. 1.7(b), point $b$ is $-9 \mathrm{~V}$ above point $a$. We may say that in Fig. 1.7(a), there is a 9-V voltage drop from $a$ to $b$ or equivalently a 9-V voltage rise from $b$ to $a$. In other words, a voltage drop from $a$ to $b$ is equivalent to a voltage rise from $b$ to $a$.

## 电子代写|电路基础代写Circuit Fundamentals代考|Power and Energy

Although current and voltage are the two basic variables in an electric circuit, they are not sufficient by themselves. For practical purposes, we need to know how much power an electric device can handle. We all know from experience that a 100-watt bulb gives more light than a 60-watt bulb. We also know that when we pay our bills to the electric utility companies, we are paying for the electric energy consumed over a certain period of time. Thus, power and energy calculations are important in circuit analysis.

To relate power and energy to voltage and current, we recall from physics that:
Power is the time rate of expending or absorbing energy, measured in watts (W).
We write this relationship as
$$p \triangleq \frac{d w}{d t}$$

where $p$ is power in watts (W), $w$ is energy in joules (J), and $t$ is time in seconds (s). From Eqs. (1.1), (1.3), and (1.5), it follows that
$$p=\frac{d w}{d t}=\frac{d w}{d q} \cdot \frac{d q}{d t}=v i$$
or
$$p=v i$$
The power $p$ in Eq. (1.7) is a time-varying quantity and is called the $\mathrm{in}$ stantaneous power. Thus, the power absorbed or supplied by an element is the product of the voltage across the element and the current through it. If the power has a $+$ sign, power is being delivered to or absorbed by the element. If, on the other hand, the power has a – sign, power is being supplied by the element. But how do we know when the power has a negative or a positive sign?

Current direction and voltage polarity play a major role in determining the sign of power. It is therefore important that we pay attention to the relationship between current $i$ and voltage $v$ in Fig. 1.8(a). The voltage polarity and current direction must conform with those shown in Fig. 1.8(a) in order for the power to have a positive sign. This is known as the passive sign convention. By the passive sign convention, current enters through the positive polarity of the voltage. In this case, $p=+v i$ or $v i>0$ implies that the element is absorbing power. However, if $p=-v i$ or $v i<0$, as in Fig. 1.8(b), the element is releasing or supplying power.

## 电子代写|电路基础代写Circuit Fundamentals代考|Voltage

.输出说明

$$v_{a b} \triangleq \frac{d w}{d q}$$
where $w$ 能量单位是焦耳(J)和 $q$ 电荷单位为库仑(C)。电压 $v_{a b}$ 或者简单地说 $v$ 伏特(V)是为了纪念意大利物理学家亚历山德罗·安东尼奥·伏特(1745-1827)，他发明了第一个伏打电池。由式(1.3)可知，
1伏 $=1$ 焦耳/库仑 $=1$ 因此，电压(或电位差)是将一个单位电荷从参考点( $-)$ 到另一个点(+)，单位为伏特(V)。
$1.6$ 显示通过连接到点的元素(由矩形块表示)的电压 $a$ 和 $b$。+(+)和- ( $-)$ 符号用于定义参考方向或电压极性。。 $v_{a b}$ 可以用两种方式解释:(1)点 $a$ 它的势是 $v_{a b}$ 电压高于点 $b$，或(2)点的势 $a$ 关于点 $b$ 是 $v_{a b}$。从逻辑上讲，一般
$$v_{a b}=-v_{b a}$$例如，在图1.7中，我们有相同电压的两种表示。图1.7(a)中，点 $a$ 是 $+9 \mathrm{~V}$ 以上一点 $b$;图1.7(b)中，点 $b$ 是 $-9 \mathrm{~V}$ 以上一点 $a$。我们可以说，在图1.7(a)中，有一个9 v的电压降 $a$ 到 $b$ 或相当于9v电压上升 $b$ 到 $a$。换句话说，电压降 $a$ 到 $b$ 相当于电压从 $b$ 到 $a$.

## 电子代写|电路基础代写Circuit Fundamentals代考|Power and Energy

.

$$p \triangleq \frac{d w}{d t}$$

$$p=\frac{d w}{d t}=\frac{d w}{d q} \cdot \frac{d q}{d t}=v i$$

$$p=v i$$

## Matlab代写

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