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4차 산업혁명에 걸맞는 인터넷 기반 고등교육기관
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근처)
라는 것을 알아냈으며, 우주배경복사의
미세한 온도 변화가 있음을 알아냈다. 그리고 더욱 정밀한 측정을 하기 위해 WMAP 위성을 지구궤도에 올리게 되었다.On electronic circuits, they contain many components such as transistors or operational amplifiers. However, these components can be classified in to either an active element or a passive element. Passive elements are elements which cannot supply energy by themselves, whereas a battery is an active element which generate energy. Resistors, capacitors and inductors are passive elements which we are going to discuss in this chapter.
Before we begin our discussion, let’s see <Table 1> and briefly think about how Resistors, capacitors and inductors work.
| Relation | Resistor | Capacitor | Inductor |
|---|---|---|---|
| Schematic |  |
 |
 |
| Voltage | \(v_R(t)=i_R(t)R\) | \(\frac{1}{C}\int_{t_0}^{t}idt+v(t_0)\) | \(v_L(t)= \frac{di_l(L)}{dt}\) |
| Current | \(i_R(t)=\frac{v_R(t)}{R}\) | \(i=c\frac{dv_c(t)}{dt}\) | \(\frac{1}{L}\int_{t_0}^{t}vdt+i(t_0)\) |
| Power or Energy | \(p=i^2R=\frac{v^2}{R}\) | \(w=\frac{1}{2}cv^2\) | \(w=\frac{1}{2}li^2\) |
| Series | \(R_{eq}=R_1+R_2+... R_n\) | \(C_{eq}=\frac{C_1C_2}{C_1+C_2}\) | \(L_{eq}=L_1+L_2+... L_n\) |
| Parallel | \(R_{eq}=\frac{R_1R_2}{R_1+R_2}\) | \(C_{eq}=C_1+C_2+... C_n\) | \(L_{eq}=\frac{L_1L_2}{L_1+L_2}\) |
| At DC Voltage | Same | Open Circuit | Short Circuit |
Capacitor is composed of two conducting plates separated by a dielectric. When voltage is applied between the plates, +q charges accumulate in one plate while -q charge on the other. Capacitance, which is ability to store an electrical charge, depends on the geometry of the conductors and permittivity of the insulating medium in between the plates. Capacitance of ideal parallel plate capacitor is defined by following equation. \[C=\frac{\epsilon A}{d}\]
Where plate area \(A\) separated by distance \(d\) with a permittivity of \(\epsilon\).
The relation between charge and capacitance is \[q=Cv\] By looking at this equation, we can know that the plate charge increases as the voltage increases. when we take a derivative for both sides, we can get \[i=\frac{Cdv}{dt}\] where \(i=\frac{dq}{dt}\)
when \(v\) is a constant voltage(which is a DC voltage), then \(i=0\). This means when DC voltage is applied to capacitor, it looks as if it is an open circuit. Meanwhile, abrupt change in it voltage \(v\) is not allowable.
If we integrate equation over time, \[\int_{-\infty}^{t}idt=\int_{-\infty}^{t}C\frac{dv}{dt}dt\] \[v=\frac{1}{C}\int_{-infty}^{t}idt\] \[v=\frac{1}{C}\int_{0}^{t}idt + v(0)\]
\(v(0)\) repr
4차 산업혁명에 걸맞는 인터넷 기반 고등교육기관
이메일: ilkmooc@ilkmooc.kr
홈페이지주소: http://ilkmooc.kr
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Problem 2: Prove \(-\nabla \frac{1}{|\vec{r}-\vec{r}\prime|}=\frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}\)
Let \[\vec{r}=x_1\hat{x} + y_1\hat{y} + z_1\hat{z}\] \[\vec{r}\prime=x_2\hat{x} + y_2\hat{y} + z_2\hat{z}\] \[\zeta =(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2\]
\[\nabla \frac{1}{|\vec{r}-\vec{r}\prime|} = \frac{\partial}{\partial x}\hat{x} + \frac{\partial}{\partial y}\hat{y} + \frac{\partial}{\partial z}\hat{z}\frac{1}{|\vec{r}-\vec{r}\prime|}\]
Let us consider only \(\hat{x}\) components. \[\begin{aligned} \frac{\partial}{\partial x}\hat{x}\frac{1}{\zeta^{\frac{1}{2}}} &= \frac{1}{2}\zeta^{-\frac{3}{2}}\zeta\prime \nonumber \\ &=\frac{x_1 - x_2}{\zeta^{\frac{3}{2}}}\hat{x} \nonumber \\ &=\frac{x_1 - x_2}{|\vec{r}-\vec{r}\prime|^3}\hat{x} \nonumber\end{aligned}\]
Same procedure occur for the other two component. When we add all, final result become \[-\nabla \frac{1}{|\vec{r}-\vec{r}\prime|}=\frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}\]
4차 산업혁명에 걸맞는 인터넷 기반 고등교육기관
이메일: ilkmooc@ilkmooc.kr
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Problem 1: Prove \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)
Answer:
\[\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
\frac{x_1-x_2}{|\vec{\eta}^3|} & \frac{y_1-y_2}{|\vec{\eta}^3|} & \frac{z_1-z_2}{|\vec{\eta}^3|}
\end{vmatrix}\] Where \[\vec{r}=x_1\hat{x} + y_1\hat{y} + z_1\hat{z}\] \[\vec{r}\prime=x_2\hat{x} + y_2\hat{y} + z_2\hat{z}\] \[\vec{\eta} = \vec{r}-\vec{r}\prime\], \[|\vec{\eta}|=\sqrt{(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2}\] \[\zeta =(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2\] \[\begin{aligned}
\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3} &= (\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\eta}^3|}))\vec{x} \nonumber \\
& -(\frac{\partial}{\partial x}(\frac{z_1-z_2}{|\vec{\eta}^3|}) - \frac{\partial}{\partial z}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{y} \nonumber \\
& +(\frac{\partial}{\partial x}(\frac{y_1-y_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial y}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{z} \nonumber\end{aligned}\]
Since \[\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(z_1 - z_2)(y_1 - y_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]
\[\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(y_1 - y_2)(z_1 - z_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]
They cancel out each others. Same procedure occur for the other two components. Hence, \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)
4차 산업혁명에 걸맞는 인터넷 기반 고등교육기관
이메일: ilkmooc@ilkmooc.kr
홈페이지주소: http://ilkmooc.kr
산동일크무크란? https://goo.gl/FnvqXd
