Steady Current¶

Current Density Vector¶

The current strength is $\(i = \frac{dq}{dt}$\)

The current density vector is

\[ \begin{gathered} \begin{aligned} i &= \iint_A \vec{j} d\vec{A} \\ &= \iint_A j\cos\theta dA \end{aligned} \\\\ \vec{j} = \frac{d\vec{i}}{dA} \end{gathered} \]

Consider a closed surface, we have the current continuity equation $\(\iint_A \vec{j} d\vec{A} = - \frac{dq}{dt}$\)

If the current is steady, then we have $\(\iint_A \vec{j} d\vec{A} = 0$\)

Ohm Law¶

The Ohm Law is $\(R = \frac{dV}{dI}$\)

Set the resistivity $\rho$, the conductivity $\sigma = \frac{1}{\rho}$, we have $\(R = \rho \frac{L}{A} = \int \rho \frac{dl}{A}$\)

Consider a sphere with half buried underground, its grounding resistance is $\(R = \int_a^{\infty} \rho \frac{dr}{2\pi r^2} = \frac{\rho}{2\pi a}$\)

The differential form of Ohm’s Law is

\[ \begin{gathered} \Delta i = \frac{\Delta V}{R} \\\\ \Rightarrow j\Delta A = \frac{E\Delta l}{\rho (\Delta l / \Delta A)} \\\\ \Rightarrow j = \frac{E}{\rho} = \sigma E \end{gathered} \]

Below is the microscopic explanation of Ohm Law.

Let $\lambda$ be the mean free path, $\tau$ be the mean free time, $v_t$ be the mean thermal speed, $u$ be the mean drift speed. Then

\[ \begin{gathered} \vec{a} = -\frac{e}{m} \vec{E} \\\\ \vec{u}_1 = \vec{a} \tau = - \frac{e}{m} \vec{E} \tau \\\\ \vec{u} = \frac{\vec{u}_0 + \vec{u}_1}{2} = - \frac{e}{2m} \vec{E} \tau = - \frac{e}{2m} \frac{\lambda}{v_t} \vec{E} \end{gathered} \]

Let $n$ be the electron number per volume. Then

\[ \begin{gathered} \Delta q = ne \cdot u\Delta t\Delta A \\\\ \Rightarrow \Delta i = neu\Delta A \\\\ \Rightarrow j = neu \end{gathered} \]

Therefore, we can get $\(\sigma = \frac{ne^2\lambda}{2mv_t}$\)

Source and Electromotive Force¶

For non-electrostatic force, we can set $\(K = \frac{F}{q}$\)

Then the electromotive force can be defined as $\(\epsilon = \int K dl$\)

The terminal voltage of a seat is

\[ \begin{aligned} \Delta V &= V_{+} - V_{-} \\ &= \int_{+}^{-} \vec{E} d\vec{l} \\ &= \int_{+}^{-} -\vec{K} + \frac{\vec{j}}{\sigma} d\vec{l} \\ &= \int_{-}^{+} \vec{K} d\vec{l} - \int_{-}^{+} \rho j dl \cos\theta \\ \end{aligned} \]

When the seat is discharging, $\cos\theta = 1, \Delta V = \epsilon - ir$. When the seat is charging, $\cos\theta = -1, \Delta V = \epsilon + ir$.

So the output power is $\(P_{out} = iV_{AB} = (\frac{\epsilon}{r + R})^2 R$\)

And we have $\(\frac{dP_{out}}{dR} = \epsilon^2 \frac{r - R}{(r + R)^2}$\)

So when $R = r$, output power becomes maximum.

Simple Circuit¶

For a voltmeter, we have $\(R_m = \frac{U - I_gR_g}{I_g} = \frac{U}{I_g} - R_g$\)

For a ampermeter, we have

\[ \begin{gathered} I_gR_g = (I - I_g)R_s \\\\ \Rightarrow R_s = \frac{I_g}{I - I_g}R_g \end{gathered} \]

Complicated Circuit¶

Kirchhoff’s Equations¶

In a complicated circuit consisting of many seats and resistors, we can use Kirchhoff’s equations.

The first Kirchhoff’s equations: for each node, $\sum I = 0$. The second Kirchhoff’s equations: for each loop, $\int E dl = 0$.

Equivalent Source Law¶

Thevenin’s theorem: At a pair of terminals of a network, it can be replaced by a voltage source $V_{\mathrm{th}}$ and a single resistance $R_{\mathrm{th}}$ in series. - The equivalent voltage $V_{\mathrm{th}}$ is the voltage obtained at $A–B$ of the network with terminals $A–B$ open circuited. - The equivalent resistance $R_{\mathrm{th}}$ is the resistance that the circuit between terminals $A$ and $B$ would have if all ideal voltage sources in the circuit were replaced by a short circuit and all ideal current sources were replaced by an open circuit. - If terminals $A$ and $B$ are connected to one another, the current flowing from $A$ and $B$ will be $V_{\mathrm{th}}/R_{\mathrm{th}}$ This means that $R_{\mathrm{th}}$ could alternatively be calculated as $V_{\mathrm{th}}$ divided by the short-circuit current between $A$ and $B$ when they are connected together.

Norton's theorem: At a pair of terminals of a network, it can be replaced by a current source $I_{\mathrm{no}}$ and a single resistance $R_{\mathrm{no}}$ in parallel. - the Norton current $I_{\mathrm{no}}$ is calculated as the current flowing at the terminals into a short circuit (zero resistance between $A$ and $B$). - The Norton resistance $R_{\mathrm{no}}$ is found by calculating the output voltage produced with no resistance connected at the terminals; equivalently, this is the resistance between the terminals with all (independent) voltage sources short-circuited and independent current sources open-circuited. This is equivalent to calculating the Thevenin resistance.

Superposition Principle¶

Superposition Principle shows that if there are multiple power sources in the circuit, the current of any branch is equal to the sum of the currents generated in that branch when each source exists separately.

Substitution Between Y-$\Delta$ Circuit¶

The substitution between $Y-\Delta $ circuit is

\[ \begin{cases} R_1 = \dfrac{R_{31}R_{12}}{R_{12} + R_{23} + R_{31}} \\\\ R_2 = \dfrac{R_{12}R_{23}}{R_{12} + R_{23} + R_{31}} \\\\ R_3 = \dfrac{R_{23}R_{31}}{R_{12} + R_{23} + R_{31}} \\ \end{cases} ,\quad \begin{cases} R_{12} = \dfrac{R_1R_2 + R_2R_3 + R_3R_1}{R_3} \\\\ R_{23} = \dfrac{R_1R_2 + R_2R_3 + R_3R_1}{R_1} \\\\ R_{31} = \dfrac{R_1R_2 + R_2R_3 + R_3R_1}{R_2} \\ \end{cases} \]

If $R_1 = R_2 = R_3 = R$, then $R_{12} = R_{23} = R_{31} = 3R$.

Proof:

\[ \begin{gathered} \begin{cases} R_1 + R_2 = \dfrac{R_{12}(R_{23} + R_{31})}{R_{12} + R_{23} + R_{31}} \\\\ R_2 + R_3 = \dfrac{R_{23}(R_{31} + R_{12})}{R_{12} + R_{23} + R_{31}} \\\\ R_3 + R_1 = \dfrac{R_{31}(R_{12} + R_{23})}{R_{12} + R_{23} + R_{31}} \\ \end{cases} \quad \Rightarrow \quad \begin{cases} R_1 = \dfrac{R_{31}R_{12}}{R_{12} + R_{23} + R_{31}} \\\\ R_2 = \dfrac{R_{12}R_{23}}{R_{12} + R_{23} + R_{31}} \\\\ R_3 = \dfrac{R_{23}R_{31}}{R_{12} + R_{23} + R_{31}} \end{cases} \\\\ \Rightarrow R_1R_2 + R_2R_3 + R_3R_1 = \frac{R_{12}R_{23}R_{31}}{R_{12} + R_{23} + R_{31}} \\\\ \Rightarrow \begin{cases} R_{12} = \dfrac{R_1R_2 + R_2R_3 + R_3R_1}{R_3} \\\\ R_{23} = \dfrac{R_1R_2 + R_2R_3 + R_3R_1}{R_1} \\\\ R_{31} = \dfrac{R_1R_2 + R_2R_3 + R_3R_1}{R_2} \\ \end{cases} \end{gathered} \]

Thermo-electromotive Force¶

The non-electrostatic force generated by temperature difference is $\(\kappa = \sigma(T) \frac{dT}{dl}$\)

where $\sigma$ is Thompson coefficient. So $\(\epsilon(T_1, T_2) = \int_0^l \vec{\kappa} d\vec{l} = \int_{T_1}^{T_2} \sigma(T)dT$\)