\begin{equation}
\phi_d= 0; \phi_a = 6\ \text{V}; \phi_b = \dfrac{60}{17}\ \text{V}; \phi_c = \dfrac{40}{17}\ \text{V}. I_1 = \dfrac{2}{7}\ \text{A}; I_2 = \dfrac{3}{34}\ \text{A}; I = \dfrac{7}{34}\ \text{A}.
\end{equation}
============Solution:=============
To find current through the resistors, we need potential drop across each. Therefore, we start with labeling the node potential points on the circuit as shown in the figure.
As explained above, the node points represent points of unique electric potential values. There are four nodes in the given circuit, which are labeled \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\text{.}\) Node \(d\) is connected to the negative of the source and is usually set zero volt as reference, \(\phi_d = 0\text{.}\) Since voltage of the battery is 6 volts, the potential at \(a\) is 6 V, that is, \(\phi_a = 0\text{.}\) Now, we follow step-by-step procedure of simplifying the circuit to find the potentials at \(b\) and \(c\text{,}\) viz., \(\phi_b\) and \(\phi_c\text{,}\) respectively.
Step 1:
If any resistors in series, replace them by their equivalent resistors. Here only \(10\ \Omega\) and \(20\ \Omega\) are in series. Therefore, we replace them by a \(30\text{-}\Omega\) resistor. It is best to keep the same labels here as in the previous diagram.
Step 2:
If any resistors in parallel, replace them by equivalent resistors. Here only 30 \(\Omega\) and 40 \(\Omega\) are in parallel. Therefore, we replace the combination by their equivalent resistance, which is \(\dfrac{120}{7}\ \Omega\text{.}\)
Step 3:
After replacing the parallel resistors by \(\dfrac{120}{7}\ \Omega\text{,}\) we find that it is in series with the \(12\text{-}\Omega\) resistor. Therefore current through the \(\dfrac{120}{7}\ \Omega\) must be same as that goes through the \(12\text{-}\Omega\) resistor. Now the two resistors in series can be replaced by a \(\dfrac{204}{7}\text{-}\Omega\) resistor resulting in a really simple circuit. The current through the \(\dfrac{204}{7}\text{-}\Omega\) resistor must be same as the \(12\text{-}\Omega\) or \(\dfrac{120}{7}\text{-} \Omega\) resistor since it is replacing resistors in series.
Step 4:
Now, we can use Ohm's law to find the current through \(\dfrac{204}{7}\text{-}\Omega\) resistor.
\begin{equation*}
I = \dfrac{6V}{(204/7)\Omega} = \dfrac{7}{34}\ \text{A}.
\end{equation*}
Step 5:
We then trace our path backwards towards the original circuit. The current through the \(12\text{-}\Omega\) resistor is also \(7/34\text{ A}\text{.}\) Therefore, potential drop across the \(12\text{-}\Omega\) resistor i
\begin{equation*}
\phi_a-\phi_b = \dfrac{7}{34}\ \text{A} \times 12\ \Omega = \dfrac{42}{17}\ \text{V}.
\end{equation*}
The potential drop across the \(\dfrac{120}{7}\text{-} \Omega\) resistor is \((60/17)\text{V}\text{.}\)
\begin{equation*}
\phi_b-\phi_d = \dfrac{7}{34}\ \text{A} \times \dfrac{120}{7}\ \Omega = \dfrac{60}{17}\ \text{V}\ \ \Longrightarrow\ \phi_b = \dfrac{60}{17}\ \text{V}.
\end{equation*}
Going back one more step we find that potential drop across the \(30\text{-}\Omega\) resitor is same as that across the \(\dfrac{120}{7}\text{-}\Omega\) resistor we worked out. Hence, current through the \(30\text{-}\Omega\) resistor is simply
\begin{equation*}
I_1 = \dfrac{(60/17)\ \text{V}}{30\ \Omega} = \dfrac{2}{17}\ \text{A}
\end{equation*}
Therefore, voltage \(\phi_c\) is obtained from the potential drop across the \(20\text{-} \Omega\) resistor as given by
\begin{equation*}
\phi_c - \phi_d = \left( \dfrac{2}{17}\ \text{A} \right) \times 20\ \Omega = \dfrac{40}{17}\ \text{V}.
\end{equation*}
Since \(\phi_d = 0\text{,}\) we obtain \(\phi_c = \dfrac{40}{17}\ \text{V}\text{.}\) We can use the node voltages to obtain the current through various resistors. Alternately, since the total current \(I\) was split into \(I_1\) and \(I_2\text{,}\) we can find \(I_2\) by subtracting \(I_1\) from \(I\text{.}\)
\begin{equation*}
I_2 = I - I_1 = \dfrac{7}{34}\ \text{A} - \dfrac{2}{17}\ \text{A} = \dfrac{3}{34}\ \text{A}
\end{equation*}
Summarizing the answer:
\begin{align*}
\amp \phi_d= 0;\ \phi_a = 6\ \text{V};\ \phi_b = \dfrac{60}{17}\ \text{V};\ \phi_c = \dfrac{40}{17}\ \text{V}.\\
\amp I_1 = \dfrac{2}{7}\ \text{A};\ I_2 = \dfrac{3}{34}\ \text{A};\ I = \dfrac{7}{34}\ \text{A}.
\end{align*}
The emfs and resistances in the circuit of the figure have the following values:\[ \mathscr{E}_1= 4.4 V, \mathscr{E}_2= 2.1 V, \] \[r_1 = 2.3 \Omega, r_2 = 1.8 \Omega, R= 5.5 \Omega .\] (a) What is the current \(i\) in the circuit? (b) What is the potential difference between the terminals of battery 1 in the figure? 
The figure shows a multiloop circuit containing one ideal battery and four resistances with the following values: \[R_1= 20 \, \Omega, \quad R_2= 20 \, \Omega,\] \[ \mathscr{E}=12 \,V, \quad R_3 = 30\, \Omega, \quad R_4= 8.0 \, \Omega.\] (a) What is the current through the battery? (b) What is the current \(i_2\) through \(R_2\)? (c) What is the current \(i_3\) through \(R_3\)? 
The figure shows a circuit whose elements have the following values: \[ \mathscr{E}_1= 3.0 \,V, \quad \mathscr{E}_2= 6.0 \, V, \] \[ R_1 = 2.0 \, \Omega, \quad R_2 = 4.0 \, \Omega.\] The three batteries are ideal batteries. Find the magnitude and direction of the current in each of the three branches. 
假設我們已經有一個充了電的電容器如圖所示，若要使用充電電容器上的電荷，可以通過電阻 \(R\) 的裝置和開關 \(S\) 連接電容器的末端，如圖所示。當開關閉合時，電流在電路中流動，直到來自負極板的電子中和正極板上的正電荷。放電速率取決於板上電荷的當前值，這導致電荷呈指數衰減。令 \(Q_0\) 為 \(t = 0\) 時的起始電荷。然後，使用微分方程我們可以推導出電容上的電荷量隨著時間變化的函數關係\(Q(t)\)，我們在下面會推導，在此先把最後的指數函數呈現如下：
\begin{equation*}
Q(t) = Q_0 e^{ - t/RC}.
\end{equation*}
而相對應的圖形如右圖所示。
電容放電的速率就是電容上的電荷量對時間的微分(\(-\dfrac{dQ}{dt}=i\))也就是放電的電流其公式如下：
\[i=-\dfrac{dQ}{dt}=\dfrac{Q}{RC}=\dfrac{Q}{\tau}\]
電路中的電流隨時間變化，並且與瞬間板上的電荷成正比。達到原始電荷的 \(1/e\) 的時間(\(\tau=RC\))，其中 \(e\) 是歐拉數，值 \(e = 2.718...\)，是電路的一個重要特徵。這個時間稱為電路的時間常數。時間常數通常用希臘字母\(\tau\)表示。當然，在這段時間內，大約\(2/3\)的電荷已經通過電路放電。電路的時間常數是電路的電阻和電容的乘積：\(\tau=RC\)。通過調整 \(R\)，我們可以控制電容器放電的速率。下面我們將研究充電過程，我們會在那裡發現充電速率也具有相同的時間常數\(RC\)。
在這個電阻和電容的網路當中並沒有電池，但是起初的時候兩個並聯的電容上面都有電荷，電容上的電位差都是45伏特。在時間\(t=0\)的時候將開關接上，電容上的電荷開始放電。要我們計算電容上的電位差下降到10伏特，需要經過多久的時間，並且計算在這個時刻流過電阻的電流。
As a car rolls along pavement, electrons move from the pavement first onto the tires and then onto the car body. The car stores this excess charge and the associated electric potential energy as if the car body were one plate of a capacitor and the pavement were the other plate. When the car stops, it discharges its excess charge and energy through the tires, just as a capacitor can discharge through a resistor. If a conducting object comes within a few centimeters of the car before the car is discharged, the remaining energy can be suddenly transferred to a spark between the car and the object. Suppose the conducting object is a fuel dispenser. The spark will not ignite the fuel and cause a fire if the spark energy is less than the critical value \(U_{fire}=50\) mJ. When the car of the figure stops at time \(t = 0\), the car-ground potential difference is \(V_0 = 30\) kV. The car-ground capacitance is \(C= 500\) pF, and the resistance of each tire is \(R_{tire}=100\) G\(\Omega\). How much time does the car take to discharge through the tires to drop below the critical value \(U_{fire}\)?