# 電路學 作業一

:::info
計算過程的各步驟以計算到小數點(但不包括”0”)以下兩位為原則。
(e.g. 0.232 $\rightarrow$ 0.23； 0.0232 $\rightarrow$ 0.023)，在單位正確下，數值+/-10%正確給分
(Finial Answer正確即給分；但若無Computation Process (或亂寫) >> 扣一半分)
:::

## 1. 

### (a) What is the total resistance, $R_T$? (in formula by $R_1$ to $R_6$) (5%) (可無計算過程)

#### (Final Answer)
$R_T = ( ( ( R_4 + R_5 + R_6 ) // R_3 ) + R_2 ) // R_1$
### (b) What is the value of $R_T$? (unit: Ω) (5%)
#### (Final Answer) ( 188.74 Ω )
#### (Computation Process)

$$
\begin{align*}
R_T & = ( ( ( R4 + R5 + R6 ) // R3 ) + R2 ) // R1 \\
& = ( ( 300Ω // R3 ) + R2 ) // R1 \\
& = ( 157.14 + 470 ) // 270 Ω \\
& = 627.14 // 270 Ω \\
& \simeq 188.74 Ω
\end{align*}
$$

### \(c\) What is the total current, $I_T$? (unit: mA) (5%)
#### (Final Answer) ( 26.5 mA )
#### (Computation Process)

$$
\begin{align*}
& V = IR \\
& 5 = I_T \times 188.74 \\
& I_T = 0.026491 A \simeq 26.5 mA
\end{align*}
$$

### (d) What is the current through $R_2$ (i.e. $I_2$)? (unit: mA) (5%)
#### (Final Answer) ( 7.98 mA )
#### (Computation Process)

$$
\begin{align*}
I_2 & = I_T \frac{ R_1 } { R_1 + R_{2-6} } \\
& = 26.5 mA \times \frac{270 Ω} { 270 Ω + R_{2-6} } \\
& = 26.5 mA \times \frac{270} {270 + 627.14} \\
& \simeq 7.98 mA
\end{align*}
$$

### (e) What is the current through $R_3$ (i.e. $I_3$)? (unit: mA) (5%)
#### (Final Answer) ( -3.79 mA )
#### (Computation Process)

$$
\begin{align*}
I_2 & = -I_3 + I_{4-6} \\
V_3 & = V_s\frac{R_{3-6}}{(R_2+R_{3-6})} \\
& = 5 \times \frac{157.14}{627.14} \\
& \simeq 1.25V \\
I3 & = -\frac{1.25V}{0.33kΩ} \simeq -3.79 mA
\end{align*}
$$

### (f) What is the voltage drop of $R_5$ (i.e. $V_5$)? (unit: V) (5%)
#### (Final Answer) ( -0.41V )
#### (Computation Process)

$$
\begin{align*}
& -V_s + V_2 + V_4 – V_5 + V_6 = 0 \\
V_2 & = I_2R_2 = 7.98 mA \times 0.47 kΩ \simeq 3.75 V \\
V_{3-6} & = V_s – V_2 = 1.25 V \\
V_4 & = 1.25 \times \frac{R_4}{R_{4-6}} \simeq 0.42 V \\
V_6 & = 0.42 V \\
V_5 & = -5 + 3.75 + 0.42 + 0.42 = -0.41 V
\end{align*}
$$

---

## 2. 

### (a) What is the total resistance, $R_T$? (in formula by $R_1$ to $R_8$) (5%) (可無計算過程)
#### (Final Answer)
$(((R_5 // R_6)+R_7+R_2+R_8)//(R_3+R_4))+R_1$

### (b) What is the value of $R_T$? (unit: KΩ) (5%)
#### (Final Answer) ( 1.64 kΩ )
#### (Computation Process)

$$
\begin{align*}
& (((R5 // R6)+R7+R2+R8)//(R3+R4))+R1 \\
& = ((0.6+0.68+0.33+0.1)//(0.47+0.56))+1 \\
& = (1.71 // 1.03)+1 \\
& \simeq 0.64 + 1 \\
& = 1.64 kΩ
\end{align*}
$$

### \(c\) What is the total current, $I_T$? (unit: mA) (5%)
#### (Final Answer) ( 60.98 mA )
#### (Computation Process)
$$
\begin{align*}
V & = IR \\
I_T & = \frac{V_s}{R_T} \\
& = \frac{100 V }{ 1.64kΩ} \\
& = \frac{100 V }{ 1640 Ω} \\
& \simeq 0.060975 A \\
& \simeq 60.98 mA
\end{align*}
$$

### (d) What is the current through $R_5$ (i.e. $I_5$)? (unit: mA) (6%)
#### (Final Answer) ( 13.75 mA )
#### (Computation Process)
$$
\begin{align*}
I_5 & = I_2 \frac{R_6}{R_5+R_6} \\
I_2 & = I_T \frac{R_3+R_4}{R_{2,5-8}+R_3+R_4} \\
& = 60.98 mA \times \frac{0.47+0.56}{1.71+0.47+0.56} \\
& \simeq 22.92mA \\
I5 & = 22.92mA \times \frac{1.5}{2.5} \simeq 13.75mA
\end{align*}
$$

### (e) What is the value of $V_A$? (unit: V) (4%)
#### (Final Answer) ( 21.23V )
#### (Computation Process)
$$
\begin{align*}
V_{2,5-8} & = V_{3-4} = V_s \frac{R_{2,5-8}}{R_{2,5-8}+R_1} \\
& = 100 \times \frac{0.64}{1.64} \\
& \simeq 39.024V \\
V_3 & = V_{3-4} \frac{R_3}{R_3+R_4} \\
& = 39.024 \times \frac{0.47}{0.47+0.56} \\
& \simeq 17.807 V \\
V_4 & = 39.024 – 17.81 \simeq 21.23 V \\
V_1 & = I_TR_1 = 60.98V \\
100V & = V_1 + V_3 + V_4(V_A) = 60.98 + 17.807 + 21.23V \ (KVL)\\
V_A & \simeq 21.23V
\end{align*}
$$

### (f) What is the value of $V_B$? (unit: V) (4%)
#### (Final Answer) ( 17.8 V )
#### (Computation Process)
$$
\begin{align*}
V_5 & = V_6 \\
& = V_{2,5-8}\frac{R_{5,6}}{R_{2,5-8}} \\
& = 39.024 \times \frac{0.6}{1.71} \\
& \simeq 13.69 V \\
V_2 & = V_{2,5-8} \frac{R_2}{R_{2,5-8}} \simeq 7.53V \\
V_7 & = V_{2,5-8}\frac{R_7}{R_{2,5-8}} \simeq 15.52 V \\
100V & = V_1 + V_2 + V_5 + V_B = 82.2 + V_B \\
V_B & = 17.8 V
\end{align*}
$$

### (g) What is the value of $V_{BA}$? (unit: V) (5%)
#### (Final Answer) ( -3.43V )
#### (Computation Process)

$V_{BA} = V_B – V_A = 17.8 – 21.23 = -3.43V$

---

# 3. 

### (a) What is the equivalent resistance of $R_4, R_5, R_8, R_9$ and $R_{12}$ ? (unit: Ω) (6%)
#### (Final Answer) ( 14.704 Ω )
#### (Computation Process)

$$
\begin{align*}
& R_{Upper}: ((((R_9 + R_{12})//R_8)+R_5)//R_4)+R_1 \\
& R_{Lower}: ((((R_{11}+R_{10})//R_7)+R_6)//R_3)+R_2 \\
& ((((R9 + R12)//R8)+R5)//R4) \\
& = ((24 // 18)+22)//27 \\
& = (10.29+22)//27 \\
& = 32.29//27 \\
& \simeq 14.704 Ω
\end{align*}
$$

### (b) What is the equivalent resistance of upper loop? (unit: Ω) (6%)
#### (The Final Answer) ( 78.704 Ω )
#### (Computation Process)

$$
\begin{align*}
& ((((R_9 + R_{12})//R_8)+R_5)//R_4)+R_1 \\
& = 14.704 + 64 \\
& = 78.704 Ω
\end{align*}
$$

### \(c\) What is the equivalent resistance of lower loop? (unit: Ω) (6%)
#### (The Final Answer) ( 61.704 Ω )
#### (Computation Process)

$$
\begin{align*}
& ((((R_{11}+R_{10})//R_7)+R_6)//R_3)+R_2 \\
& = (((24 // 18)+22)//27)+47 \\
& = ((10.29+22)//27)+47 \\
& = (32.29//27)+47 \\
& = 14.704+47 \\
& = 61.704 Ω
\end{align*}
$$

### (d) What is the total resistance of $R_T$? (unit: Ω) (6%)
#### (The Final Answer) ( 34.59 Ω )
#### (Computation Process)

$$
R_T = R_{Upper} // R_{Lower} = 78.70 // 61.70 \simeq 34.59 Ω
$$

### (e) What is the total current, $I_T$? (unit: mA) (4%)
#### (The Final Answer) ( 867.303 mA )
#### (Computation Process)
$$
I_T = \frac{V_s}{R_T} = \frac{30}{34.59} \simeq 0.867303 A = 867.303 mA
$$

### (f) What is the value of $V_{out}$? (unit: V) (8%)
#### (The Final Answer) ( 1.55V )
#### (Computation Process)
$$
\begin{align*}
V_{Upper} & = V_{Lower} = V_s = 30 V \\
V_1 & = V_{Upper} \frac{R_1}{R_{4,5,8,9,12} + R_1} \\
& = 30 \times \frac{64}{14.704+64} \\
& \simeq 24.4V \\
V_2 & = V_{Lower} \frac{R_2}{R_{3,6,7,10,11} + R_2} \\
& = 30 \times \frac{47}{14.7+47} \\
& \simeq 22.85 V \\
V_{out} & = (V_s-V_1) – (V_s-V_2) = 22.85 – 24.4 = -1.55 V
\end{align*}
$$