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fix LaTeX passthrough

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JamesFlare1212
2025-05-04 15:35:08 -04:00
parent 68ffcf4956
commit 6a996f1c09
5 changed files with 177 additions and 172 deletions
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168
content/en/posts/ecse-1010/lab01/index.md
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@@ -71,8 +71,8 @@ repost:
We know that, the Ohm's Law, KCL, and KVL can be shown as these formulas:
$$
V = IR \\\
\textstyle \sum I_{in} = \sum I_{out} \\\
V = IR \\
\textstyle \sum I_{in} = \sum I_{out} \\
\textstyle \sum V_{n} = 0
$$
@@ -82,11 +82,11 @@ Based on $V = IR$, the total $I$ should be
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
I_{total} &= \frac{5}{10K + \cfrac{1}{\frac{1}{1K} + \frac{1}{1K}} + 10K} \\\
I_{total} &= \frac{5}{10000 + 500 + 10000} \\\
I_{total} &= 0.000243902439 \\\
V &= IR \\
I &= \frac{V}{R} \\
I_{total} &= \frac{5}{10K + \cfrac{1}{\frac{1}{1K} + \frac{1}{1K}} + 10K} \\
I_{total} &= \frac{5}{10000 + 500 + 10000} \\
I_{total} &= 0.000243902439 \\
\end{align*}
$$
@@ -94,8 +94,8 @@ And $I(R2) = I(R3)$ should be
$$
\begin{align*}
I(R2) = I(R3) &= I_{total} \times \frac{R2}{R2 + R3} \\\
I(R2) = I(R3) &= 0.000243902439 \times \frac{1000}{1000 + 1000} \\\
I(R2) = I(R3) &= I_{total} \times \frac{R2}{R2 + R3} \\
I(R2) = I(R3) &= 0.000243902439 \times \frac{1000}{1000 + 1000} \\
I(R2) = I(R3) &= 0.0001219512195
\end{align*}
$$
@@ -106,16 +106,16 @@ To find $V(R1) = V(R4)$ and $V(R2) = V(R3)$, we can use
$$
\begin{align*}
V(R1) = V(R4) &= V_{total} \times \frac{R1}{R1 + R2 \Vert R3 + R4} \\\
V(R1) = V(R4) &= 5 \times \frac{10000}{10000 + 500 + 10000} \\\
V(R1) = V(R4) &= V_{total} \times \frac{R1}{R1 + R2 \Vert R3 + R4} \\
V(R1) = V(R4) &= 5 \times \frac{10000}{10000 + 500 + 10000} \\
V(R1) = V(R4) &= 2.4390244
\end{align*}
$$
$$
\begin{align*}
V(R2) = V(R3) &= V_{total} \\\
V(R2) = V(R3) &= (5 - 2.4390244 - 2.4390244) \\\
V(R2) = V(R3) &= V_{total} \\
V(R2) = V(R3) &= (5 - 2.4390244 - 2.4390244) \\
V(R2) = V(R3) &= 0.1219512
\end{align*}
$$
@@ -187,24 +187,24 @@ and the voltage is the difference in potential. Based on that, we can calculate
$$
\begin{align*}
V(R1) &= V(n001) - V(n002) \\\
V(R1) &= 5 - 2.56098 \\\
V(R1) &= V(n001) - V(n002) \\
V(R1) &= 5 - 2.56098 \\
V(R1) &= \boxed{2.43902}
\end{align*}
$$
$$
\begin{align*}
V(R2) = V(R3) &= V(n002) - V(n003) \\\
V(R2) = V(R3) &= 2.56098 - 2.43902 \\\
V(R2) = V(R3) &= V(n002) - V(n003) \\
V(R2) = V(R3) &= 2.56098 - 2.43902 \\
V(R2) = V(R3) &= \boxed{0.12196}
\end{align*}
$$
$$
\begin{align*}
V(R4) &= V(n003) - V(\text{GND}) \\\
V(R4) &= 2.43902 - 0 \\\
V(R4) &= V(n003) - V(\text{GND}) \\
V(R4) &= 2.43902 - 0 \\
V(R4) &= \boxed{2.43902}
\end{align*}
$$
@@ -240,8 +240,8 @@ Using the expectation $I(R1) = I(R2) + I(R3)$ from Analysis. We can check
$$
\begin{align*}
& \quad \thickspace I(R1) + I(R2) + I(R3) \\\
&= -0.000243902 + 0.000121951 + 0.000121951 \\\
& \quad \thickspace I(R1) + I(R2) + I(R3) \\
&= -0.000243902 + 0.000121951 + 0.000121951 \\
&= \boxed{0}
\end{align*}
$$
@@ -256,8 +256,8 @@ We can use the result from previous part.
$$
\begin{align*}
V(R1) &= 2.43902 \\\
V(R2) = V(R3) &= 0.12196 \\\
V(R1) &= 2.43902 \\
V(R2) = V(R3) &= 0.12196 \\
V(R4) &= 2.43902
\end{align*}
$$
@@ -266,9 +266,9 @@ Use the expectation $V(n001) - V(n002) - V(n003) = 0$ from Analysis. We can chec
$$
\begin{align*}
& \quad \thickspace V(n001) - V(n002) - V(n003) \\\
&= 2.43902 - 0.12196 - 0.12196 \\\
&= 0 \\\
& \quad \thickspace V(n001) - V(n002) - V(n003) \\
&= 2.43902 - 0.12196 - 0.12196 \\
&= 0 \\
& \boxed{\text{True}}
\end{align*}
$$
@@ -292,24 +292,24 @@ to calculate $I$ based on Ohm's Law.
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
V &= IR \\
I &= \frac{V}{R} \\
I(R1) &= \frac{2.4963}{10000} = \boxed{0.00024963}
\end{align*}
$$
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
V &= IR \\
I &= \frac{V}{R} \\
I(R2) = I(R3) &= \frac{0.1665}{1000} = \boxed{0.0001665}
\end{align*}
$$
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
V &= IR \\
I &= \frac{V}{R} \\
I(R4) &= \frac{2.4616}{10000} = \boxed{0.00024616}
\end{align*}
$$
@@ -407,16 +407,16 @@ V(n002): 2.5 voltage
$$
\begin{align*}
V(R1) &= V(n001) - V(n002) \\\
V(R1) &= 5 - 2.5 \\\
V(R1) &= V(n001) - V(n002) \\
V(R1) &= 5 - 2.5 \\
V(R1) &= \boxed{2.5}
\end{align*}
$$
$$
\begin{align*}
V(R2) &= V(n002) - \text{GND} \\\
V(R2) &= 2.5 - 0 \\\
V(R2) &= V(n002) - \text{GND} \\
V(R2) &= 2.5 - 0 \\
V(R2) &= \boxed{2.5}
\end{align*}
$$
@@ -469,7 +469,7 @@ Also, we know that $R_1 = R_2 = 10K$ and the voltage across the resistor can be
$$
\begin{align*}
\frac{V_1}{V_2} &= \frac{R_1}{R_2} \\\
\frac{V_1}{V_2} &= \frac{R_1}{R_2} \\
\frac{V_1}{V_2} &= \frac{10K}{10K} = \frac{1}{1}
\end{align*}
$$
@@ -480,8 +480,8 @@ Using these values, we can find out $I(R1)$ and $I(R2)$ by
$$
\begin{align*}
I(R1) = I(R2) &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{2.5}{10K} \\\
I(R1) = I(R2) &= \frac{V}{R} \\
I(R1) = I(R2) &= \frac{2.5}{10K} \\
I(R1) = I(R2) &= \boxed{0.00025}
\end{align*}
$$
@@ -539,16 +539,16 @@ Based on the Ohm's Law - the relationship we got in Analysis $I = \frac{V}{R}$.
$$
\begin{align*}
I(R1) &= \frac{V}{R} \\\
I(R1) &= \frac{2.5539}{10K} \\\
I(R1) &= \frac{V}{R} \\
I(R1) &= \frac{2.5539}{10K} \\
I(R1) &= 0.00025539
\end{align*}
$$
$$
\begin{align*}
I(R2) &= \frac{V}{R} \\\
I(R2) &= \frac{2.5204}{10K} \\\
I(R2) &= \frac{V}{R} \\
I(R2) &= \frac{2.5204}{10K} \\
I(R2) &= 0.00025204
\end{align*}
$$
@@ -589,8 +589,8 @@ Also, the voltage is potential difference between the component.
$$
\begin{align*}
V(R1) = V(R2) &= n001 - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= n001 - \text{GND} \\
V(R1) = V(R2) &= 5 - 0 \\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
@@ -637,8 +637,8 @@ Let's put values into equation
$$
\begin{align*}
I_{total} &= \frac{5}{\cfrac{1}{\frac{1}{10K} + \frac{1}{10K}}} \\\
I_{total} &= \frac{5}{5K} \\\
I_{total} &= \frac{5}{\cfrac{1}{\frac{1}{10K} + \frac{1}{10K}}} \\
I_{total} &= \frac{5}{5K} \\
I_{total} &= \boxed{0.001}
\end{align*}
$$
@@ -689,8 +689,8 @@ and the voltage is the difference in potential. Based on that, we can caculate t
$$
\begin{align*}
V(R1) = V(R2) &= V(n001) - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= V(n001) - \text{GND} \\
V(R1) = V(R2) &= 5 - 0 \\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
@@ -733,9 +733,9 @@ So, we get
$$
\begin{align*}
I_{total} &= I(R2) + I(R1) \\\
I_{total} &= 0.0005 + 0.0005 \\\
I_{total} &= 0.001 \\\
I_{total} &= I(R2) + I(R1) \\
I_{total} &= 0.0005 + 0.0005 \\
I_{total} &= 0.001 \\
& 0.001 = 0.001 \\; \boxed{\text{True}}
\end{align*}
$$
@@ -764,9 +764,9 @@ Then, we can find out $I(R1)$ and $I(R2)$ by
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\\
V &= IR \\
I &= \frac{V}{R} \\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\
I(R1) = I(R2) &= \boxed{0.0005305}
\end{align*}
$$
@@ -808,8 +808,8 @@ Also, the voltage is potential difference between the component.
$$
\begin{align*}
V(R1) = V(R2) &= n001 - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= n001 - \text{GND} \\
V(R1) = V(R2) &= 5 - 0 \\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
@@ -830,16 +830,16 @@ and get
$$
\begin{align*}
I(R1) &= \frac{V(R1)}{R1} \\\
I(R1) &= \frac{5}{10K} \\\
I(R1) &= \frac{V(R1)}{R1} \\
I(R1) &= \frac{5}{10K} \\
I(R1) &= \boxed{0.0005}
\end{align*}
$$
$$
\begin{align*}
I(R2) &= \frac{V(R2)}{R2} \\\
I(R2) &= \frac{5}{10K} \\\
I(R2) &= \frac{V(R2)}{R2} \\
I(R2) &= \frac{5}{10K} \\
I(R2) &= \boxed{0.0005}
\end{align*}
$$
@@ -848,10 +848,10 @@ the relationship between $I(R1)$ and $I(R2)$ can be express as
$$
\begin{align*}
\frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{V(R1)}{R1}}{\cfrac{V(R2)}{R2}} \\\
\because V(R1) &= V(R2) \\\
\therefore \frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{\cancel{V(R1)}}{R1} \times \cfrac{1}{\cancel{V(R1)}}}{\cfrac{\cancel{V(R2)}}{R2} \times \cfrac{1}{\cancel{V(R2)}}} \\\
\frac{I(R1)}{I(R2)} &= \frac{\frac{1}{R1}}{\frac{1}{R2}} \\\
\frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{V(R1)}{R1}}{\cfrac{V(R2)}{R2}} \\
\because V(R1) &= V(R2) \\
\therefore \frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{\cancel{V(R1)}}{R1} \times \cfrac{1}{\cancel{V(R1)}}}{\cfrac{\cancel{V(R2)}}{R2} \times \cfrac{1}{\cancel{V(R2)}}} \\
\frac{I(R1)}{I(R2)} &= \frac{\frac{1}{R1}}{\frac{1}{R2}} \\
&\boxed{\frac{I(R1)}{I(R2)} = \frac{R2}{R1}}
\end{align*}
$$
@@ -866,9 +866,9 @@ as we get the $I_{total}$ by
$$
\begin{align*}
I_{total} &= \frac{V_{total}}{R_{total}} \\\
I_{total} &= \frac{5}{\cfrac{1}{\cfrac{1}{10K} + \cfrac{1}{10K}}} \\\
I_{total} &= \frac{5}{5K} \\\
I_{total} &= \frac{V_{total}}{R_{total}} \\
I_{total} &= \frac{5}{\cfrac{1}{\cfrac{1}{10K} + \cfrac{1}{10K}}} \\
I_{total} &= \frac{5}{5K} \\
I_{total} &= 0.001
\end{align*}
$$
@@ -883,8 +883,8 @@ We can get $I(R1)$ and $I(R2)$ by
$$
\begin{align*}
I(R1) = I(R2) &= I_{total} \times \frac {R1}{R1 + R2} \\\
I(R1) = I(R2) &= 0.001 \times \frac {10K}{10K + 10K} \\\
I(R1) = I(R2) &= I_{total} \times \frac {R1}{R1 + R2} \\
I(R1) = I(R2) &= 0.001 \times \frac {10K}{10K + 10K} \\
I(R1) = I(R2) &= \boxed{0.0005}
\end{align*}
$$
@@ -893,12 +893,12 @@ At this point, our logic is consistent, which
$$
\begin{align*}
& \because R1 = R2 = 10K \\\
& \because I(R1) = I(R2) = 0.0005 \\\
& \because V(R1) = V(R2) = 5 \\\
& \because I_{total} = I(R1) + I(R2) = 0.001 \\\
& \because \frac{I(R1)}{I(R2)} = \frac{R2}{R1} \\\
& \therefore I(R1) = I(R2) = I_{total} \times \frac {R1}{R1 + R2} \\\
& \because R1 = R2 = 10K \\
& \because I(R1) = I(R2) = 0.0005 \\
& \because V(R1) = V(R2) = 5 \\
& \because I_{total} = I(R1) + I(R2) = 0.001 \\
& \because \frac{I(R1)}{I(R2)} = \frac{R2}{R1} \\
& \therefore I(R1) = I(R2) = I_{total} \times \frac {R1}{R1 + R2} \\
\end{align*}
$$
@@ -958,8 +958,8 @@ and the voltage is the difference in potential. Based on that, we can calculate
$$
\begin{align*}
V(R1) = V(R2) &= V(n001) - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= V(n001) - \text{GND} \\
V(R1) = V(R2) &= 5 - 0 \\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
@@ -999,9 +999,9 @@ Then, we can find out $I(R1)$ and $I(R2)$ by
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\\
V &= IR \\
I &= \frac{V}{R} \\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\
I(R1) = I(R2) &= \boxed{0.0005305}
\end{align*}
$$
@@ -1038,7 +1038,7 @@ We are going to use NTC 100K as our thermistor, and we will compare the reading
NTC thermistor uses Beta formula to calculate the resistance under a specific temperature. The formula is like
$$
\frac{1}{T_1} = \frac{1}{T_0} + \frac{1}{\beta} \ln\left(\frac{R_1}{R_0}\right) \\\
\frac{1}{T_1} = \frac{1}{T_0} + \frac{1}{\beta} \ln\left(\frac{R_1}{R_0}\right) \\
$$
We can move $R_1$ to the left side to get
@@ -1054,7 +1054,7 @@ Since we want to find out the resistance of this thermistor under a specific tem
The thermistor we are using is NTC 100K. Which means it has $100k \Omega$ at the reference temperature $25 \degree C$
$$
T_0 = 298.15K \\\
T_0 = 298.15K \\
R_0 = 100k \Omega
$$
@@ -1107,7 +1107,7 @@ We used the math function in Scope with
to get the temperature in $\degree C$. This is from
$$
\frac{1}{T_1} = \frac{1}{T_0} + \frac{1}{\beta} \ln\left(\frac{R_1}{R_0}\right) \\\
\frac{1}{T_1} = \frac{1}{T_0} + \frac{1}{\beta} \ln\left(\frac{R_1}{R_0}\right) \\
$$
which $T_1$ is the temperature reading we want.

22
content/en/posts/ecse-1010/lab02/index.md
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@@ -518,11 +518,11 @@ $$
Substituting known values. Given $V_A = 5$ and $V_D = 0$, the equations become:
$$
2.5V_B - V_C = 5 \\\
2.5V_B - V_C = 5 \\
2V_C - V_B = 0
$$
Matrix form: $\begin{bmatrix} 2.5 & -1 \\\ -1 & 2 \end{bmatrix}$ $\begin{bmatrix} V_B \\\ V_C \end{bmatrix} =$ $\begin{bmatrix} 5 \\\ 0 \end{bmatrix}$
Matrix form: $\begin{bmatrix} 2.5 & -1 \\ -1 & 2 \end{bmatrix}$ $\begin{bmatrix} V_B \\ V_C \end{bmatrix} =$ $\begin{bmatrix} 5 \\ 0 \end{bmatrix}$
Solve them "by hand"
@@ -547,7 +547,7 @@ The solution is:
1.2500
```
Thus, $\begin{bmatrix} V_B \\\ V_C \end{bmatrix} =$ $\begin{bmatrix}2.5 \\\ 1.25 \end{bmatrix}$
Thus, $\begin{bmatrix} V_B \\ V_C \end{bmatrix} =$ $\begin{bmatrix}2.5 \\ 1.25 \end{bmatrix}$
### Simulation
@@ -666,8 +666,8 @@ A non-inverted comparator has a transfer function of
$$
\begin{equation*}
V_{out}=\begin{cases}
\text{if} \; V_{in} < V_{ref}, V_{out} = V_s - \\\
\text{if} \; V_{in} > V_{ref}, V_{out} = V_s + \\\
\text{if} \; V_{in} < V_{ref}, V_{out} = V_s - \\
\text{if} \; V_{in} > V_{ref}, V_{out} = V_s + \\
\end{cases}
\end{equation*}
$$
@@ -677,8 +677,8 @@ In our case, we got
$$
\begin{equation*}
V_{out}=\begin{cases}
\text{if} \; V_{in} < 0V, V_{out} = -5V \\\
\text{if} \; V_{in} > 0V, V_{out} = 5V \\\
\text{if} \; V_{in} < 0V, V_{out} = -5V \\
\text{if} \; V_{in} > 0V, V_{out} = 5V \\
\end{cases}
\end{equation*}
$$
@@ -723,8 +723,8 @@ In our case, we want to use $50K \Omega$ potentiometer as the resistors, so it c
$$
\begin{align*}
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\\
V_{out} &= - V1 - V2 \\\
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\
V_{out} &= - V1 - V2 \\
\end{align*}
$$
@@ -782,8 +782,8 @@ In our case, we want to use $50K \Omega$ potentiometer as the resistors, so it c
$$
\begin{align*}
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\\
V_{out} &= - V1 - V2 \\\
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\
V_{out} &= - V1 - V2 \\
\end{align*}
$$

4
content/en/posts/ecse-1010/lab03/index.md
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@@ -227,7 +227,7 @@ Plugging in the Values
$$
\begin{align*}
f_c &= \frac{510}{2 \times \pi \times 0.001} \\\
f_c &= \frac{510}{2 \times \pi \times 0.001} \\
f_c &= \frac{510}{0.00628318} \approx 81,000 \text{ Hz}
\end{align*}
$$
@@ -329,7 +329,7 @@ we can get the capacitance of the target LP filter.
$$
\begin{align*}
300 &= \frac{1}{2 \pi \cdot 1K \cdot C} \\\
300 &= \frac{1}{2 \pi \cdot 1K \cdot C} \\
C &= 5.6 \times 10^{-7}
\end{align*}
$$

148
content/zh-cn/posts/ecse-1010/lab01/index.md
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@@ -71,8 +71,8 @@ repost:
我们知道，欧姆定律、KCL 和 KVL 可以表示成以下公式：
$$
V = IR \\\
\sum I_{in} = \sum I_{out} \\\
V = IR \\
\sum I_{in} = \sum I_{out} \\
\sum V_n = 0
$$
@@ -82,11 +82,11 @@ $$
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
I_{total} &= \frac{5}{10K + \cfrac{1}{\frac{1}{1K} + \frac{1}{1K}} + 10K} \\\
I_{total} &= \frac{5}{10000 + 500 + 10000} \\\
I_{total} &= 0.000243902439 \\\
V &= IR \\
I &= \frac{V}{R} \\
I_{total} &= \frac{5}{10K + \cfrac{1}{\frac{1}{1K} + \frac{1}{1K}} + 10K} \\
I_{total} &= \frac{5}{10000 + 500 + 10000} \\
I_{total} &= 0.000243902439 \\
\end{align*}
$$
@@ -94,8 +94,8 @@ $$
$$
\begin{align*}
I(R2) = I(R3) &= I_{total} \times \frac{R2}{R2 + R3} \\\
I(R2) = I(R3) &= 0.000243902439 \times \frac{1000}{1000 + 1000} \\\
I(R2) = I(R3) &= I_{total} \times \frac{R2}{R2 + R3} \\
I(R2) = I(R3) &= 0.000243902439 \times \frac{1000}{1000 + 1000} \\
I(R2) = I(R3) &= 0.0001219512195
\end{align*}
$$
@@ -106,16 +106,16 @@ $$
$$
\begin{align*}
V(R1) = V(R4) &= V_{total} \times \frac{R1}{R1 + R2 \Vert R3 + R4} \\\
V(R1) = V(R4) &= 5 \times \frac{10000}{10000 + 500 + 10000} \\\
V(R1) = V(R4) &= V_{total} \times \frac{R1}{R1 + R2 \Vert R3 + R4} \\
V(R1) = V(R4) &= 5 \times \frac{10000}{10000 + 500 + 10000} \\
V(R1) = V(R4) &= 2.4390244
\end{align*}
$$
$$
\begin{align*}
V(R2) = V(R3) &= V_{total} - (V(R1) + V(R4)) \\\
V(R2) = V(R3) &= 5 - 2.4390244 - 2.4390244 \\\
V(R2) = V(R3) &= V_{total} - (V(R1) + V(R4)) \\
V(R2) = V(R3) &= 5 - 2.4390244 - 2.4390244 \\
V(R2) = V(R3) &= 0.1219512
\end{align*}
$$
@@ -185,24 +185,24 @@ V(n003): 2.43902 voltage
$$
\begin{align*}
V(R1) &= V(n001) - V(n002) \\\
V(R1) &= 5 - 2.56098 \\\
V(R1) &= V(n001) - V(n002) \\
V(R1) &= 5 - 2.56098 \\
V(R1) &= \boxed{2.43902}
\end{align*}
$$
$$
\begin{align*}
V(R2) = V(R3) &= V(n002) - V(n003) \\\
V(R2) = V(R3) &= 2.56098 - 2.43902 \\\
V(R2) = V(R3) &= V(n002) - V(n003) \\
V(R2) = V(R3) &= 2.56098 - 2.43902 \\
V(R2) = V(R3) &= \boxed{0.12196}
\end{align*}
$$
$$
\begin{align*}
V(R4) &= V(n003) - V(\text{GND}) \\\
V(R4) &= 2.43902 - 0 \\\
V(R4) &= V(n003) - V(\text{GND}) \\
V(R4) &= 2.43902 - 0 \\
V(R4) &= \boxed{2.43902}
\end{align*}
$$
@@ -238,8 +238,8 @@ I(V1): -0.000243902 device_current
$$
\begin{align*}
& \quad \thickspace I(R1) + I(R2) + I(R3) \\\
&= -0.000243902 + 0.000121951 + 0.000121951 \\\
& \quad \thickspace I(R1) + I(R2) + I(R3) \\
&= -0.000243902 + 0.000121951 + 0.000121951 \\
&= \boxed{0}
\end{align*}
$$
@@ -254,8 +254,8 @@ KCL 很可能为真。
$$
\begin{align*}
V(R1) &= 2.43902 \\\
V(R2) = V(R3) &= 0.12196 \\\
V(R1) &= 2.43902 \\
V(R2) = V(R3) &= 0.12196 \\
V(R4) &= 2.43902
\end{align*}
$$
@@ -264,9 +264,9 @@ $$
$$
\begin{align*}
& \quad \thickspace V(n001) - V(n002) - V(n003) \\\
&= 2.43902 - 0.12196 - 0.12196 \\\
&= 0 \\\
& \quad \thickspace V(n001) - V(n002) - V(n003) \\
&= 2.43902 - 0.12196 - 0.12196 \\
&= 0 \\
& \boxed{\text{True}}
\end{align*}
$$
@@ -290,24 +290,24 @@ V(R4) = 2.4616V
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
V &= IR \\
I &= \frac{V}{R} \\
I(R1) &= \frac{2.4963}{10000} = \boxed{0.00024963}
\end{align*}
$$
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
V &= IR \\
I &= \frac{V}{R} \\
I(R2) = I(R3) &= \frac{0.1665}{1000} = \boxed{0.0001665}
\end{align*}
$$
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
V &= IR \\
I &= \frac{V}{R} \\
I(R4) &= \frac{2.4616}{10000} = \boxed{0.00024616}
\end{align*}
$$
@@ -405,16 +405,16 @@ V(n002): 2.5 voltage
$$
\begin{align*}
V(R1) &= V(n001) - V(n002) \\\
V(R1) &= 5 - 2.5 \\\
V(R1) &= V(n001) - V(n002) \\
V(R1) &= 5 - 2.5 \\
V(R1) &= \boxed{2.5}
\end{align*}
$$
$$
\begin{align*}
V(R2) &= V(n002) - \text{GND} \\\
V(R2) &= 2.5 - 0 \\\
V(R2) &= V(n002) - \text{GND} \\
V(R2) &= 2.5 - 0 \\
V(R2) &= \boxed{2.5}
\end{align*}
$$
@@ -467,7 +467,7 @@ $$
$$
\begin{align*}
\frac{V_1}{V_2} &= \frac{R_1}{R_2} \\\
\frac{V_1}{V_2} &= \frac{R_1}{R_2} \\
\frac{V_1}{V_2} &= \frac{10K}{10K} = 1
\end{align*}
$$
@@ -478,8 +478,8 @@ $$
$$
\begin{align*}
I(R1) = I(R2) &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{2.5}{10K} \\\
I(R1) = I(R2) &= \frac{V}{R} \\
I(R1) = I(R2) &= \frac{2.5}{10K} \\
I(R1) = I(R2) &= \boxed{0.00025}
\end{align*}
$$
@@ -537,16 +537,16 @@ $V(R2) = 2.5204V$
$$
\begin{align*}
I(R1) &= \frac{V}{R} \\\
I(R1) &= \frac{2.5539}{10K} \\\
I(R1) &= \frac{V}{R} \\
I(R1) &= \frac{2.5539}{10K} \\
I(R1) &= 0.00025539
\end{align*}
$$
$$
\begin{align*}
I(R2) &= \frac{V}{R} \\\
I(R2) &= \frac{2.5204}{10K} \\\
I(R2) &= \frac{V}{R} \\
I(R2) &= \frac{2.5204}{10K} \\
I(R2) &= 0.00025204
\end{align*}
$$
@@ -587,7 +587,7 @@ $$
$$
\begin{align*}
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= 5 - 0 \\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
@@ -634,8 +634,8 @@ $$
$$
\begin{align*}
I_{total} &= \frac{5}{\cfrac{1}{\frac{1}{10K} + \frac{1}{10K}}} \\\
I_{total} &= \frac{5}{5K} \\\
I_{total} &= \frac{5}{\cfrac{1}{\frac{1}{10K} + \frac{1}{10K}}} \\
I_{total} &= \frac{5}{5K} \\
I_{total} &= \boxed{0.001}
\end{align*}
$$
@@ -686,7 +686,7 @@ V(n001): 5 voltage
$$
\begin{align*}
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= 5 - 0 \\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
@@ -729,9 +729,9 @@ $$
$$
\begin{align*}
I_{total} &= I(R2) + I(R1) \\\
I_{total} &= 0.0005 + 0.0005 \\\
I_{total} &= 0.001 \\\
I_{total} &= I(R2) + I(R1) \\
I_{total} &= 0.0005 + 0.0005 \\
I_{total} &= 0.001 \\
& 0.001 = 0.001 \\; \boxed{\text{True}}
\end{align*}
$$
@@ -760,8 +760,8 @@ R2 = 10K
$$
\begin{align*}
I &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\\
I &= \frac{V}{R} \\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\
I(R1) = I(R2) &= \boxed{0.0005305}
\end{align*}
$$
@@ -803,7 +803,7 @@ $$
$$
\begin{align*}
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= 5 - 0 \\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
@@ -824,16 +824,16 @@ $$
$$
\begin{align*}
I(R1) &= \frac{V(R1)}{R1} \\\
I(R1) &= \frac{5}{10K} \\\
I(R1) &= \frac{V(R1)}{R1} \\
I(R1) &= \frac{5}{10K} \\
I(R1) &= \boxed{0.0005}
\end{align*}
$$
$$
\begin{align*}
I(R2) &= \frac{V(R2)}{R2} \\\
I(R2) &= \frac{5}{10K} \\\
I(R2) &= \frac{V(R2)}{R2} \\
I(R2) &= \frac{5}{10K} \\
I(R2) &= \boxed{0.0005}
\end{align*}
$$
@@ -842,10 +842,10 @@ $I(R1)$ 和 $I(R2)$ 的关系可以表示为：
$$
\begin{align*}
\frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{V(R1)}{R1}}{\cfrac{V(R2)}{R2}} \\\
\because V(R1) &= V(R2) \\\
\therefore \frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{\cancel{V(R1)}}{R1} \times \cfrac{1}{\cancel{V(R1)}}}{\cfrac{\cancel{V(R2)}}{R2} \times \cfrac{1}{\cancel{V(R2)}}} \\\
\frac{I(R1)}{I(R2)} &= \frac{\frac{1}{R1}}{\frac{1}{R2}} \\\
\frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{V(R1)}{R1}}{\cfrac{V(R2)}{R2}} \\
\because V(R1) &= V(R2) \\
\therefore \frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{\cancel{V(R1)}}{R1} \times \cfrac{1}{\cancel{V(R1)}}}{\cfrac{\cancel{V(R2)}}{R2} \times \cfrac{1}{\cancel{V(R2)}}} \\
\frac{I(R1)}{I(R2)} &= \frac{\frac{1}{R1}}{\frac{1}{R2}} \\
&\boxed{\frac{I(R1)}{I(R2)} = \frac{R2}{R1}}
\end{align*}
$$
@@ -860,9 +860,9 @@ $$
$$
\begin{align*}
I_{total} &= \frac{V_{total}}{R_{total}} \\\
I_{total} &= \frac{5}{\cfrac{1}{\cfrac{1}{10K} + \cfrac{1}{10K}}} \\\
I_{total} &= \frac{5}{5K} \\\
I_{total} &= \frac{V_{total}}{R_{total}} \\
I_{total} &= \frac{5}{\cfrac{1}{\cfrac{1}{10K} + \cfrac{1}{10K}}} \\
I_{total} &= \frac{5}{5K} \\
I_{total} &= 0.001
\end{align*}
$$
@@ -877,8 +877,8 @@ $$
$$
\begin{align*}
I(R1) = I(R2) &= I_{total} \times \frac {R1}{R1 + R2} \\\
I(R1) = I(R2) &= 0.001 \times \frac {10K}{10K + 10K} \\\
I(R1) = I(R2) &= I_{total} \times \frac {R1}{R1 + R2} \\
I(R1) = I(R2) &= 0.001 \times \frac {10K}{10K + 10K} \\
I(R1) = I(R2) &= \boxed{0.0005}
\end{align*}
$$
@@ -887,11 +887,11 @@ $$
$$
\begin{align*}
& \because R1 = R2 = 10K \\\
& \because I(R1) = I(R2) = 0.0005 \\\
& \because V(R1) = V(R2) = 5 \\\
& \because I_{total} = I(R1) + I(R2) = 0.001 \\\
& \because \frac{I(R1)}{I(R2)} = \frac{R2}{R1} \\\
& \because R1 = R2 = 10K \\
& \because I(R1) = I(R2) = 0.0005 \\
& \because V(R1) = V(R2) = 5 \\
& \because I_{total} = I(R1) + I(R2) = 0.001 \\
& \because \frac{I(R1)}{I(R2)} = \frac{R2}{R1} \\
& \therefore I(R1) = I(R2) = I_{total} \times \frac {R1}{R1 + R2}
\end{align*}
$$
@@ -952,7 +952,7 @@ V(n001): 5 voltage
$$
\begin{align*}
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= 5 - 0 \\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
@@ -992,8 +992,8 @@ R2 = 10K
$$
\begin{align*}
I &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\\
I &= \frac{V}{R} \\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\
I(R1) = I(R2) &= \boxed{0.0005305}
\end{align*}
$$