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improve katex style

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JamesFlare1212
2024-11-19 16:54:06 -05:00
parent fc70b1776c
commit 47358d25a8
4 changed files with 180 additions and 169 deletions
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7
assets/css/_custom.scss
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@@ -46,4 +46,9 @@ details summary strong {
margin-top: 0;
margin-bottom: 0;
}
}
}
.katex-display {
overflow-x: auto;
overflow-y: clip;
}

326
content/en/posts/ecse-1010/lab01/index.md
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@@ -82,11 +82,11 @@ Based on $V = IR$, the total $I$ should be
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
I_{total} &= \frac{5}{10K + \cfrac{1}{\frac{1}{1K} + \frac{1}{1K}} + 10K} \\\
I_{total} &= \frac{5}{10000 + 500 + 10000} \\\
I_{total} &= 0.000243902439 \\\
V &= IR \\\
I &= \frac{V}{R} \\\
I_{total} &= \frac{5}{10K + \cfrac{1}{\frac{1}{1K} + \frac{1}{1K}} + 10K} \\\
I_{total} &= \frac{5}{10000 + 500 + 10000} \\\
I_{total} &= 0.000243902439 \\\
\end{align*}
$$
@@ -94,9 +94,9 @@ And $I(R2) = I(R3)$ should be
$$
\begin{align*}
I(R2) = I(R3) &= I_{total} \times \frac{R2}{R2 + R3} \\\
I(R2) = I(R3) &= 0.000243902439 \times \frac{1000}{1000 + 1000} \\\
I(R2) = I(R3) &= 0.0001219512195
I(R2) = I(R3) &= I_{total} \times \frac{R2}{R2 + R3} \\\
I(R2) = I(R3) &= 0.000243902439 \times \frac{1000}{1000 + 1000} \\\
I(R2) = I(R3) &= 0.0001219512195
\end{align*}
$$
@@ -106,17 +106,17 @@ To find $V(R1) = V(R4)$ and $V(R2) = V(R3)$, we can use
$$
\begin{align*}
V(R1) = V(R4) &= V_{total} \times \frac{R1}{R1 + R2 \Vert R3 + R4} \\\
V(R1) = V(R4) &= 5 \times \frac{10000}{10000 + 500 + 10000} \\\
V(R1) = V(R4) &= 2.4390244
V(R1) = V(R4) &= V_{total} \times \frac{R1}{R1 + R2 \Vert R3 + R4} \\\
V(R1) = V(R4) &= 5 \times \frac{10000}{10000 + 500 + 10000} \\\
V(R1) = V(R4) &= 2.4390244
\end{align*}
$$
$$
\begin{align*}
V(R2) = V(R3) &= V_{total} \\\
V(R2) = V(R3) &= (5 - 2.4390244 - 2.4390244) \\\
V(R2) = V(R3) &= 0.1219512
V(R2) = V(R3) &= V_{total} \\\
V(R2) = V(R3) &= (5 - 2.4390244 - 2.4390244) \\\
V(R2) = V(R3) &= 0.1219512
\end{align*}
$$
@@ -135,14 +135,14 @@ We will check if the experimental results fit these expectations.
```text
--- Operating Point ---
V(n001): 5 voltage
V(n002): 2.56098 voltage
V(n003): 2.43902 voltage
I(R1): -0.000243902 device_current
I(R2): 0.000121951 device_current
I(R3): 0.000121951 device_current
I(R4): 0.000243902 device_current
I(V1): -0.000243902 device_current
V(n001): 5 voltage
V(n002): 2.56098 voltage
V(n003): 2.43902 voltage
I(R1): -0.000243902 device_current
I(R2): 0.000121951 device_current
I(R3): 0.000121951 device_current
I(R4): 0.000243902 device_current
I(V1): -0.000243902 device_current
```
### Measurement
@@ -178,40 +178,40 @@ To find out the theoretic values of $V(R1)$, $V(R2) = V(R3)$ and $V(R4)$. We nee
We know the simulation output is
```text
V(n001): 5 voltage
V(n002): 2.56098 voltage
V(n003): 2.43902 voltage
V(n001): 5 voltage
V(n002): 2.56098 voltage
V(n003): 2.43902 voltage
```
and the voltage is the difference in potential. Based on that, we can caculate the theoretic values of $V(R1)$, $V(R2) = V(R3)$ and $V(R4)$.
and the voltage is the difference in potential. Based on that, we can calculate the theoretic values of $V(R1)$, $V(R2) = V(R3)$ and $V(R4)$.
$$
\begin{align*}
V(R1) &= V(n001) - V(n002) \\\
V(R1) &= 5 - 2.56098 \\\
V(R1) &= \boxed{2.43902}
V(R1) &= V(n001) - V(n002) \\\
V(R1) &= 5 - 2.56098 \\\
V(R1) &= \boxed{2.43902}
\end{align*}
$$
$$
\begin{align*}
V(R2) = V(R3) &= V(n002) - V(n003) \\\
V(R2) = V(R3) &= 2.56098 - 2.43902 \\\
V(R2) = V(R3) &= \boxed{0.12196}
V(R2) = V(R3) &= V(n002) - V(n003) \\\
V(R2) = V(R3) &= 2.56098 - 2.43902 \\\
V(R2) = V(R3) &= \boxed{0.12196}
\end{align*}
$$
$$
\begin{align*}
V(R4) &= V(n003) - V(\text{GND}) \\\
V(R4) &= 2.43902 - 0 \\\
V(R4) &= \boxed{2.43902}
V(R4) &= V(n003) - V(\text{GND}) \\\
V(R4) &= 2.43902 - 0 \\\
V(R4) &= \boxed{2.43902}
\end{align*}
$$
Let's make a table to compare the results
|Iteams|Analysis|Simulation|Experiement|diff|$\%$diff|
|Items|Analysis|Simulation|Experiment|diff|$\%$diff|
|:-:|:-:|:-:|:-:|:-:|-:|
|$V(R1)$|$2.4390V$|$2.4390V$|$2.4963V$|$57.28mV$|$2.3\%$|
|$V(R2)$|$0.1219V$|$0.1219V$|$0.1665V$|$44.54mV$|$26.8\%$|
@@ -229,20 +229,20 @@ Now, let's check KCL.
We got the simulations data like
```text
I(R1): -0.000243902 device_current
I(R2): 0.000121951 device_current
I(R3): 0.000121951 device_current
I(R4): 0.000243902 device_current
I(V1): -0.000243902 device_current
I(R1): -0.000243902 device_current
I(R2): 0.000121951 device_current
I(R3): 0.000121951 device_current
I(R4): 0.000243902 device_current
I(V1): -0.000243902 device_current
```
Using the expectation $I(R1) = I(R2) + I(R3)$ from Analysis. We can check
$$
\begin{align*}
& I(R1) + I(R2) + I(R3) \\\
=& -0.000243902 + 0.000121951 + 0.000121951 \\\
=& \; \boxed{0}
& \quad \thickspace I(R1) + I(R2) + I(R3) \\\
&= -0.000243902 + 0.000121951 + 0.000121951 \\\
&= \boxed{0}
\end{align*}
$$
@@ -256,9 +256,9 @@ We can use the result from previous part.
$$
\begin{align*}
V(R1) &= 2.43902 \\\
V(R2) = V(R3) &= 0.12196 \\\
V(R4) &= 2.43902
V(R1) &= 2.43902 \\\
V(R2) = V(R3) &= 0.12196 \\\
V(R4) &= 2.43902
\end{align*}
$$
@@ -266,10 +266,10 @@ Use the expectation $V(n001) - V(n002) - V(n003) = 0$ from Analysis. We can chec
$$
\begin{align*}
& V(n001) - V(n002) - V(n003) \\\
=& \; 2.43902 - 0.12196 - 0.12196\\\
=& \; 0 \\\
& \; 0 = 0 \; \boxed{\text{True}}
& \quad \thickspace V(n001) - V(n002) - V(n003) \\\
&= 2.43902 - 0.12196 - 0.12196 \\\
&= 0 \\\
\boxed{\text{True}}
\end{align*}
$$
@@ -292,31 +292,31 @@ to calculate $I$ based on Ohm's Law.
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R1) &= \frac{2.4963}{10000} = \boxed{0.00024963}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R1) &= \frac{2.4963}{10000} = \boxed{0.00024963}
\end{align*}
$$
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R2) = I(R3) &= \frac{0.1665}{1000} = \boxed{0.0001665}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R2) = I(R3) &= \frac{0.1665}{1000} = \boxed{0.0001665}
\end{align*}
$$
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R4) &= \frac{2.4616}{10000} = \boxed{0.00024616}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R4) &= \frac{2.4616}{10000} = \boxed{0.00024616}
\end{align*}
$$
Then, we can check these current result with simulation data.
|Iteams|Analysis|Simulation|Experiement|diff|$\%$diff|
|Items|Analysis|Simulation|Experiment|diff|$\%$diff|
|:-:|:-|:-|:-|:-:|-:|
|$I(R1)$|$0.2439mA$|$0.2439mA$|$0.2496mA$|$0.005728mA$|$2.3\%$|
|$I(R2)$|$0.1665mA$|$0.1665mA$|$0.1219mA$|$0.044549mA$|$26.8\%$|
@@ -366,11 +366,11 @@ We know that $V_1 + V_2 = 5$ and $1 \cdot V_1 = 1 \cdot V_2$. So, we should expe
```text
--- Operating Point ---
V(n001): 5 voltage
V(n002): 2.5 voltage
I(R1): -0.00025 device_current
I(R2): -0.00025 device_current
I(V1): -0.00025 device_current
V(n001): 5 voltage
V(n002): 2.5 voltage
I(R1): -0.00025 device_current
I(R2): -0.00025 device_current
I(V1): -0.00025 device_current
```
### Measurement
@@ -401,29 +401,29 @@ and the voltage is the difference in potential. Based on that, we can calculate
We know the simulation output is
```text
V(n001): 5 voltage
V(n002): 2.5 voltage
V(n001): 5 voltage
V(n002): 2.5 voltage
```
$$
\begin{align*}
V(R1) &= V(n001) - V(n002) \\\
V(R1) &= 5 - 2.5 \\\
V(R1) &= \boxed{2.5}
V(R1) &= V(n001) - V(n002) \\\
V(R1) &= 5 - 2.5 \\\
V(R1) &= \boxed{2.5}
\end{align*}
$$
$$
\begin{align*}
V(R2) &= V(n002) - \text{GND} \\\
V(R2) &= 2.5 - 0 \\\
V(R2) &= \boxed{2.5}
V(R2) &= V(n002) - \text{GND} \\\
V(R2) &= 2.5 - 0 \\\
V(R2) &= \boxed{2.5}
\end{align*}
$$
Let's make a table to compare the results
|Iteams|Analysis|Simulation|Experiement|diff|$\%$diff|
|Items|Analysis|Simulation|Experiment|diff|$\%$diff|
|:-:|:-:|:-:|:-:|:-:|-:|
|$V(R1)$|$2.5V$|$2.5V$|$2.5539V$|$0.0539V$|$2.1\%$|
|$V(R2)$|$2.5V$|$2.5V$|$2.5204V$|$0.0204V$|$0.8\%$|
@@ -469,8 +469,8 @@ Also, we know that $R_1 = R_2 = 10K$ and the voltage across the resistor can be
$$
\begin{align*}
\frac{V_1}{V_2} &= \frac{R_1}{R_2} \\\
\frac{V_1}{V_2} &= \frac{10K}{10K} = \frac{1}{1}
\frac{V_1}{V_2} &= \frac{R_1}{R_2} \\\
\frac{V_1}{V_2} &= \frac{10K}{10K} = \frac{1}{1}
\end{align*}
$$
@@ -480,9 +480,9 @@ Using these values, we can find out $I(R1)$ and $I(R2)$ by
$$
\begin{align*}
I(R1) = I(R2) &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{2.5}{10K} \\\
I(R1) = I(R2) &= \boxed{0.00025}
I(R1) = I(R2) &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{2.5}{10K} \\\
I(R1) = I(R2) &= \boxed{0.00025}
\end{align*}
$$
@@ -495,11 +495,11 @@ We expect $I(R1)$ and $I(R2)$ to be $0.00025A$.
```text
--- Operating Point ---
V(n001): 5 voltage
V(n002): 2.5 voltage
I(R1): -0.00025 device_current
I(R2): -0.00025 device_current
I(V1): -0.00025 device_current
V(n001): 5 voltage
V(n002): 2.5 voltage
I(R1): -0.00025 device_current
I(R2): -0.00025 device_current
I(V1): -0.00025 device_current
```
### Measurement
@@ -519,8 +519,8 @@ $V(R2) = 2.5204V$
From the Simulation Result,
```text
I(R1): -0.00025 device_current
I(R2): -0.00025 device_current
I(R1): -0.00025 device_current
I(R2): -0.00025 device_current
```
It proves $I(R1)=I(R2)$, which
@@ -539,17 +539,17 @@ Based on the Ohm's Law - the relationship we got in Analysis $I = \frac{V}{R}$.
$$
\begin{align*}
I(R1) &= \frac{V}{R} \\\
I(R1) &= \frac{2.5539}{10K} \\\
I(R1) &= 0.00025539
I(R1) &= \frac{V}{R} \\\
I(R1) &= \frac{2.5539}{10K} \\\
I(R1) &= 0.00025539
\end{align*}
$$
$$
\begin{align*}
I(R2) &= \frac{V}{R} \\\
I(R2) &= \frac{2.5204}{10K} \\\
I(R2) &= 0.00025204
I(R2) &= \frac{V}{R} \\\
I(R2) &= \frac{2.5204}{10K} \\\
I(R2) &= 0.00025204
\end{align*}
$$
@@ -589,9 +589,9 @@ Also, the voltage is potential difference between the component.
$$
\begin{align*}
V(R1) = V(R2) &= n001 - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= \boxed{5}
V(R1) = V(R2) &= n001 - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
@@ -637,9 +637,9 @@ Let's put values into equation
$$
\begin{align*}
I_{total} &= \frac{5}{\cfrac{1}{\frac{1}{10K} + \frac{1}{10K}}} \\\
I_{total} &= \frac{5}{5K} \\\
I_{total} &= \boxed{0.001}
I_{total} &= \frac{5}{\cfrac{1}{\frac{1}{10K} + \frac{1}{10K}}} \\\
I_{total} &= \frac{5}{5K} \\\
I_{total} &= \boxed{0.001}
\end{align*}
$$
@@ -652,10 +652,10 @@ We can check this to double confirm our result.
```text
--- Operating Point ---
V(n001): 5 voltage
I(R2): 0.0005 device_current
I(R1): 0.0005 device_current
I(V1): -0.001 device_current
V(n001): 5 voltage
I(R2): 0.0005 device_current
I(R1): 0.0005 device_current
I(V1): -0.001 device_current
```
### Measurement
@@ -682,22 +682,22 @@ To find out the theoretic values of $V(R1)$ and $V(R2)$. We need to do some math
We know the simulation output is
```text
V(n001): 5 voltage
V(n001): 5 voltage
```
and the voltage is the difference in potential. Based on that, we can caculate the theoretic values of $V(R1)$ and $V(R2)$.
$$
\begin{align*}
V(R1) = V(R2) &= V(n001) - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= \boxed{5}
V(R1) = V(R2) &= V(n001) - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
Let's make a table to compare the results
|Iteams|Analysis|Simulation|Experiement|diff|$\%$diff|
|Items|Analysis|Simulation|Experiment|diff|$\%$diff|
|:-:|:-:|:-:|:-:|:-:|-:|
|$V(R1)$|$5V$|$5V$|$5.0305V$|$0.0305V$|$0.6\%$|
|$V(R2)$|$5V$|$5V$|$5.0305V$|$0.0305V$|$0.6\%$|
@@ -713,8 +713,8 @@ Secondly, we can check to $I_{total}$ as double confirm.
We know the simulation output is
```text
I(R2): 0.0005 device_current
I(R1): 0.0005 device_current
I(R2): 0.0005 device_current
I(R1): 0.0005 device_current
```
and from Analysis, we expect
@@ -733,10 +733,10 @@ So, we get
$$
\begin{align*}
I_{total} &= I(R2) + I(R1) \\\
I_{total} &= 0.0005 + 0.0005 \\\
I_{total} &= 0.001 \\\
& 0.001 = 0.001 \; \boxed{\text{True}}
I_{total} &= I(R2) + I(R1) \\\
I_{total} &= 0.0005 + 0.0005 \\\
I_{total} &= 0.001 \\\
& 0.001 = 0.001 \\; \boxed{\text{True}}
\end{align*}
$$
@@ -764,10 +764,10 @@ Then, we can find out $I(R1)$ and $I(R2)$ by
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\\
I(R1) = I(R2) &= \boxed{0.0005305}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\\
I(R1) = I(R2) &= \boxed{0.0005305}
\end{align*}
$$
@@ -808,9 +808,9 @@ Also, the voltage is potential difference between the component.
$$
\begin{align*}
V(R1) = V(R2) &= n001 - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= \boxed{5}
V(R1) = V(R2) &= n001 - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
@@ -830,17 +830,17 @@ and get
$$
\begin{align*}
I(R1) &= \frac{V(R1)}{R1} \\\
I(R1) &= \frac{5}{10K} \\\
I(R1) &= \boxed{0.0005}
I(R1) &= \frac{V(R1)}{R1} \\\
I(R1) &= \frac{5}{10K} \\\
I(R1) &= \boxed{0.0005}
\end{align*}
$$
$$
\begin{align*}
I(R2) &= \frac{V(R2)}{R2} \\\
I(R2) &= \frac{5}{10K} \\\
I(R2) &= \boxed{0.0005}
I(R2) &= \frac{V(R2)}{R2} \\\
I(R2) &= \frac{5}{10K} \\\
I(R2) &= \boxed{0.0005}
\end{align*}
$$
@@ -848,11 +848,11 @@ the relationship between $I(R1)$ and $I(R2)$ can be express as
$$
\begin{align*}
\frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{V(R1)}{R1}}{\cfrac{V(R2)}{R2}} \\\
\because V(R1) &= V(R2) \\\
\therefore \frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{\cancel{V(R1)}}{R1} \times \cfrac{1}{\cancel{V(R1)}}}{\cfrac{\cancel{V(R2)}}{R2} \times \cfrac{1}{\cancel{V(R2)}}} \\\
\frac{I(R1)}{I(R2)} &= \frac{\frac{1}{R1}}{\frac{1}{R2}} \\\
&\boxed{\frac{I(R1)}{I(R2)} = \frac{R2}{R1}}
\frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{V(R1)}{R1}}{\cfrac{V(R2)}{R2}} \\\
\because V(R1) &= V(R2) \\\
\therefore \frac{I(R1)}{I(R2)} &= \cfrac{\cfrac{\cancel{V(R1)}}{R1} \times \cfrac{1}{\cancel{V(R1)}}}{\cfrac{\cancel{V(R2)}}{R2} \times \cfrac{1}{\cancel{V(R2)}}} \\\
\frac{I(R1)}{I(R2)} &= \frac{\frac{1}{R1}}{\frac{1}{R2}} \\\
&\boxed{\frac{I(R1)}{I(R2)} = \frac{R2}{R1}}
\end{align*}
$$
@@ -866,10 +866,10 @@ as we get the $I_{total}$ by
$$
\begin{align*}
I_{total} &= \frac{V_{total}}{R_{total}} \\\
I_{total} &= \frac{5}{\cfrac{1}{\cfrac{1}{10K} + \cfrac{1}{10K}}} \\\
I_{total} &= \frac{5}{5K} \\\
I_{total} &= 0.001
I_{total} &= \frac{V_{total}}{R_{total}} \\\
I_{total} &= \frac{5}{\cfrac{1}{\cfrac{1}{10K} + \cfrac{1}{10K}}} \\\
I_{total} &= \frac{5}{5K} \\\
I_{total} &= 0.001
\end{align*}
$$
@@ -883,9 +883,9 @@ We can get $I(R1)$ and $I(R2)$ by
$$
\begin{align*}
I(R1) = I(R2) &= I_{total} \times \frac {R1}{R1 + R2} \\\
I(R1) = I(R2) &= 0.001 \times \frac {10K}{10K + 10K} \\\
I(R1) = I(R2) &= \boxed{0.0005}
I(R1) = I(R2) &= I_{total} \times \frac {R1}{R1 + R2} \\\
I(R1) = I(R2) &= 0.001 \times \frac {10K}{10K + 10K} \\\
I(R1) = I(R2) &= \boxed{0.0005}
\end{align*}
$$
@@ -893,12 +893,12 @@ At this point, our logic is consistent, which
$$
\begin{align*}
& \because R1 = R2 = 10K \\\
& \because I(R1) = I(R2) = 0.0005 \\\
& \because V(R1) = V(R2) = 5 \\\
& \because I_{total} = I(R1) + I(R2) = 0.001 \\\
& \because \frac{I(R1)}{I(R2)} = \frac{R2}{R1} \\\
& \therefore I(R1) = I(R2) = I_{total} \times \frac {R1}{R1 + R2} \\\
& \because R1 = R2 = 10K \\\
& \because I(R1) = I(R2) = 0.0005 \\\
& \because V(R1) = V(R2) = 5 \\\
& \because I_{total} = I(R1) + I(R2) = 0.001 \\\
& \because \frac{I(R1)}{I(R2)} = \frac{R2}{R1} \\\
& \therefore I(R1) = I(R2) = I_{total} \times \frac {R1}{R1 + R2} \\\
\end{align*}
$$
@@ -921,10 +921,10 @@ $$
```text
--- Operating Point ---
V(n001): 5 voltage
I(R2): 0.0005 device_current
I(R1): 0.0005 device_current
I(V1): -0.001 device_current
V(n001): 5 voltage
I(R2): 0.0005 device_current
I(R1): 0.0005 device_current
I(V1): -0.001 device_current
```
### Measurement
@@ -951,16 +951,16 @@ To find out the theoretic values of $V(R1)$ and $V(R2)$. We need to do some math
We know the simulation output is
```text
V(n001): 5 voltage
V(n001): 5 voltage
```
and the voltage is the difference in potential. Based on that, we can caculate the theoretic values of $V(R1)$ and $V(R2)$.
and the voltage is the difference in potential. Based on that, we can calculate the theoretic values of $V(R1)$ and $V(R2)$.
$$
\begin{align*}
V(R1) = V(R2) &= V(n001) - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= \boxed{5}
V(R1) = V(R2) &= V(n001) - \text{GND} \\\
V(R1) = V(R2) &= 5 - 0 \\\
V(R1) = V(R2) &= \boxed{5}
\end{align*}
$$
@@ -999,14 +999,14 @@ Then, we can find out $I(R1)$ and $I(R2)$ by
$$
\begin{align*}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\\
I(R1) = I(R2) &= \boxed{0.0005305}
V &= IR \\\
I &= \frac{V}{R} \\\
I(R1) = I(R2) &= \frac{5.0305}{10000} \\\
I(R1) = I(R2) &= \boxed{0.0005305}
\end{align*}
$$
|Iteams|Analysis|Simulation|Experiement|diff|$\%$diff|
|Items|Analysis|Simulation|Experiment|diff|$\%$diff|
|:-:|:-:|:-:|:-:|:-:|-:|
|$I(R1)$|$0.5mA$|$0.5mA$|$0.50305mA$|$0.00305mA$|$0.6\%$|
|$I(R2)$|$0.5mA$|$0.5mA$|$0.50305mA$|$0.00305mA$|$0.6\%$|
@@ -1146,12 +1146,14 @@ But even we look the thermometer's reading, both of them shows around $24 \degre
The wheatstone bridge is better than a normal voltage divider because it is more sensitive than a voltage divider. A voltage divider relies on the ratio of resistances between two resistors, so even if the ratio of the resistors change, as long as the change isn't massive, the output voltage stays the same.
When a wheatstone bridge is balanced, meaning R1/R2=R3/R4, the current flowing through the galvanometer in the center of the wheatstone bridge is 0. When current is zero, the calculated resistance is no longer affected by innate resistance of wires, resistors, and voltameters. This allows measurements with the wheatstone bridge to be more accurate than a voltage divider.
When a wheatstone bridge is balanced, meaning R1/R2=R3/R4, the current flowing through the galvanometer in the center of the wheatstone bridge is 0. When current is zero, the calculated resistance is no longer affected by innate resistance of wires, resistors, and voltameters. This allows measurements with the wheatstone bridge to be more accurate than a voltage divider.
Advantages:
- Wheatstone bridge is more accurate than voltage divider
- Voltage source does not need to be calibrated to measure resistance
Disadvantages:
- Voltage divider is easier and cheaper to make
Disadvantages:
- Voltage divider is easier and cheaper to make
- Lower power consumption

6
content/zh-cn/posts/csci-1100/hw-7/index.md
View File

@@ -192,9 +192,9 @@ print(ratings['3520029'])
在这个家庭作业中，假设你有两个名为 `movies.json` 和 `ratings.json` 的文件。请从这些文件中读取数据，并让用户输入一个年份范围（最小和最大年份）以及两个权重：`w1` 和 `w2`。找到在 min 到 max 年之间制作的所有电影（包含 min 和 max）。对于每部电影，计算其综合评分为：
$$
(w1 \times imdb_rating + w2 \times average_twitter_rating) / (w1 + w2)
$$
```python
(w1 * imdb_rating + w2 * average_twitter_rating) / (w1 + w2)
```
其中 `imdb_rating` 来自于 movies 文件，而 `average_twitter_rating` 是 ratings 文件中的平均评分。

10
content/zh-cn/posts/ecse-1010/lab01/index.md
View File

@@ -238,7 +238,8 @@ I(V1): -0.000243902 device_current
$$
\begin{align*}
I(R1) + I(R2) + I(R3) &= -0.000243902 + 0.000121951 + 0.000121951 \\\
& \quad \thickspace I(R1) + I(R2) + I(R3) \\\
&= -0.000243902 + 0.000121951 + 0.000121951 \\\
&= \boxed{0}
\end{align*}
$$
@@ -263,7 +264,8 @@ $$
$$
\begin{align*}
V(n001) - V(n002) - V(n003) &= 2.43902 - 0.12196 - 0.12196 \\\
& \quad \thickspace V(n001) - V(n002) - V(n003) \\\
&= 2.43902 - 0.12196 - 0.12196 \\\
&= 0 \\\
\boxed{\text{True}}
\end{align*}
@@ -730,7 +732,7 @@ $$
I_{total} &= I(R2) + I(R1) \\\
I_{total} &= 0.0005 + 0.0005 \\\
I_{total} &= 0.001 \\\
& 0.001 = 0.001 \; \boxed{\text{True}}
& 0.001 = 0.001 \\; \boxed{\text{True}}
\end{align*}
$$
@@ -1137,9 +1139,11 @@ $V = 4.441V, T = 30.25 \degree C$
惠斯通电桥比普通分压器更灵敏，因为它的输出电压受电阻比例变化的影响较小。当惠斯通电桥平衡时（即 $R1/R2 = R3/R4$），中心的电流表中的电流为零。此时，计算出的阻值不再受到导线、电阻和伏特计固有电阻的影响，从而使得测量结果更准确。
优点：
- 惠斯通电桥比分压器更准确
- 电压源不需要校准即可测量阻值
缺点：
- 分压器更容易且成本更低制作
- 功耗较低