feat: finished ac started capacitors
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Areas/electricity/ac.md.md
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Areas/electricity/ac.md.md
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# AC (Alternating Current)
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A few important characteristics of an AC Signal:
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## Time Period
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How much time passes during one cycle of the signal (until it reaches the same point it started twice)
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## Frequency
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How many cycles the signal completes in a specified time frame.
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## Amplitute
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The maximum voltage of the signal compared to 0
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## Peak-Peak Voltage
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The difference between the highest and lowest peak.
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## Root Mean Square Values
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This helps calculating the current an equivalent DC Signal would need to provide the same amount of power.
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$$
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\begin{flalign}
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V_{RMS} = 0.7 * V_{PEAK} &&\\
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V_{PEAK} = 1.4 * V_{RMS}
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\end{flalign}
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$$
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|
||||
id="tspan3637">Conductive plates</tspan></text>
|
||||
<text
|
||||
xml:space="preserve"
|
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|
||||
<path
|
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||||
x="17"
|
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y="62">A</tspan></text>
|
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<path
|
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id="path2800-2" />
|
||||
</svg>
|
||||
|
After Width: | Height: | Size: 13 KiB |
BIN
Areas/electricity/assets/ceramic-capacitor.webp
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Areas/electricity/assets/ceramic-capacitor.webp
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|
After Width: | Height: | Size: 5.4 KiB |
1
Areas/electricity/assets/kirchhoffs-law-1.svg
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1
Areas/electricity/assets/kirchhoffs-law-1.svg
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|
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BIN
Areas/electricity/assets/rms.gif
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Areas/electricity/assets/rms.gif
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|
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BIN
Areas/electricity/assets/tantalum-capacitor.jpg
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BIN
Areas/electricity/assets/tantalum-capacitor.jpg
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|
After Width: | Height: | Size: 83 KiB |
BIN
Areas/electricity/assets/wave.gif
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Areas/electricity/assets/wave.gif
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|
After Width: | Height: | Size: 5.2 KiB |
@ -32,27 +32,26 @@ $$
|
||||
|
||||
$$
|
||||
\begin{flalign}
|
||||
&\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} ... &\\
|
||||
|
||||
\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} ... &&\\
|
||||
\\
|
||||
&\textit{For two resistors in parallel:} &\\
|
||||
\textit{For two resistors in parallel:} &&\\
|
||||
\\
|
||||
&R = \frac{R1 * R2}{R1 + R2}
|
||||
R = \frac{R1 * R2}{R1 + R2} &&\\
|
||||
\end{flalign}
|
||||
$$
|
||||
|
||||
*Tip:*
|
||||
If two resistors of the same value are in parallel the total resistance is half the value of a single resistor
|
||||
***Tip:***
|
||||
If resistors of the same value are in parallel the total resistance is a single resistor divided by the amount if resistors.
|
||||
|
||||
# Kirchhoff's Law
|
||||
|
||||
### Conservation of Charge
|
||||
## Conservation of Charge (First Law)
|
||||
|
||||
All current entering a node in a circuit must also leave that node
|
||||
All current entering a node must also leave that node
|
||||
|
||||
$$
|
||||
\begin{flalign}
|
||||
& \sum{I_{IN}} = \sum{I_{OUT}} &
|
||||
\sum{I_{IN}} = \sum{I_{OUT}}&&
|
||||
\end{flalign}
|
||||
$$
|
||||
|
||||
@ -63,54 +62,15 @@ $$
|
||||
For this circuit kirchhoffs law states that:
|
||||
$$
|
||||
\begin{flalign}
|
||||
&i1 = i2 + i3 + i4 &
|
||||
i1 = i2 + i3 + i4 &&
|
||||
\end{flalign}
|
||||
$$
|
||||
|
||||
### Conservation of Energy
|
||||
## Conservation of Energy (Second Law)
|
||||
All the potential differences around the loop must sum to zero.
|
||||
$$
|
||||
\begin{flalign}
|
||||
& \sum{V} = 0 &
|
||||
\sum{V} = 0 &&
|
||||
\end{flalign}
|
||||
$$
|
||||
|
||||
**Example:**
|
||||
|
||||
|
||||
|
||||
For this circuit this means.
|
||||
|
||||
$$
|
||||
\begin{flalign}
|
||||
V_{R1} = -(I*R1) && \text{Voltage Drop across R}_{1} \\
|
||||
V_{R2} = -(I*R2) && \text{Voltage Drop across R}_{2}\\
|
||||
\\
|
||||
|
||||
V_{S} - IR_{1} - IR_{2} &= 0 & | + IR_{1} + IR_{2}\\
|
||||
V_{S} &= IR_{1} + IR_{2} \\
|
||||
\\
|
||||
V_{S} &= I(R_{1}+R_{2}) \\
|
||||
V_{S} &= I*R_{t} \\
|
||||
|
||||
\end{flalign}
|
||||
$$
|
||||
|
||||
We can change that formula around to find out the current of the entire circuit, aswell as current through single resistors
|
||||
|
||||
$$
|
||||
\begin{flalign}
|
||||
& V_{S} = I*R_{t} & | \div R_{t} \\
|
||||
\\
|
||||
& I = \frac{V_{S}}{R_{t}} = \frac{V_{S}}{R_{1}+R_{2}} \\
|
||||
\\
|
||||
& VR_{1} = V_{S}(\frac{R1}{R1+R2}) \\
|
||||
& VR_{2} = V_{S}(\frac{R2}{R1+R2}) \\
|
||||
\\
|
||||
& VR_{1} = 5(\frac{500}{500+1500}) \\
|
||||
& VR_{1} = 1.25v
|
||||
\\
|
||||
& VR_{2} = 5(\frac{1500}{500+1500}) \\
|
||||
& VR_{2} = 3.75v
|
||||
|
||||
\end{flalign}
|
||||
$$
|
||||
135
Areas/electricity/formulas/kirchhoffs-law.md.md
Normal file
135
Areas/electricity/formulas/kirchhoffs-law.md.md
Normal file
@ -0,0 +1,135 @@
|
||||
# Kirchhoffs Law
|
||||
|
||||
### Example 1
|
||||
**Example:**
|
||||
|
||||
|
||||
|
||||
For this circuit this means.
|
||||
|
||||
$$
|
||||
\begin{flalign}
|
||||
V_{R1} = -(I*R1) && \text{Voltage Drop across R}_{1} \\
|
||||
V_{R2} = -(I*R2) && \text{Voltage Drop across R}_{2}\\
|
||||
\\
|
||||
|
||||
V_{S} - IR_{1} - IR_{2} &= 0 & | + IR_{1} + IR_{2}\\
|
||||
V_{S} &= IR_{1} + IR_{2} \\
|
||||
\\
|
||||
V_{S} &= I(R_{1}+R_{2}) \\
|
||||
V_{S} &= I*R_{t} \\
|
||||
|
||||
\end{flalign}
|
||||
$$
|
||||
|
||||
We can change that formula around to find out the current of the entire circuit, aswell as current through single resistors
|
||||
|
||||
$$
|
||||
\begin{flalign}
|
||||
I &= I*R_{t} = \frac{V_{S}}{R_{t}} = \frac{V_{S}}{R_{1}+R_{2}}\\
|
||||
\\
|
||||
VR_{1} &= V_{S}(\frac{R1}{R1+R2})\\
|
||||
VR_{2} &= V_{S}(\frac{R2}{R1+R2})\\
|
||||
\\
|
||||
VR_{1} &= 5(\frac{500}{500+1500})\\
|
||||
VR_{1} &= 1.25v&\\
|
||||
\\
|
||||
VR_{2} &= 5(\frac{1500}{500+1500})\\
|
||||
VR_{2} &= 3.75v
|
||||
|
||||
\end{flalign}
|
||||
$$
|
||||
|
||||
### Example 2
|
||||
|
||||
|
||||
```circuitjs
|
||||
$ 1 0.000005 10.20027730826997 50 5 43 5e-11
|
||||
v 192 288 192 112 0 0 40 10 0 0 1
|
||||
x 125 208 167 211 4 24 10V
|
||||
r 192 112 336 112 0 10
|
||||
r 336 112 480 112 0 20
|
||||
w 480 112 480 288 0
|
||||
r 336 112 336 288 0 50
|
||||
w 480 288 336 288 0
|
||||
w 336 288 192 288 0
|
||||
x 304 204 319 207 4 12 R3
|
||||
x 256 140 271 143 4 12 R1
|
||||
x 400 138 415 141 4 12 R2
|
||||
x 332 104 340 107 4 12 A
|
||||
x 332 304 340 307 4 12 B
|
||||
x 205 102 215 105 4 12 I1
|
||||
x 467 105 477 108 4 12 I2
|
||||
x 346 276 356 279 4 12 I3
|
||||
```
|
||||
|
||||
In this circuit we have three major loops we can apply [[formulas#Conservation of Energy Second Law|Kirchhoffs Second Law]] to, the one on the left, the one on the right and the most outer one. We can also use [[formulas#Conservation of Charge First Law | Kirchhoffs First Law]] for the node title **A**.
|
||||
|
||||
$$
|
||||
\begin{flalign}
|
||||
&\textit{Node A:} \\\
|
||||
& I_{1} = I_{2}+I_{3} \\
|
||||
\\
|
||||
&\textit{Left Loop:} \circlearrowright &\\
|
||||
& 10_{v} - I_{1}*R_{1} - I_{3}*R_{3} = 0& \\
|
||||
\\
|
||||
&\textit{Right Loop} \circlearrowleft &\\
|
||||
&I_{2}*R_{2} - I_{3}*R_{3} = 0; &\\
|
||||
\\
|
||||
&\textit{Outer Loop:} \circlearrowright &\\
|
||||
& 10_{v} - I_{1}*R_{1} - I_{2}*R_{2} = 0& \\
|
||||
\\
|
||||
\end{flalign}
|
||||
$$
|
||||
|
||||
Now if we would like to find out I2 for example we can use the Right Loop Formula to do so:
|
||||
|
||||
$$
|
||||
\begin{flalign}
|
||||
& \textit{Right Loop} \circlearrowleft &\\
|
||||
& I_{2}*R_{2} - I_{3}*R_{3} = 0 &&| + I_{3}*R_{3}\\\
|
||||
\\
|
||||
& I_{2}*R_{2} = I_{3}*R_{3} &&| \div R_{2} \\
|
||||
\\
|
||||
& I_{2} = \frac{I_{3}*R_{3}}{R_{2}} &&| \text{ Input Numbers } \\
|
||||
\\
|
||||
& I_{2} = \frac{I_{3}*5}{2} &&| \text{ Simplify} \\
|
||||
\\
|
||||
& I_{2} = \frac{5I_{3}}{2} &&| \text{ Simplify} \\
|
||||
\\
|
||||
&\textit{LeftLoop:} \\
|
||||
& 10 - I_{1}*R_{1} - I_{3}*R_{3} = 0&&| \text{ Replace I1 with node a} \\
|
||||
& 10 - (I_{2}+I_{3})*R_{1} - I_{3}*R_{3} = 0 &&| \text{ Replace I2 with previous} \\
|
||||
\\
|
||||
& 10 - (\frac{5I_{3}}{2}+I_{3})*R_{1} - I_{2}*R_{2} = 0
|
||||
\end{flalign}
|
||||
$$
|
||||
|
||||
On the last formula we only have one unknown variable, which is $I_{3}$ so lets solve for that.
|
||||
|
||||
$$
|
||||
\begin{flalign}
|
||||
& 10 - (\frac{5I_{3}}{2}+I_{3})R_{1} - I_{3}R_{3} = 0 &&| \text{ Replace Variables}\\\
|
||||
\\
|
||||
& 10 - (2.5I_{3}+I_{3})10 - 50I_{3} = 0 && \\
|
||||
\\
|
||||
& 10 - 35I_{3} - 50I_{3} = 0 &&\\
|
||||
& 10 - 85I_{3} = 0 && | +85I_{3}\\
|
||||
& 10 = 85I_{3} && | \div 85 \\
|
||||
& 0.117647059 \approx I_{3} \\
|
||||
& 117.64mA = I_{3}
|
||||
\end{flalign}
|
||||
$$
|
||||
|
||||
Now lets found out $I_{2}$ aswell
|
||||
|
||||
$$
|
||||
\begin{flalign}
|
||||
I_{2} = \frac{I_{3}*5}{2} &&\\
|
||||
I_{2} = \frac{0.11764 * 5}{2} &&\\
|
||||
\\
|
||||
I_{2} \approx 0.2941 && \\
|
||||
\\
|
||||
I_2 \approx 294mA && \\
|
||||
\end{flalign}
|
||||
$$
|
||||
@ -9,6 +9,31 @@ Current
|
||||
## Ohms
|
||||
Resistance
|
||||
|
||||
## Watt (Power)
|
||||
|
||||
$Power = V * I = \frac{V^{2}}{R} = I^{2}R$
|
||||
|
||||
Joules per Second
|
||||
|
||||
Term | Symbol | Weight
|
||||
-----------|----|------
|
||||
Nanowatt | nW | 10-9
|
||||
Microwatt | $\micro$W | $10^{-6}$
|
||||
Milliwatt | mW | $10^{-3}$
|
||||
Watt | W | $10^{0}$
|
||||
Kilowatt | kW | $10^{3}$
|
||||
Megawatt | MW | $10^{6}$
|
||||
Gigawatt | GW | $10^{9}$
|
||||
|
||||
**Examples:**
|
||||
|
||||
Device | Power
|
||||
-------|----------
|
||||
Arduino| 167mW
|
||||
Laptop | 1.5W
|
||||
House | 2.2kW
|
||||
|
||||
|
||||
## Ohms Law
|
||||
$$
|
||||
V = \frac{I}{R}
|
||||
@ -25,6 +50,16 @@ Means if a component is symmetric or not
|
||||
Polarised means that a component is not symmetric
|
||||
|
||||
## Voltage Divider
|
||||
|
||||
## Farad
|
||||
Term | Symbol | Weight
|
||||
-----------|----|------
|
||||
Picofarad | pW | $10^{-12}$
|
||||
Nanofarad | nF | $10^{-9}$
|
||||
Microfarad | $\micro$F | $10^{-6}$
|
||||
Milifarad | mF | $10^{-3}$
|
||||
Kilofarad | kF | $10^{3}$
|
||||
|
||||
## LED
|
||||
|
||||
Anode - The shorter Leg
|
||||
|
||||
65
Areas/electricity/parts/capacitors.md.md
Normal file
65
Areas/electricity/parts/capacitors.md.md
Normal file
@ -0,0 +1,65 @@
|
||||
# Capacitors
|
||||
|
||||
Capacity is measured in [[glossary#Farad|Farads]].
|
||||
|
||||
Capacity is calculated as follows:
|
||||
|
||||
$$
|
||||
\begin{flalign}
|
||||
& C = \epsilon r \frac{A}{4\pi d} &&\\\
|
||||
\\
|
||||
& \epsilon r = \text{Dielectrics relative permittivity} &&\\
|
||||
& A = \text{Amount of Area the plates overlap} &&\\
|
||||
& d = \text{Distance between plates} &&\\
|
||||
\end{flalign}
|
||||
$$
|
||||
|
||||
|
||||
|
||||
|
||||
### Important Metrics
|
||||
|
||||
**Size:**
|
||||
Larger Capacity $\approx$ Larger Size
|
||||
|
||||
**Maximum Voltage**
|
||||
Each capacitor has a maximum voltage that can be dropped across it.
|
||||
|
||||
**Leakage Current**
|
||||
Capacitors are not perfect, and leak some current across the terminals.
|
||||
|
||||
**Equivalent series Resistance (ESR)**
|
||||
The terminals are not 100% conductive, so the will have some very small resistance, (usually less than $0.01\ohm$)
|
||||
|
||||
**Tolerance**
|
||||
The capacity is not always exact, the tolerance describes how much it could vary, usually about $\mp 1\%$ to $\mp 20\%$
|
||||
|
||||
## Ceramic Capacitors
|
||||
- least expansive
|
||||
- relative small usually $< 10\micro F$
|
||||
- low current leakage and ESR
|
||||
- best for high frequency coupling
|
||||
|
||||
|
||||
|
||||
|
||||
## Aluminium and Tantalum Electrolytic
|
||||
- Usually polarized
|
||||
- Capacity usuially $1\micro F - 1mF$
|
||||
- Good for high voltage
|
||||
|
||||
|
||||
|
||||
## Super Capacitors
|
||||
|
||||
- Usually can handle only low voltage
|
||||
- Capacity in the range of farads
|
||||
|
||||
## Film Capacitor
|
||||
- usually low ESR
|
||||
|
||||
## Mica Capacitor
|
||||
- Can work in hot environments > $200\deg$
|
||||
- Low ESR
|
||||
- High Precision
|
||||
- High Cost
|
||||
@ -30,7 +30,7 @@ Our power source must have more voltage than the voltage drop, otherwise we cant
|
||||
|
||||
First we need to find the datasheet of the specific component, it can be easily found by googling it
|
||||
|
||||
|
||||
|
||||
|
||||
Now lets put that LED into a test circuit and calculate the resistance for it:
|
||||
|
||||
3
Areas/electricity/parts/resistors.md.md
Normal file
3
Areas/electricity/parts/resistors.md.md
Normal file
@ -0,0 +1,3 @@
|
||||
# Resistors
|
||||
|
||||
Resistors transform voltage into heat.
|
||||
Loading…
Reference in New Issue
Block a user