feat: add some kirchhoffs law

Areas/electricity/assets/led-example.svg

@@ -0,0 +1 @@
+<svg version="1.1" xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" width="480" height="228"><defs><linearGradient id="XfyXafdTXvsW" x1="0px" x2="32px" y1="0px" y2="0px" gradientUnits="userSpaceOnUse"><stop offset="0" stop-color="#807f7f"/><stop offset="1" stop-color="#807f7f"/></linearGradient></defs><g><rect fill="#000000" stroke="none" x="0" y="0" width="480" height="228"/><g transform="scale(1,1) translate(-22,-78)"><path fill="none" stroke="#807f7f" paint-order="fill stroke markers" d=" M 144 256 L 144 196" stroke-linecap="round" stroke-miterlimit="10" stroke-width="3" stroke-dasharray=""/><path fill="none" stroke="#01fd01" paint-order="fill stroke markers" d=" M 144 188 L 144 128" stroke-linecap="round" stroke-miterlimit="10" stroke-width="3" stroke-dasharray=""/><path fill="none" stroke="#807f7f" paint-order="fill stroke markers" d=" M 134 196 L 154 196" stroke-linecap="round" stroke-miterlimit="10" stroke-width="3" stroke-dasharray=""/><path fill="none" stroke="#01fd01" paint-order="fill stroke markers" d=" M 128 188 L 160 188" stroke-linecap="round" stroke-miterlimit="10" stroke-width="3" stroke-dasharray=""/><path fill="none" stroke="#807f7f" paint-order="fill stroke markers" d=" M 304 256 L 144 256" stroke-linecap="round" stroke-miterlimit="10" stroke-width="3" stroke-dasharray=""/><text fill="#ffffff" stroke="none" font-family="sans-serif" font-size="12px" font-style="normal" font-weight="normal" text-decoration="normal" x="224" y="250" text-anchor="start" dominant-baseline="alphabetic"></text><path fill="none" stroke="#01fd01" paint-order="fill stroke markers" d=" M 304 128 L 304 180" stroke-linecap="round" stroke-miterlimit="10" stroke-width="3" stroke-dasharray=""/><path fill="none" stroke="#807f7f" paint-order="fill stroke markers" d=" M 304 204 L 304 256" stroke-linecap="round" stroke-miterlimit="10" stroke-width="3" stroke-dasharray=""/><path fill="none" stroke="#808080" paint-order="fill stroke markers" d=" M 315.76 192 A 11.76 11.76 0 1 1 315.7599941200005 191.98824000195998" stroke-linecap="round" stroke-miterlimit="10" stroke-width="3" stroke-dasharray=""/><path fill="#ff0000" stroke="none" paint-order="stroke fill markers" d=" M 312 192 A 8 8 0 1 1 311.99999999988734 191.9999575425633 Z"/><g><text fill="#c0c0c0" stroke="none" font-family="sans-serif" font-size="24px" font-style="normal" font-weight="normal" text-decoration="normal" x="92" y="202" text-anchor="start" dominant-baseline="alphabetic">9V</text></g><path fill="none" stroke="#807f7f" paint-order="fill stroke markers" d=" M 144 128 L 208 128" stroke-linecap="round" stroke-miterlimit="10" stroke-width="3" stroke-dasharray=""/><path fill="none" stroke="#807f7f" paint-order="fill stroke markers" d=" M 240 128 L 304 128" stroke-linecap="round" stroke-miterlimit="10" stroke-width="3" stroke-dasharray=""/><g transform="matrix(1,0,0,1,208,128)"><path fill="none" stroke="url(#XfyXafdTXvsW)" paint-order="fill stroke markers" d=" M 0 0 L 2 6 L 6 -6 L 10 6 L 14 -6 L 18 6 L 22 -6 L 26 6 L 30 -6 L 32 0" stroke-linecap="round" stroke-miterlimit="10" stroke-width="3" stroke-dasharray=""/></g><g><text fill="#c0c0c0" stroke="none" font-family="sans-serif" font-size="24px" font-style="normal" font-weight="normal" text-decoration="normal" x="321" y="179" text-anchor="start" dominant-baseline="alphabetic">Vf = 2V</text></g><g><text fill="#c0c0c0" stroke="none" font-family="sans-serif" font-size="24px" font-style="normal" font-weight="normal" text-decoration="normal" x="329" y="208" text-anchor="start" dominant-baseline="alphabetic">If = 20mA</text></g></g></g></svg>

Areas/electricity/formulas.md

@@ -0,0 +1,116 @@
+# Ohms Law
+Solve for voltage:
+
+$$
+\begin{flalign}
+V &= \frac{I}{R}&
+\end{flalign}
+$$
+*Solve for resistance:*
+
+$$
+\begin{flalign}
+R &= \frac{V}{I} &
+\end{flalign}
+$$
+_Solve for current_
+$$
+\begin{flalign}
+I & = \frac{V}{R} &
+\end{flalign}
+$$
+
+# Resistors in Series
+
+$$
+\begin{flalign}
+R &= R1 + R2 + R3 ... &
+\end{flalign}
+$$
+
+# Resistors in Parallel
+
+$$
+\begin{flalign}
+&\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} ... &\\
+
+\\
+&\textit{For two resistors in parallel:} &\\
+\\
+&R = \frac{R1 * R2}{R1 + R2}
+\end{flalign}
+$$
+
+*Tip:*
+If two resistors of the same value are in parallel the total resistance is half the value of a single resistor
+
+# Kirchhoff's Law
+
+### Conservation of Charge
+
+All current entering a node in a circuit must also leave that node
+
+$$
+\begin{flalign}
+& \sum{I_{IN}} = \sum{I_{OUT}} &
+\end{flalign}
+$$
+
+**Example:**
+
+![](./assets/kirchhoffs-law-01.svg)
+
+For this circuit kirchhoffs law states that:
+$$
+\begin{flalign}
+&i1 = i2 + i3 + i4 &
+\end{flalign}
+$$
+
+### Conservation of Energy
+$$
+\begin{flalign}
+& \sum{V} = 0 &
+\end{flalign}
+$$
+
+**Example:**
+
+![](./assets/kirchhoffs-law-02.svg)
+
+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
+
+$$
+\begin{flalign}
+& V_{S} = I*R_{t} & | \div R_{t} \\
+\\
+& I = \frac{V_{S}}{R_{t}} = \frac{V_{S}}{R_{1}+R_{2}} \\
+\\
+& VR_{1} = V_{S}(\frac{R1}{R1+R2}) \\
+& VR_{2} = V_{S}(\frac{R2}{R1+R2}) \\
+\\
+& VR_{1} = 5(\frac{500}{500+1500}) \\
+& VR_{1} = 1.25v
+\\
+& VR_{2} = 5(\frac{1500}{500+1500}) \\
+& VR_{2} = 3.75v
+
+\end{flalign}
+$$

Areas/electricity/glossary.md

@@ -15,6 +15,11 @@ V = \frac{I}{R}
 $$
 
 ## Impedance
+= Resistance for Nerds
+
+## Current
+How many electrons flow through a circuit in  a second
+
 ## Polarity
 Means if a component is symmetric or not
 Polarised means that a component is not symmetric
@@ -28,5 +33,9 @@ Cathode -  The longer Leg
 ## Diode
 ## Anode
 The positive end of a diode
+
 ## Cathode
-The negative end of a diode
+The negative end of a diode
+
+## Conventional Current Flow
+When electricity was discovered people thought the electrons flow from the positive terminal to the negative, in actuality they flow in the opposite direction, but it is still possible to calculate the flow with the old way.

Areas/electricity/index.md

@@ -1,6 +1,8 @@
 # Learning Analog Hardware
 
+[Glossary](./glossary.md)
+[Formulas](./formulas.md)
 
-## Volts
-## Watts
-## Amps
+## Building Blocks
+- [LED](./led)
+- [Voltage Divider](./voltage-divider)

Areas/electricity/led.md.md

@@ -4,8 +4,50 @@ Leds have polarity, that means they only allow current to flow in one direction.
 
 While in resistors the relations of voltage to current is linear:
 
-![](./current-voltage-characteristic.svg)
+![](current-voltage-characteristic.svg)
 
-that is not the case for leds
+that is not the case for leds:
 
-![](./diode-voltage-graph.svg)
+
+![](diode-voltage-graph.svg)
+
+Each led has a specific Vd or voltage drop, it needs that amount of voltage to turn on, after that amount is reached the current will increase exponentially.
+
+Because of this we will always need a resistor infront of the LED, because when the current is too high the LED will burn out.
+
+Values that are important for the led:
+$$
+\begin{flalign}
+&Vd = \text{Voltage Drop} \\
+&Vf = \text{Forward Voltage}\\
+&If = \text{Forward Current}
+\end{flalign}
+$$
+
+Our power source must have more voltage than the voltage drop, otherwise we cant turn on the led.
+
+## Example (TLUR6400)
+
+First we need to find the datasheet of the specific component, it can be easily found by googling it
+
+![TLUR DataSheet](./datasheets/tlur6400.pdf)
+
+Now lets put that LED into a test circuit and calculate the resistance for it:
+
+![](led-example.svg)
+
+Because we now the Voltage Drop of the entire circuit must be 9v and the LED already drops 2V, we now know that the resistor must drop the remaining 7v.
+
+$$
+\begin{align}
+R &= \frac{V}{I} &\text{Ohms Law} \\
+R &= \frac{7}{0.02} &\text{Replace the values} \\
+R &= 350 \text{Ω}
+\end{align}
+$$
+
+If we would pack everything into one formulare it would look like this
+
+$$
+R = \frac{Vs - Vf}{i}
+$$

Areas/electricity/voltage-dividers.md

@@ -4,7 +4,7 @@
 
 This is sort of unrealistic because there is no current flowing out of the voltage divider on the right side:
 
-![Voltage Divider](./voltage-divider.svg)
+![Voltage Divider](voltage-divider.svg)
 
 #### Equation
 
@@ -18,7 +18,7 @@ $$
 
 When the output of the voltage divider is connected to something the current drops on the output, as that something uses some of it.
 
-![Voltage Divider](./voltage-divider-load.svg)
+![Voltage Divider](voltage-divider-load.svg)
 
 The load is connected in parallel to R2, so we can calculate it as a parallel resistor.
 
@@ -34,7 +34,7 @@ $$
 
 Lets calculate the current in this circuit:
 
-![Voltage Divider Load Example](./voltage-divider-load-example.svg)
+![Voltage Divider Load Example](voltage-divider-load-example.svg)
 
 1. We calculate the Resistance in the subcircuit (R2 and RL) as they are connected in parallel which means
 
@@ -50,7 +50,7 @@ $$
 
 The simplified circuit now looks like this;
 
-![Voltage Divider Load Example](./voltage-divider-load-example-2.svg)
+![Voltage Divider Load Example](voltage-divider-load-example-2.svg)
 
 Now we can easily calculate the Resistance in the circuit