====== Block 04 — Kirchhoff’s laws ======

===== Learning objectives =====
<callout>
After this 90-minute block, you can
  * Identify nodes, branches, and (essential) loops in DC circuits and draw consistent reference arrows for $U$ and $I$ (passive/active sign conventions).
  * State and apply **Kirchhoff’s Current Law (KCL)** at an arbitrary node and **Kirchhoff’s Voltage Law (KVL)** around a loop. 
  * Translate between verbal circuit descriptions and node/loop equations; check signs and units; solve for unknown currents/voltages in small networks.
  * Use KCL/KVL as the foundation for upcoming techniques (series/parallel, dividers, bridges, source equivalents). 
</callout>

===== Preparation at Home =====

As always: 
  * Please read through the following chapter.
  * Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting). 

For checking your understanding please do the following exercises:
  * 2.3.3 (1) 
  * 2.4.3

===== 90-minute plan =====
  - Warm-up (10 min): What is a “node”? what is a “mesh”? quick sketch-and-label drill.
  - Core concepts (40 min): reference arrows & sign conventions → KCL at a node → KVL in a loop → dimensional check.
  - Worked examples (20 min): one KCL node solve; one KVL loop solve with a source and two resistors.
  - Guided practice (15–20 min): short tasks (incl. sims/drawings below).
  - Wrap-up (5 min): checklist, common pitfalls, outlook to Block 05 (series/parallel, dividers, bridge). 

~~PAGEBREAK~~ ~~CLEARFIX~~
===== Core Content  =====

==== Nodes, Branches, and Loops ====

Electrical circuits typically have the structure of networks. Networks consist of two elementary structural elements:
  - <fc #cd5c5c>**Branches**</fc> (German: Zweige): Connections between two nodes. 
  - <fc #6495ed>**Node**</fc> (German: Knoten): Connection "point" of several branches. 

<WRAP>
<imgcaption BildNr0 | circuitry and mesh>
</imgcaption> \\
{{drawio>Stromkreise_Stromnetze.svg}}
</WRAP>

Please note in the case of electrical circuits, we will use the following definition:

  - <fc #cd5c5c>**Branches**</fc> contain at least one component.
  - <fc #6495ed>**Nodes**</fc> connect __more than two branches__. Since the wire in a circuit diagram is an ideal conductor, all connected wires to a node are at the same voltage level. Therefore the node in the circuit diagram can also be spatially extended by the wires. 

<WRAP>
<imgcaption BildNr8 | nodes, branches and loops>
</imgcaption> \\
{{drawio>KnotenZweigeMaschen.svg}}
</WRAP>

Sometimes there is a differentiation between "simple nodes" (only connecting 2 branches) and "principal nodes" (connecting more than 2 branches). We will in the following often only mark the connection of more than two branches with a node.

Branches in electrical networks are also called two-terminal networks.
Their behavior is described by current-voltage characteristics and explained in more detail in [[block06]].

In addition, another term is to be explained: \\

A loop begins and ends at the same node and runs over at least one further node.

Since a voltmeter can also be present as a component between two nodes, it is also possible to close a loop by a drawn voltage arrow (cf. $U_1$ in <imgref BildNr8>).

~~PAGEBREAK~~ ~~CLEARFIX~~
Please keep in mind, that usually the entire behavior of networked circuits almost always changes when a change occurs in one branch or at one node. This is in contrast to other cause-effect relationships, but comparable to changes in other larger networks, e.g. a traffic jam in the road network, due to which other roads experience a higher load. For electrical engineering, this means that in the case of changing circuits, the focus is often on determining the interrelationships (formulas, current-voltage characteristics) and not on a single numerical value.

~~PAGEBREAK~~ ~~CLEARFIX~~

==== Reshaping Circuits ====

With the knowledge of nodes, branches, and meshes, circuits can be simplified.
Circuits can be reshaped arbitrarily as long as all branches remain at the same nodes after reshaping
The <imgref BildNr9> shows how such a transformation is possible.

<WRAP>
<imgcaption BildNr9 | Example of circuit conversion>
{{elektrotechnik_1:umwandlungeinerschaltung.gif}}
</imgcaption>
</WRAP>

For practical tasks, repeated trial and error can be useful.
It is important to check afterward that the same components are connected to each node as before the transformation.

~~PAGEBREAK~~ ~~CLEARFIX~~

==== Kirchhoff’s Current Law (KCL) ====

<WRAP right>
<imgcaption BildNr10 | Kirchhoff's current law>
</imgcaption>
{{drawio>Knotensatz.svg}}
</WRAP>

In any node, the algebraic sum of currents is zero. All reference arrows are drawn either **into** or **out of** the node.  
\begin{align*}
\boxed{\sum_{\nu=1}^{n} I_\nu = 0} \quad
\end{align*}

Interpretation: the sum of currents flowing **into** a node equals the sum flowing **out of** that node → no net charge accumulation in steady DC. 

**Sign rule used here.** If you write currents with reference arrows **toward** the node as positive, and **away** from the node as negative, KCL is $\sum I_x=0$. (Any one consistent choice is fine.) 

<panel type="info" title="Example / micro-exercise">
At node $N$, suppose $I_1=2.00~{\rm A}$ and $I_2=0.50~{\rm A}$ flow **into** the node, and $I_3$ flows **out** of the node. With “into” positive:
\begin{align*}
I_1 + I_2 - I_3 = 0 \Rightarrow I_3 = 2.50~{\rm A}.
\end{align*}

Units check: $[I]={\rm A}$ on every term, so the sum is consistent.
</panel>

<callout color="gray" icon="fa fa-check">
**Dimensional check (new formula).** In the **current divider** relation $I_1/I_2=G_1/G_2$, both sides are ratios → dimensionless. Since $[G]={\rm S}$, the unit cancels in a ratio. 
</callout>

~~PAGEBREAK~~ ~~CLEARFIX~~
==== Parallel circuit of resistors ====

From Kirchhoff's current law, the total resistance for resistors connected in parallel can be derived (<imgref BildNr11>):

<WRAP>
<imgcaption BildNr11 | Parallel circuit>
</imgcaption>
{{drawio>Parallelschaltung.svg}}
</WRAP>

Since the same voltage $U_{\rm ab}$ is dropped across all resistors, using Kirchhoff's current law:

$\large{{U_{\rm ab}}\over{R_1}}+ {{U_{\rm ab}}\over{R_2}}+ ... + {{U_{\rm ab}}\over{R_n}}= {{U_{\rm ab}}\over{R_{\rm eq}}}$

$\rightarrow \large{{{1}\over{R_1}}+ {{1}\over{R_2}}+ ... + {{1}\over{R_n}}= {{1}\over{R_{\rm eq}}} = \sum_{x=1}^{n} {{1}\over{R_x}}}$

Thus, for resistors connected in parallel, the equivalent conductance $G_{\rm eq}$ (German: //Ersatzleitwert//) is the sum of the individual conductances: $G_{\rm eq} = \sum_{x=1}^{n} {G_x}$

__In general__: the equivalent resistance of a parallel circuit is always smaller than the smallest resistance.

Especially for two parallel resistors $R_1$ and $R_2$ applies: 

\begin{align*}
\boxed{R_{\rm eq}= R_1 || R_2 = \large{{R_1 \cdot R_2}\over{R_1 + R_2}}  }
\end{align*}


==== Current divider ====

The current divider rule shows in which way an incoming current on a node will be divided into two outgoing branches.
The rule states that the currents $I_1, ... I_n$ on parallel resistors $R_1, ... R_n$ behave just like their conductances $G_1, ... G_n$ through which the current flows. \\

\begin{align*}
\large{{I_1}\over{I_{\rm res}}} = {{G_1}\over{G_{\rm res}}}
\large{{I_1}\over{I_2}} = {{G_1}\over{G_2}}
\end{align*}

The rule also be derived from Kirchhoff's current law: \\
  - The voltage drop $U$ on parallel resistors $R_1, ... R_n$ is the same. 
  - When $U_1 = U_2 = ... = U$, then the following equation is also true: $R_1 \cdot I_1 = R_2 \cdot I_2 = ... = R_{\rm eq} \cdot I_{\rm res}$. \\
  - Therefore, we get with the conductance: ${{I_1} \over {G_1}} = {{I_2} \over {G_2}}= ... = {{I_{\rm eq}} \over {G_{\rm res}}}$

~~PAGEBREAK~~ ~~CLEARFIX~~
==== Kirchhoff’s Voltage Law (KVL) ====

Around any closed loop, the algebraic sum of voltages is zero:
\begin{align*}
\boxed{\sum_{\nu=1}^{n} U_\nu = 0} \quad 
\end{align*}

Equivalently: the sum of rises equals the sum of drops along the chosen loop direction. The result does **not** depend on the specific path between two nodes. 

<WRAP>
<imgcaption BildNr12 | loop law>
</imgcaption>
{{drawio>Maschensatz.svg}}
</WRAP> 

**Sign rule used here.** Choose a loop direction (often clockwise). A voltage arrow **aligned** with the loop direction is added; **opposed** to it is subtracted. (Source polarities must respect the active/passive conventions fixed earlier.) 


<panel type="info" title="Example / micro-exercise">
Series loop with source $U_{\rm s}=12.0~{\rm V}$, resistors $R_1=3.0~\Omega$, $R_2=5.0~\Omega$. With passive sign convention across both resistors and loop direction from the $+$ of the source:
\begin{align*}
-U_{\rm s} + U_{R_1} + U_{R_2} = 0, \quad U_{R_k} = R_k I.
\end{align*}

Thus  \\
$I = \frac{U_{\rm s}}{R_1+R_2} = \frac{12.0~{\rm V}}{8.0~\Omega}=1.50~{\rm A}$ \\
$U_{R_1}=4.50~{\rm V}$, $U_{R_2}=7.50~{\rm V}$ \\
$-12.0~{\rm V}+4.50~{\rm V}+7.50~{\rm V}=0$. (Check: volts add algebraically to $0$.) 
</panel>

~~PAGEBREAK~~ ~~CLEARFIX~~
=== Series circuit of resistors ===

<WRAP>
<imgcaption BildNr13 | series circuit>
</imgcaption>
{{drawio>Reihenschaltung.svg}}
</WRAP>

Using Kirchhoff's voltage law, the total resistance of a series circuit (in German: //Reihenschaltung//, see <imgref BildNr13>) can be easily determined:

\begin{align*}
U_1 + U_2 + ... + U_n = U_{\rm res}
R_1 \cdot I_1 + R_2 \cdot I_2 + ... + R_n \cdot I_n = R_{\rm eq} \cdot I 
\end{align*}

Since in a series circuit, the current through all resistors must be the same - i.e. $I_1 = I_2 = ... = I$ - it follows that:

\begin{align*}
R_1 + R_2 + ... + R_n = R_{\rm eq} =  \sum_{x=1}^{n} R_x 
\end{align*}


__In general__: The equivalent resistance of a series circuit is always greater than the greatest resistance.

===== From laws to tools (preview) =====

KCL and KVL immediately yield:
  * **Series**: same current through all series elements $\Rightarrow$ voltages add, $R_{\rm eq}=R_1+R_2+\dots$.
  * **Parallel**: same voltage across all parallel elements $\Rightarrow$ currents add, $G_{\rm eq}=G_1+G_2+\dots$.
  * **Dividers** and **bridge** behavior follow from the same laws. (We will formalize these in [[Block05]].) 

For orientation, the short slides you cross-check with present the same sequence: KCL/KVL → resistive networks → (later) real sources and two-port models.

~~PAGEBREAK~~ ~~CLEARFIX~~

===== Common pitfalls =====
  * **Mixed conventions**: do not swap passive/active mid-solution. Fix reference arrows once, then stick to them.
  * **Sign slips**: in KVL, mark loop direction on the drawing; in KCL, decide “in is $+$” (or “out is $+$”) and keep it for *all* terms.
  * **Units**: always write ${\rm A}$ for currents and ${\rm V}$ for voltages in interim and final results.

~~PAGEBREAK~~ ~~CLEARFIX~~
===== Exercises =====

<panel type="info" title="Exercise 2.3.1 Branches and Nodes"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
<WRAP>
<imgcaption BildNr70 | Branches and Nodes>
</imgcaption>
{{drawio>ZweigeundKnoten.svg}}
</WRAP>

For the markings in the circuits in <imgref BildNr70> indicate whether it is a branch, a node, or neither.

</WRAP></WRAP></panel>

<panel type="info" title="Exercise 2.3.2 Branches and Nodes (with explanation)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>GNumiT_Y4B8}}
</WRAP></WRAP></panel>


<panel type="info" title="Exercise 2.3.3 Reshaping circuits"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
<WRAP>
<imgcaption BildNr71 | more Branches and Nodes>
</imgcaption>
{{drawio>SchaltungenVereinfachen.svg}}
</WRAP>

Reshape the circuits in <imgref BildNr71>. 
</WRAP></WRAP></panel>


 


<wrap anchor #exercise_2_4_1 />
<panel type="info" title="Exercise 2.4.1 Current divider"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP>
<imgcaption BildNr85| Current divider>
</imgcaption> \\
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsBMA2AzAgnAdjBgBxhgK7q64gKTXWQBQATiCgmuLh2xxpACxRwcBmHKt2nbpLApCQqkiS0V0JADUA9gBsALgEMA5gFMGhibwzTeCFFAYB3CyD6CeLq+GYvbUn3dchMHhHfxcBMIxPMDNIzywA30YnQLEbOzT7J3dM1K4s50ywNFpMxhZi2hQUQUrnUpCnOtSSjw4Y82bo1oSCupyeiMYAZ3BW6trWsvAQADN9bWHTIA noborder}}
</WRAP>

In the simulation in <imgref BildNr85> a current divider can be seen. The resistances are just inversely proportional to the currents flowing through it.

  - What currents would you expect in each branch if the input voltage were lowered from $5~\rm V$ to $3.3V~\rm $? __After__ thinking about your result, you can adjust the ''Voltage'' (bottom right of the simulation) accordingly by moving the slider.
  - Think about what would happen if you flipped the switch __before__ you flipped the switch. \\ After you flip the switch, how can you explain the current in the branch?

</WRAP></WRAP></panel>

<panel type="info" title="Exercise 2.4.2 two resistors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Two resistors of $18~\Omega$ and $2~\Omega$ are connected in parallel. The total current of the resistors is $3~\rm A$. \\
Calculate the total resistance and how the currents are split to the branches.


<button size="xs" type="link" collapse="Loesung_2_4_2_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_2_4_2_1_Lösungsweg" collapsed="true">
The substitute resistor can be calculated to
\begin{equation*}
R_{eq} = \frac{R_1R_2}{R_1+R_2} = \frac{18~\Omega \cdot 2~\Omega}{18~\Omega+2~\Omega}
\end{equation*}
The current through resistor $R_1$ is
\begin{equation*}
I_1 = \frac{R_{eq}}{R_1} I =\frac{1.8~\Omega}{18~\Omega} \cdot 3~\rm A
\end{equation*}
The current through resistor $R_2$ is
\begin{equation*}
I_2 = \frac{R_{eq}}{R_2}I = \frac{1.8~\Omega}{2~\Omega} \cdot 3~\rm A
\end{equation*}
</collapse>
<button size="xs" type="link" collapse="Loesung_2_4_1_4_2_Lösungsweg">{{icon>eye}} Final result</button><collapse id="Loesung_2_4_1_4_2_Lösungsweg" collapsed="true">
The values of the substitute resistor and the currents in the branches are
\begin{equation*}
R_{eq} = 1.8~\Omega \qquad I_1 = 0.3~{\rm A} \qquad I_2 = 2.7~\rm A
\end{equation*}
</collapse>
</WRAP></WRAP></panel>
\\ 

<panel type="info" title="Exercise 4.3 KVL in a loop (source + two resistors)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
Sketch a loop with $U_{\rm s}=5.00~{\rm V}$, $R_1=1.00~{\rm k\Omega}$, $R_2=1.00~{\rm k\Omega}$. Draw passive arrows across both resistors and choose clockwise loop direction.  \\
Write KVL, solve $I$, then compute $U_{R_1}$ and $U_{R_2}$. Confirm that algebraic sum equals $0~{\rm V}$.  \\
*Expected:* $I=2.50~{\rm mA}$, $U_{R_1}=2.50~{\rm V}$, $U_{R_2}=2.50~{\rm V}$.
</WRAP></WRAP></panel>


<panel type="info" title="Exercise 2.4.3 Three Resistors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Three equal resistors of $20~k\Omega$ each are given. \\
Which values are realizable by the arbitrary interconnection of one to three resistors?\\
<button size="xs" type="link" collapse="Loesung_2_4_3_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_2_4_3_1_Lösungsweg" collapsed="true">
The resistors can be connected in series:
\begin{equation*}
R_{\rm series} = 3\cdot R = 3\cdot20~k\Omega
\end{equation*}
The resistors can also be connected in parallel:
\begin{equation*}
R_{\rm parallel} = \frac{R}{3} = \frac{20~k\Omega}{3}
\end{equation*}
On the other hand, they can also be connected in a way that two of them are in parallel and those are in series to the third one:
\begin{equation*}
R_{\rm res} = R + \frac{R\cdot R}{R+R} = \frac{3}{2}R = \frac{3}{2} \cdot 20~k\Omega
\end{equation*}
</collapse>
<button size="xs" type="link" collapse="Loesung_2_4_3_2_Lösungsweg">{{icon>eye}} Final values</button><collapse id="Loesung_2_4_3_2_Lösungsweg" collapsed="true">
\begin{equation*}
R_{series} = 60~k\Omega\qquad R_{\rm parallel} = 6.7~k\Omega\qquad R_{\rm res} = 30~k\Omega
\end{equation*}
</collapse>
</WRAP></WRAP></panel>

~~PAGEBREAK~~ ~~CLEARFIX~~

===== Embedded resources =====

<WRAP column half>
Explanation of the different network structures \\
(Graphs and trees are only needed in later chapters)
{{youtube>-82UNytyrCQ}}
</WRAP>

<WRAP column half>
Reshaping circuits
{{youtube>PnzijvMQmE8}}
</WRAP>

<WRAP column half>
{{wp>Kirchhoff's circuit laws}}
{{youtube>d0O-KUKP4nM}}
</WRAP> 

<WRAP column half>
Derivation of the current divider with examples
{{youtube>VojwBoSHc8U}}
</WRAP>

~~PAGEBREAK~~ ~~CLEARFIX~~
===== Summary =====
  * **KCL:** sum of signed currents at any node is $0$ (charge does not pile up in steady DC).  
  * **KVL:** sum of signed voltages around any loop is $0$ (potential differences are path-independent).  
  * **Conventions matter:** fix passive/active sign conventions once, then *stay consistent*.  
  * **Next:** apply KCL/KVL to build series/parallel laws, dividers, and bridges (Block 05); then model real sources and two-port equivalents. 

~~PAGEBREAK~~ ~~CLEARFIX~~