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2. Simple DC circuits

So far, only simple circuits consisting of a source and a load connected by wires have been considered.
In the following, more complicated circuit arrangements will be analysed. These initially contain only one source, but several lines and many ohmic loads (cf. Abbildung 1).

2.1 ideal components

Every electrical circuit consists of three elements:

1. Consumers: consumers convert electrical energy into energy that is not purely electrical.
   e.g.
   1. into electrostatic energy (capacitor)
   2. into magnetostatic energy (magnet)
   3. into electromagnetic energy (LED, light bulb)
   4. into mechanical energy (loudspeaker, motor)
   5. into chemical energy (charging an accumulator)
2. sources (generators): sources convert energy from another form of energy into electrical energy. (e.g. generator, battery, photovoltaic).
3. wires (interconnections): the wires of interconnection lines link consumers to sources.

These elements will be considered in more detail below.

Consumer

• The colloquial term 'consumer' in electrical engineering stands for an electrical consumer - i.e. a component which converts electrical energy into another form of energy.
• A resistor is often also referred to as a consumer. In addition to pure ohmic consumers, however, there are also ohmic-inductive consumers (e.g. coils in a motor) or ohmic-capacitive consumers (e.g. various power supplies using capacitors at the output). Correspondingly the equation „consumer is a resistor“ is wrong.
• Current-voltage characteristics (vgl. Abbildung 2)
  • Current-voltage characteristics of a load always run through the origin, because without current there is no voltage and vice versa.
  • Ohmic loads have a linear current-voltage characteristic which can be described by a single numerical value.
    The slope in the $U$-$I$-characteristic is the conductance: $I = G \cdot U = {{U}\over{R}}$

Sources

• Sources act as generators of electrical energy
• A distinction is made between ideal and real sources.
  The real sources are described in the following chapter (non-ideal_sources_and_two_pole_networks).

The ideal voltage source generates a defined constant output voltage $U_s$ (in German often $U_q$ for Quellenspannung). In order to maintain this voltage, it can supply any current. The current-voltage characteristic also represents this (see Abbildung 3).
The circuit symbol shows a circle with two terminals. In the circuit, the two terminals are short-circuited.
Another circuit symbol shows the negative terminal of the voltage source as a „thick minus symbol“, the positive terminal is drawn wider.

The ideal current source produces a defined constant output current $I_s$ (in German often $I_q$ for Quellenstrom). For this current to flow, any voltage can be applied to its terminals. The current-voltage characteristic also represents this (see Abbildung 4).
The circuit symbol shows again a circle with two connections. This time the two connections are left open in the circle and a line is drawn perpendicular to them.

wire connection

• The ideal connection line is resistance-free and transmits current and voltage instantaneously.
• Real existing influences (e.g. voltage drop) of connections are considered via separately drawn components (e.g. ohmic resistance).

2.2 Reference-arrow Systems and first consideration of a DC circuit

In the chapter 1. Preparation and Proportions the conventional directional sense of currents and voltages has already been discussed. Unfortunately, with meshed networks it is often not clear ahead of the calculation in which direction the conventional sense of direction of all currents and voltages runs.

In Abbildung 5 such a meshed net is shown. In this circuit a switch $S_1$ and a current $I_2$ are marked. Once the state of the switch is swapped, the direction of the current changes.

Generator and Load (Reference) Arrow System

For the direction of the arrows different conventions are available. Here (and quite often in Germany) the convention of power engineering is used. This convention is

Generator Reference Arrow System

2.3 Nodes, Branches and Loops

Electrical circuits typically have the structure of networks. Networks consist of two elementary structural elements:

1. Branches (German: Zweige): Connections between two nodes
2. Node (German: Knoten): Connection points of several branches

Please note in the case of electrical circuits:

1. Branches contain at least one component.
2. Nodes connect more than two branches and can also be spatially extended.

Branches in electrical networks are also called two-pole. Their behaviour is described by current-voltage characteristics and explained in more detail in the chapter non-ideal_sources_and_two_pole_networks .

In addition, another term is to be explained:

A loop is a closed path in the mesh. This means that 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 Abbildung 10).

Please keep in mind, that usually the entire behaviour 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.

Simplifications

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 Abbildung 11 shows how such a transformation is possible.

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

Further examples can be found in the following video

2.4 Kirchhoff's circuit laws

Kirchhoff's current law (Kirchhoff's first law)

The Kirchhoff's current law (Kirchhoff's nodal rule, in German: Knotensatz) formulates in the language of mathematics the experience that no charge „accumulations“ occur in electrical wires. This is of particular relevance at a network node (Abbildung 14). To formulate the equation at this node, the reference arrows of the currents are all set in the same way. That means: all point away from or towards the node.

Parallel circuit of resistors

From the Kirchhoff's current law, the total resistance for resistors connected in parallel can be derived (Abbildung 15):

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

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

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

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

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: $R_{eq}= \large{{R_1 \cdot R_2}\over{R_1 + R_2}}$

Current divider

The current divider rule can also be derived from the Kirchhoff's current law.
This states that, for resistors $R_1, ... R_n$ their currents $I_1, ... I_n$ behave just like the conductances $G_1, ... G_n$ through which they flow.

$\large{{I_1}\over{I_g}} = {{G_1}\over{G_g}}$

$\large{{I_1}\over{I_2}} = {{G_1}\over{G_2}}$

Kirchhoff's voltage law (Kirchhoff's second law)

Also the Kirchhoff's voltage law describes in mathematical language another practical experience: Between two points $1$ and $2$ of a network there is only one potential difference. Thus the potential difference is in particular independent of the way a network is traversed between the two points $1$ and $2$. This can be described by considering the meshes.

Proof of Kirchhoff's voltage law

If one expresses the voltage in Abbildung 17 dby the potentials in the nodes, we get: $U_{12}= \varphi_1 - \varphi_2 $
$U_{23}= \varphi_2 - \varphi_3 $
$U_{34}= \varphi_3 - \varphi_4 $
$U_{41}= \varphi_4 - \varphi_1 $

If these voltages are added, this leads to Kirchhoff's voltage law.

$U_{12}+U_{23}+U_{34}+U_{41} = 0$

Series circuit of resistors

Using Kirchhoff's voltage law, the total resistance of a series circuit (in German: Reihenschaltung, see <imref FigNo13>) can be easily determined:

$U_1 + U_2 + ... + U_n = U_g$

$R_1 \cdot I_1 + R_2 \cdot I_2 + ... + R_n \cdot I_n = R_{ersatz} \cdot I $

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

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

In general: The equivalent resistance of a series circuit is always greater than the greatest resistance..

2.5 Unloaded and loaded voltage divider

The unloaded voltage divider

Especially the series ciruit of two resistors $R_1$ and $R_2$ shall be considered now. This situation occurs in many practical applications (e.g. potentiometer). In Abbildung 21 this circuit is shown.

Via the Kirchhoff's voltage law we get

$\boxed{ {{U_1}\over{U}} = {{R_1}\over{R_1 + R_2}} }$

The ratio $k={{R_1}\over{R_1 + R_2}}$ also corresponds to the position on a potentiometer.

The loaded voltage divider

If - in contrast to the above-mendtioned, unloaded voltage divider - a load $R_L$ is connected to the output terminals (Abbildung 23), this load influences the output voltage.

A circuit analysis yields:

$ U_1 = \LARGE{{U} \over {1 + {{R_2}\over{R_L}} + {{R_2}\over{R_1}} }} }$

or on a potentiometer with $k$ and the sum of resistors $R_s = R_1 + R_2$:

$ U_1 = \LARGE{{k \cdot U} \over { 1 + k \cdot (1-k) \cdot{{R_s}\over{R_L}} }} }$

Abbildung 24 shows the ratio of the output voltage $U_1$ to the input voltage $U$ (y-axis), in relation to the ratio $k={{R_1}\over{R_1 + R_2}}$. In principle, this is similar to Abbildung 21, but here it has another dimension: multiple graphs are plotted. These differ by the ratio ${{R_s}\over{R_L}}$.

What does this diagram tell us? This shall be investigated by an example. First, assume an unloaded voltage divider with $R_2 = 4 k\Omega$ and $R_1 = 6 k\Omega$, and an input voltage of $10V$. Thus $k = 0.6$, $R_s = 10k\Omega$ and $U_1 = 6V$. Now this voltage divider is loaded with a load resistor. If this is at $R_L = R_1 = 10 k\Omega$, $k$ reduces to about $0.48$ and $U_1$ reduces to $4.8V$ - so the output voltage drops. For $R_L = 4k\Omega$, $k$ becomes even smaller to $k=0.375$ and $U_1 = 3.75V$. If the load $R_L$ is only one tenth of the resistor $R_s=R_1 + R_2$, the result is $k=0.18$ and $U_1=1.8V$. The output voltage of the unloaded voltage divider ($6V$) thus became less than one third.

What is the practical use of the (loaded) voltage divider?
Here some examples:

• Voltage dividers are in use for controlling the output of power supply IC's (see Voltage Dividers in Power Supplies). In order not to create a loaded voltage divider, a range for the resistance is given here.
• Another „invisible“ voltage divider is for example in the electical system of a car. As we will learn in the next chapters, voltage supplies have in internal resistance (and therefore batteries, too). The other consumer in the car also represent a resistance. By this, the electical system states an unloaded voltage divider. Given another, additional low-resistance load (e.g. the spark of the starter system) one can understand that there will be a voltage drop when starting the car.

2.6 Star-delta circuit

At the beginning of the chapter an example of a network was shown (Abbildung 1). Here, however, one does not come directly to the solution with the set of nodes and loops. It is visible, that there are many triangle-shaped loops resp. star-shaped nodes (Abbildung 28). These will be discussed in more detail now.

First of all a summary of the previous findings. Using the node and loop rule it became clear that an equivalent resistance can be determined from a series as well as from a parallel circuit. If one considers the equivalent resistance as a black box - i.e. the internals are unknown - it could be interpreted by both types of circuit (Abbildung 29).

Now how does this help us in the case of a triangular loop?

Also in this case one can provide a black box. However, this should always behave in the same way as the triangular loop, i.e. any voltages applied should produce the same currents as the known structure.
In other words: The resistances measured between two terminals must be identical for the blackbox and the known circuit.

For this purpose, the different resistances between the individual nodes $a$, $b$ and $c$ are now to be considered, see Abbildung 30. The aim is to find out how a delta circuit can be developed from a star circuit (and vice versa).

Delta circuit

In the delta connection, the 3 resistors $R_{ab}^1$, $R_{bc}^1$ and $R_{ca}^1$ are connected in a loop this terminals on each node.

For the resistors between the two terminals (e.g. $a$ and $b$), the third one ($c$) is considered as not connected to anything outside. This results in a parallel circuit of the direct star resistor $R_{ab}^1$ with the series connection of the other two star resistors $R_{ca}^1 + R_{bc}^1$:

$R_{ab} = R_{ab}^1 || (R_{ca}^1 + R_{bc}^1) $
$R_{ab} = {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + (R_{ca}^1 + R_{bc}^1)}} = {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} $

The same applies to the other connections. This results in:

\begin{align*} R_{ab} = {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \\ R_{bc} = {{R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \\ R_{ca} = {{R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \tag{2.6.1} \end{align*}

Star circuit

Given the idea, that the star ciruict shall behave equal to the delta circuit, the resistance measured between the terminals must be similar. Also in the star connection 3 resistors are connected, but in a star shape. The star resistors are all connected with another node $0$ in the middle: $R_{a0}^1$, $R_{b0}^1$ and $R_{c0}^1$.

Again, the procedure is the same as for the delta connection: the resistance between two terminals (e.g. $a$ and $b$) is determined, and the further terminal ($c$) is considered open. The resistance of the further terminal ($R_{c0}^1$) is only connected at one node. Therefore, no current flows through it - it is thus not to be considered. It results in:

\begin{align*} R_{ab} = R_{a0}^1 + R_{b0}^1 \\ R_{bc} = R_{b0}^1 + R_{c0}^1 \\ R_{ca} = R_{c0}^1 + R_{a0}^1 \tag{2.6.2} \end{align*}

From equations $(2.6.1)$ and $(2.6.2)$ we get:

\begin{align} R_{ab} = {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} = R_{a0}^1 + R_{b0}^1 \tag{2.6.3} \end{align} \begin{align} R_{bc} = {{R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} = R_{b0}^1 + R_{c0}^1 \tag{2.6.4} \end{align} \begin{align} R_{ca} = {{R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} = R_{c0}^1 + R_{a0}^1 \tag{2.6.5} \end{align}

Die Gleichungen $(2.6.3)$ bis $(2.6.5)$ lassen sich nun so geschickt zusammenfassen, dass auf einer Seite nur noch ein Widerstand steht.
Eine Variante ist die Formeln als ${{1}\over{2}} \cdot \left( (2.6.3) + (2.6.4) - (2.6.5) \right)$ bzw. ${{1}\over{2}} \cdot \left(R_{ab} + R_{bc} - R_{ca}\right)$ zu kombinieren. Damit ergibt sich $R_{b0}^1$

\begin{align*} {{1}\over{2}} \cdot \left( {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} + {{R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} - {{R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) &= {{1}\over{2}} \cdot \left( R_{a0}^1 + R_{b0}^1 + R_{b0}^1 + R_{c0}^1 - R_{c0}^1 - R_{a0}^1 \right) \\ {{1}\over{2}} \cdot \left( {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)} + {R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)} - {R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) &= {{1}\over{2}} \cdot \left( 2 \cdot R_{b0}^1 \right) \\ {{1}\over{2}} \cdot \left( {{R_{ab}^1 R_{ca}^1 + R_{ab}^1 R_{bc}^1 + R_{bc}^1 R_{ab}^1 + R_{bc}^1 R_{ca}^1 - R_{ca}^1 R_{bc}^1 - R_{ca}^1 R_{ab}^1}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) &= R_{b0}^1 \\ {{1}\over{2}} \cdot \left( {{ 2 \cdot R_{ab}^1 R_{bc}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) &= R_{b0}^1 \\ {{ R_{ab}^1 R_{bc}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} &= R_{b0}^1 \\ \end{align*}

Auf ähnlichem Weg kann man nach $R_{a0}^1$ und $R_{c0}^1$, sowie mit etwas abgewandeltem Ansatz auch auf $R_{ab}^1$, $R_{bc}^1$ und $R_{ca}^1$ auflösen.

Stern-Dreieck-Transformation

2.7 Gruppenschaltung von Widerständen

In diesem Unterkapitel wird auf eine Methodik eingegangen, welche beim Umformen von Schaltungen helfen soll. In Unterkapitel 2.6 Stern-Dreieck-Schaltung wurde gegen Ende bereits ein Netzwerk so umgeformt, dass es keine dreieckigen Maschen mehr enthält. Nun soll dieses Vorgehen systematisiert werden. Ausgangspunkt sind Aufgaben, bei denen für ein Widerstandsnetzwerk der Gesamtwiderstand, Gesamtstrom oder die Gesamtspannung berechnet werden muss.

einfaches Beispiel

Ein Beispiel für eine solche Schaltung ist in Abbildung ## gegeben. Hier ist $I_0$ gesucht. Dieser Strom kann über die (gegebene) Spannung $U_0$ und den Gesamtwiderstand zwischen den Klemmen $a$ und $b$ ermittelt werden. Gesucht ist also $R_{ab}$.

Wie bereits in den vorherigen Unterkapitel beschrieben, können hier auch Teilschaltung schrittweise in Ersatzwiderstände umgewandelt werden. Wichtig dabei ist, dass diese Teilschaltungen zur Umwandlung in Ersatzwiderstände immer nur zwei Anschlüsse (= zwei Knoten zur „Außenwelt“) haben dürfen.

Abbildung ## zeigt die schrittweise Umwandlung der Ersatzwiderstände an diesem Beispiel.
Als Ergebnis des Ersatzwiderstands erhält man:

\begin{align*} R_g = R_{12345} &= R_{12}||R_{345} = R_{12}||(R_3+R_{45}) = (R_1||R_2)||(R_3+R_4||R_5) \\ &= {{ {{R_1 \cdot R_2}\over{R_1 + R_2}} \cdot (R_3 + {{R_4 \cdot R_5}\over{R_4 + R_5}}) }\over{ {{R_1 \cdot R_2}\over{R_1 + R_2}} +R_3 + {{R_4 \cdot R_5}\over{R_4 + R_5}} }} \quad \quad \quad \quad \quad \quad \bigg\rvert \cdot{{(R_1 + R_2) \cdot (R_4 + R_5)}\over{(R_1 + R_2) \cdot (R_4 + R_5)}} \\ &= {{ R_1 \cdot R_2 \cdot (R_3 + {{R_4 \cdot R_5}\over{R_4 + R_5}}) \cdot (R_4 + R_5) } \over { R_1 \cdot R_2\cdot(R_4 + R_5) +R_3 + R_4 \cdot R_5 \cdot (R_1 + R_2)}} \\ &= {{ R_1 \cdot R_2 \cdot (R_3 \cdot (R_4 + R_5) + R_4 \cdot R_5) } \over { R_1 \cdot R_2\cdot(R_4 + R_5) +R_3 + R_4 \cdot R_5 \cdot (R_1 + R_2)}} \\ \end{align*}

Beispiel mit Dreieck-Stern-Transformation

Mit der Dreieck-Stern-Transformation lässt sich nun auch das anfängliche Beispiel umwandeln. Bei komplizierteren Schaltungen ist die wiederholte Dreieck-Stern-Transformation mit anschließendem Zusammenfassen der Widerstände sinnvoll, solange bis die entstandene Schaltung leicht mit Knoten- und Maschensatz berechenbar wird (Abbildung ##). Hier wird auf eine Rechnung verzichtet - es empfiehlt sich hier mit Zwischenergebnissen für die transformierten Widerständen zu rechnen.

Beispiel mit Symmetrien in der Schaltung

Ein gewisser Sonderfall betrifft mögliche Symmetrien in Schaltungen. Falls dies3 vorhanden sind, kann eine weitere Vereinfachung vorgenommen werden.

Abbildung ## zeigt links ein symmetrischen Aufbau eines Netzwerks aus jeweils gleichen Widerständen $R$. Zum Verständnisgewinn ist in der Mitte in der gleichen Schaltung zusätzlich Schalter und Testpunkte (TP) verbaut, welche die Spannung gegen Masse anzeigen.

Über die Schalter kann nachgeprüft werden, ob ein Strom fließt, falls die jeweiligen Knoten verbunden werden. In der Simulation ist zu sehen, dass dies nicht der Fall ist. Im symmetrischen Aufbau sind diese Knoten jeweils auf dem gleichen Potential.

Damit lässt sich die Schaltung auch in die Form bringen, wie sie in Abbildung ## rechts zu sehen ist. Diese Schaltung ist wiederum leicht berechenbar:

\begin{align*} R_g = R || R + R || R || R || R + R || R || R || R + R || R = {{1}\over{2}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{2}}\cdot R = 1,5\cdot R \end{align*}