====== Block 05 — Resistive networks ======

===== Learning objectives =====
<callout>
After this 90-minute block, you can
  * reduce series/parallel resistor networks to an equivalent resistance $R_{\rm eq}$,
  * apply **voltage divider** (unloaded & loaded) and **current divider** rules,
  * recognize and analyze **bridge** circuits (Wheatstone; balance condition),
  * check results by units and by “sanity bounds” (e.g., $R_{\rm eq}$ in parallel is below the smallest branch).
</callout>


~~PAGEBREAK~~ ~~CLEARFIX~~
===== 90-minute plan  =====
  * 0–10 min — Recap KCL/KVL; sign conventions.
  * 10–30 min — Series & parallel; quick numeric checks.
  * 30–55 min — Voltage dividers (unloaded → loaded); potentiometer view; pitfalls.
  * 55–70 min — Current divider; typical use.
  * 70–85 min — Bridge circuits; balance; measurement idea.
  * 85–90 min — Wrap-up quiz; assign exercises.

~~PAGEBREAK~~ ~~CLEARFIX~~
===== Core Content =====
==== Unloaded voltage divider ====

The series circuit of two resistors $R_1$ and $R_2$ shall be considered now. \\
This situation occurs in many practical applications e.g. in a {{wp>potentiometer}}.

<WRAP>
<imgcaption BildNr114 | potentiometer>
</imgcaption>
{{drawio>Poti.svg}}
</WRAP>

In <imgref BildNr14> this circuit is shown.

<WRAP>
<imgcaption BildNr14 | unloaded voltage divider>
</imgcaption>
{{drawio>unbelasteterSpannungsteiler.svg}}
</WRAP>

Via Kirchhoff's voltage law, we get

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

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

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

<WRAP>
<imgcaption BildNr81| unloaded voltage divider>
</imgcaption> \\
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsAsB2ATAZgGw4QhmimJGOiApJSClpQKYC0YYAUAG7g7VEbfUwaHFFEpqSalOgI2QlCCyQFQkVgy8EIiKTjIkAJQYBnAJbGALgEMAdgGMGbAO4gNADleZFyz-0jPvBT5FNyCvfxd1XgwFLFDXGKg2AAcQoMS4sP4IKRTXSA9VfI9gnKSXMB5wYWLqkQiBX1rg-wBzQKalILcPXI00NISFEhEMRIUAHWMpmeMAVQB9AHkA9zqh9Yau9dKa-2MNot368BAAMysAG2NHIA noborder}}
</WRAP>

In the simulation in <imgref BildNr81> an unloaded voltage divider in the form of a potentiometer can be seen. The ideal voltage source provides $5~\rm V$. The potentiometer has a total resistance of $1~\rm k\Omega$. In the configuration shown, this is divided into $500 ~\Omega$ and $500 ~\Omega$.
  - What voltage $U_{\rm O}$ would you expect if the switch were closed? After thinking about your result, you can check it by closing the switch. 
  - First, think about what would happen if you would change the distribution of the resistors by moving the wiper ("intermediate terminal"). \\ You can check your assumption by using the slider at the bottom right of the simulation.
  - At which position do you get a $U_{\rm O} = 3.5~\rm V$?
</WRAP></WRAP></panel>

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==== The loaded Voltage Divider ====

If - in contrast to the abovementioned, unloaded voltage divider - a load $R_{\rm L}$ is connected to the output terminals (<imgref BildNr15>), this load influences the output voltage.

<WRAP>
<imgcaption BildNr15 | loaded voltage divider>
</imgcaption>
{{drawio>belasteterSpannungsteiler.svg}}
</WRAP>

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 $R_{\rm s} = R_1 + R_2$:

$ U_1 = \LARGE{{k \cdot U} \over { 1 + k \cdot (1-k) \cdot{{R_{\rm s}}\over{R_{\rm L}}} }}$

<imgref BildNr65> 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 <imgref BildNr14>, but here it has another dimension: multiple graphs are plotted. These differ by the ratio ${{R_{\rm s}}\over{R_{\rm L}}}$.

<WRAP>
<imgcaption BildNr65 | Voltage curve of the loaded voltage divider>
</imgcaption>
{{drawio>SpannungsverlaufBelasteterSpannungsteiler}}
</WRAP>

What does this diagram tell us? This shall be investigated by an example. First, assume an unloaded voltage divider with $R_2 = 4.0 ~\rm k\Omega$ and $R_1 = 6.0 ~\rm k\Omega$, and an input voltage of $10~\rm V$. Thus $k = 0.60$, $R_s = 10~\rm k\Omega$ and $U_1 = 6.0~\rm V$.
Now this voltage divider is loaded with a load resistor. If this is at $R_{\rm L} = R_1 = 10 ~\rm k\Omega$, $k$ reduces to about $0.48$ and $U_1$ reduces to $4.8~\rm V$ - so the output voltage drops. For $R_{\rm L} = 4.0~\rm k\Omega$, $k$ becomes even smaller to $k=0.375$ and $U_1 = 3.75~\rm V$. If the load $R_{\rm L}$ is only one-tenth of the resistor $R_{\rm s}=R_1 + R_2$, the result is $k = 0.18$ and $U_1 = 1.8~\rm V$. The output voltage of the unloaded voltage divider ($6.0~\rm V$) thus became less than one-third.

What is the practical use of the (loaded) voltage divider? \\ Here are some examples: 
  * Voltage dividers are in use for controlling the output of power supply ICs (see [[https://www.analog.com/en/technical-articles/a101121-voltage-dividers-in-power-supplies.html|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 electrical system of a car. As we will learn in the next chapters, voltage supplies have internal resistance (and therefore batteries, too). The other consumer in the car also represents a resistance. By this, the electrical system states an unloaded voltage divider. Given another, additional low-resistance load (e.g. the spark or the starter motor of the starter system) one can understand that there will be a voltage drop when starting the car.

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==== Bridge networks (Wheatstone) ====
A four-resistor bridge can be seen as **two voltage dividers in parallel**. The detector (bridge branch) sees the **difference** of the two divider node voltages. The **balance condition** (zero detector current) is
\begin{align*}
\boxed{ \frac{R_1}{R_2}=\frac{R_3}{R_4} \;\; \Leftrightarrow \;\; R_1R_4=R_2R_3 }.
\end{align*}
Use this to check sensor bridges (strain gauges etc.) or to compute unknown $R$. (Derivation via equal divider ratios.) 

<callout color="gray" icon="fa fa-lightbulb-o">
If the bridge is **nearly** balanced, the midpoint voltages differ only slightly → good for small-signal sensing. \\ In practice you’ll later buffer such nodes with high-$R_{\rm in}$ stages (see op-amps in [[Block21]]).
</callout>

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==== Strategy for network reduction ====
  * Reshape (without changing node connections), then collapse **clear** series or parallel groups.
  * If blocked by a three-terminal cluster, apply **Δ–Y** (or **Y–Δ**), then collapse again.
  * Repeat until a simple ladder remains; finish with KCL/KVL if needed. 

In this subchapter, a methodology is discussed, which should help to reshape circuits. 
The following concept works at tasks, where the total resistance, total current, or total voltage has to be calculated for a resistor network.

An example of such a circuit is given in <imgref imageNo89>. Here $I_0$ is wanted. 
This current can be found by the (given) voltage $U_0$ and the total resistance between the terminals $\rm a$ and $\rm b$. So we are looking for $R_{\rm ab}$.

<WRAP>
<imgcaption imageNo89 | example of a circuit>
</imgcaption>
{{drawio>BeispielStromkreis2.svg}}
</WRAP>

As already described in the previous subchapters, partial circuits can also be converted into equivalent resistors step by step. 
It is important to note that these partial circuits for conversion into equivalent resistors may only ever have two connections (= two nodes to the "outside world").


<WRAP>
<imgcaption imageNo88 | Step-by-step solution of the example >
</imgcaption>
{{drawio>BeispielStromkreis2Loesung.svg}}
</WRAP>
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<imgref imageNo88 > shows the step-by-step conversion of the equivalent resistors in this example. \\ 
As a result of the equivalent resistance one gets:

\begin{align*}
R_{\rm  eq} = 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*}
  
~~PAGEBREAK~~ ~~CLEARFIX~~
===== Exercises =====

<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>

<panel type="info" title="Exercise 2.5.2 loaded voltage divider I "> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Determine from the circuit in <imgref BildNr15> the equation $ U_1 = {{k \cdot U} \over { 1 + k \cdot (1-k) \cdot{{R_{\rm s}}\over{R_{\rm L}}}}}$ where $k={{R_1}\over{R_1 + R_2}}$ and $R_{\rm s} = R_1 + R_2$.\\
<button size="xs" type="link" collapse="Loesung_2_5_2_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_2_5_2_1_Lösungsweg" collapsed="true">
According to the voltage division rule, the loaded voltage is
\begin{align*}
U_1 &=\frac{\frac{R_1 R_L}{R_1+R_L}}{R_2+\frac{R_1 R_L}{R_1+R_L}}U \\
    &=\frac{R_1 R_{\rm L}}{R_2 (R_1 + R_{\rm L}) + R_1 R_{\rm L}} U \\
    &=\frac{R_1 R_{\rm L}}{R_1 R_2 + R_2 R_{\rm L} + R_1 R_{\rm L}} U \\
    &=\frac{R_1 R_{\rm L}}{R_1 R_2 + (R_1+R_2) R_{\rm L}} U
\end{align*}
The divided resistor $R_1$ and $R_2$ are put together to form $R_{\rm s}=R_1 + R_2$.
\begin{equation*}
U_1=\frac{R_1 R_{\rm L}}{R_1 R_2 + R_{\rm s} R_{\rm L}} U
\end{equation*}
With the equations given there is also $R_1=k(R_1+R_2)=k R_{\rm s}$ and $R_2 = R_{\rm s} - R_1 = R_{\rm s} - k R_{\rm s} = (1-k) R_{\rm s}$.
\begin{equation*}
U_1=\frac{k R_{\rm s} R_{\rm L}}{k R_{\rm s} (1-k) R_{\rm s} + R_{\rm s} R_{\rm L}}U
\end{equation*}
Dividing the numerator and denominator by $R_{\rm s} R_{\rm L}$ yields to
\begin{equation*}
U_1=\frac{k}{k(1-k)\frac{R_{\rm s}}{R_{\rm L}}+1}U
\end{equation*}
</collapse>

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

<panel type="info" title="Exercise 2.5.3 loaded voltage divider II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

In the simulation in <imgref BildNr82> a loaded voltage divider in the form of a potentiometer can be seen. The ideal voltage source provides $5.00~\rm V$. The potentiometer has a total resistance of $1.00~k\Omega$. In the configuration shown, this is divided into $500 ~\Omega$ and $500 ~\Omega$. The load resistance has $R_{\rm L} = 1.00 ~\rm k\Omega$.
  - What voltage ''U_O'' would you expect if the switch were closed? This is where you need to do some math! __After__ you calculated your result, you can check it by closing the switch.
  - At which position of the wiper do you get $3.50~\rm V$ as an output? Determine the result first by means of a calculation. \\ Then check it by moving the slider at the bottom right of the simulation.

<WRAP>
<imgcaption BildNr82| loaded voltage divider>
</imgcaption> \\
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3cGmkTFbrzAZCUMTSpIq06AjbDeuISJDY0fBKIhh4UWSABKDAM4BLYwBcAhgDsAxgzYB3Pvj4YBS9wMjPv4FS9hUV8XL35VBAEI0NV1FBRFKL5EqDYAB1U8XgTFbP8IaTYAJ0jo1OxkmOQ4P0rojyycxtj6lLyc1NiwHn80PH9u3uDXAZHfTP6A0SmIwrSXKZHc6bTjdtW5lSoIADMrABtjRzRybHyVmjAZ1N4AHWMHp+MAVQB9AHsAVws2AHNVJBmp4gYDeEUgA noborder}}
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</WRAP></WRAP></panel>

<panel type="info" title="Exercise 2.5.4 Application of the loaded voltage divider - motor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP>
<imgcaption BildNr16| Sketch of the setup>
</imgcaption> \\
{{drawio>MotorAmSpannungsteiler.svg}}
</WRAP>

You wanted to test a micromotor for a small robot. Using the maximum current and the internal resistance ($R_{\rm M} = 5~\Omega$) you calculate that this can be operated with a maximum of $U_{\rm M, max}=4~\rm V$. A colleague said that you can get $4~\rm V$ using the setup in <imgref BildNr16> from a $9~\rm V$ block battery.
  - First, calculate the maximum current $I_{\rm M,max}$ of the motor.
  - Draw the corresponding electrical circuit with the motor connected as an ohmic resistor.
  - At the maximum current, the motor should be able to deliver a torque of $M_{\rm max}=M(I_{\rm M, max})= 100~\rm mNm$. What torque would the motor deliver if you implement the setup like this? (Assumption: The torque of the motor increases proportionally to the motor current).
  - What might a setup with a potentiometer look like that would actually allow you to set a voltage between $0.5~\rm V$ to $4~\rm V$ on the motor? What resistance value should the potentiometer have?
  - Build and test your circuit in the simulation below. For an introduction to online simulation, see: [[circuit_design:0_tools#online_circuit_simulator]]. \\ You will essentially need the following tips for this setup:
    - Routing connections can be activated via the menu: ''Draw'' >> ''add wire''. Afterward, you have to click on the start point and then drag it to the end mode.
    - Note that connections can only ever be connected at nodes. A red-marked node (e.g. at the $5 ~\Omega$ resistor) indicates that it is not connected. This could be moved one grid step to the left, as there is a node point there.
    - Pressing the ''<ESC>'' key will disable the insertion of components.
    - With a right click on a component it can be copied or values like the resistor can be changed via ''Edit....''

<WRAP>
<imgcaption BildNr83| Simulation for motor setup>
</imgcaption> \\
{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsDsKAc64YMyRWACyRIKQhKE4BQAbiIeiHoQ0wEzuvnnEgCcUITATUATuH7sQnVu0wyuQ0QHM2zSK0bMcANijUA7upYgsG7kfWyz5G+2pgUrJ-tOvmCaRDDxkSACUAUwBnAEsQgBcAQwA7AGMgq3ZIDhRpFI4lSCswKUUXFH0bHKA noborder}}
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</WRAP></WRAP></panel>

<panel type="info" title="Exercise 2.5.5 Examples of the calculation of loaded voltage dividers"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

Exercise on the voltage divider
{{youtube>6NG4uOMDU7Y}}

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<panel type="info" title="Exercise 2.5.6 Example of a loaded voltage divider: Explanation without calculation (German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>hRVOZIb8-6g}}

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<panel type="info" title="Exercise 2.7.1 Circuit Simplification Exercise I (in German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>8nhzwwRYaUI}}

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<panel type="info" title="Exercise 2.7.2 Circuit Simplification Exercise II + III (in German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>QqUQF3ky7gk}}
{{youtube>UmvFJbS21co}}

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<panel type="info" title="Exercise 2.7.3 Circuit Simplification Exercise IV (in German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>SzXWWrPRsDU}}

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<panel type="info" title="Exercise 2.7.4 Circuit Simplification Exercise V (in German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>MhaO6kiB4dk}}

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<panel type="info" title="Exercise 2.7.5 Circuit Simplification Exercise VI (in German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>9eIRRUNba4A}}

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{{page>aufgabe_2.7.7_mit_rechnung&nofooter}}
{{page>aufgabe_2.7.8_mit_rechnung&nofooter}}
{{page>aufgabe_2.7.9&nofooter}}
{{page>aufgabe_2.7.10&nofooter}}


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

<WRAP column half>
Why are voltage dividers important? (a cutout from 0:00 to 10:56 from a full video of EEVblog, starting from 17:00 there is also a nice example for troubles with voltage dividers..)
{{youtube>xSRe_4TQbuo?end=655}}
</WRAP>


~~PAGEBREAK~~ ~~CLEARFIX~~
===== Summary (take-away) =====
  * **Series:** $R$ adds; **Parallel:** $G$ adds.  
  * **Voltage divider**: $U_1= \dfrac{R_1}{R_1+R_2}U$ (unloaded); with load, $U_1$ decreases per $\displaystyle U_1=\frac{U}{1+\frac{R_2}{R_{\rm L}}+\frac{R_2}{R_1}}$.  
  * **Current divider**: branch currents split in proportion to **conductances**.  
  * **Bridge**: balance when products of opposite arms are equal.  
  * **Δ–Y / Y–Δ**: unlocks reductions when pure series/parallel isn’t available.  
All formulas consistent with the colleague’s slides (ee1spo4.pdf ch.2). 

~~PAGEBREAK~~ ~~CLEARFIX~~

===== Preparation for next block =====
Skim **real sources & two-terminal networks** (internal resistance, Thevenin/Norton). Bring questions about **loaded dividers** and **efficiency vs. power transfer** (ties into utilization/impedance matching).