Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
| electrical_engineering_1:simple_circuits [2023/03/19 18:19] – mexleadmin | electrical_engineering_1:simple_circuits [2024/10/24 08:13] (aktuell) – mexleadmin | ||
|---|---|---|---|
| Zeile 1: | Zeile 1: | ||
| - | ====== 2. Simple DC circuits ====== | + | ====== 2 Simple DC circuits ====== |
| So far, only simple circuits consisting of a source and a load connected by wires have been considered. \\ | So far, only simple circuits consisting of a source and a load connected by wires have been considered. \\ | ||
| Zeile 376: | Zeile 376: | ||
| {{youtube> | {{youtube> | ||
| </ | </ | ||
| + | \\ \\ | ||
| + | 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. \\ | ||
| - | The current divider rule can also be derived from Kirchhoff' | + | $\large{{I_1}\over{I_{\rm res}} = {{G_1}\over{G_{\rm res}}}$ |
| - | 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. \\ | + | |
| - | + | ||
| - | $\large{{I_1}\over{I_g}} = {{G_1}\over{G_g}}$ | + | |
| $\large{{I_1}\over{I_2}} = {{G_1}\over{G_2}}$ | $\large{{I_1}\over{I_2}} = {{G_1}\over{G_2}}$ | ||
| - | This can also be derived | + | The rule also be derived |
| - | Therefore, we get with the conductance: | + | - 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: $R_1 \cdot I_1 = R_2 \cdot I_2 = ... = R_{\rm eq} \cdot I_{\rm res}$. \\ | ||
| + | | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| + | <wrap anchor # | ||
| <panel type=" | <panel type=" | ||
| Zeile 400: | Zeile 403: | ||
| In the simulation in <imgref BildNr85> | In the simulation in <imgref BildNr85> | ||
| - | - What currents would you expect in each branch if the input voltage were lowered from $5V$ to $3.3V$? __After__ thinking about your result, you can adjust the '' | + | - 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 '' |
| - 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? | - 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? | ||
| Zeile 407: | Zeile 410: | ||
| <panel type=" | <panel type=" | ||
| - | Two resistors of $18~\Omega$ and $2~\Omega$ are connected in parallel. The total current of the resistors is $3~A$. \\ | + | 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. | Calculate the total resistance and how the currents are split to the branches. | ||
| Zeile 418: | Zeile 421: | ||
| The current through resistor $R_1$ is | The current through resistor $R_1$ is | ||
| \begin{equation*} | \begin{equation*} | ||
| - | I_1 = \frac{R_{eq}}{R_1} I =\frac{1.8~\Omega}{18~\Omega} \cdot 3~A | + | I_1 = \frac{R_{eq}}{R_1} I =\frac{1.8~\Omega}{18~\Omega} \cdot 3~\rm A |
| \end{equation*} | \end{equation*} | ||
| The current through resistor $R_2$ is | The current through resistor $R_2$ is | ||
| \begin{equation*} | \begin{equation*} | ||
| - | I_2 = \frac{R_{eq}}{R_2}I = \frac{1.8~\Omega}{2~\Omega} \cdot 3~A | + | I_2 = \frac{R_{eq}}{R_2}I = \frac{1.8~\Omega}{2~\Omega} \cdot 3~\rm A |
| \end{equation*} | \end{equation*} | ||
| </ | </ | ||
| Zeile 428: | Zeile 431: | ||
| The values of the substitute resistor and the currents in the branches are | The values of the substitute resistor and the currents in the branches are | ||
| \begin{equation*} | \begin{equation*} | ||
| - | R_{eq} = 1.8~\Omega \qquad I_1 = 0.3~A \qquad I_2 = 2.7~A | + | R_{eq} = 1.8~\Omega \qquad I_1 = 0.3~{\rm A} \qquad I_2 = 2.7~\rm A |
| \end{equation*} | \end{equation*} | ||
| </ | </ | ||
| Zeile 479: | Zeile 482: | ||
| </ | </ | ||
| - | Using Kirchhoff' | + | Using Kirchhoff' |
| - | $U_1 + U_2 + ... + U_n = U_g$ | + | $U_1 + U_2 + ... + U_n = U_{\rm res}$ |
| - | $R_1 \cdot I_1 + R_2 \cdot I_2 + ... + R_n \cdot I_n = R_{eq} \cdot I $ | + | $R_1 \cdot I_1 + R_2 \cdot I_2 + ... + R_n \cdot I_n = R_{\rm eq} \cdot I $ |
| Since in a series circuit, the current through all resistors must be the same - i.e. $I_1 = I_2 = ... = I$ - it follows that: | Since in a series circuit, 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 $ | + | $R_1 + R_2 + ... + R_n = R_{\rm eq} = \sum_{x=1}^{n} R_x $ |
| __In general__: The equivalent resistance of a series circuit is always greater than the greatest resistance. | __In general__: The equivalent resistance of a series circuit is always greater than the greatest resistance. | ||
| + | |||
| + | ==== Application ==== | ||
| + | |||
| + | === Kelvin-Sensing === | ||
| + | |||
| + | Often resistors are used to measure a current $I$ via the voltage drop on the resistor $U = R \cdot I$. Applications include the measurement of motor currents in the range of $0.1 ... 500 ~\rm A$. \\ | ||
| + | Those resistors are called //shunt resistors// and are commonly in the range of some $\rm m\Omega$. | ||
| + | This measurement can be interfered by the resistor of the supply lines. | ||
| + | |||
| + | To get an accurate measurement often Kelvin sensing, also known as {{wp> | ||
| + | This is a method of measuring electrical resistance avoiding errors caused by wire resistances. \\ | ||
| + | The simulation in <imgref BildNr005> | ||
| + | |||
| + | Four-terminal sensing involves using: | ||
| + | * a pair of //current leads// or //force leads// (with the resistances $R_{\rm cl1}$ and $R_{\rm cl2}$) to supply current to the circuit and | ||
| + | * a pair of //voltage leads// or //sense leads// (with the resistances $R_{\rm vl1}$ and $R_{\rm vl2}$) to measure the voltage drop across the impedance to be measured. | ||
| + | The sense connections via the voltage leads are made immediately adjacent to the target impedance $R_{\rm s}$ at the device under test $\rm DUT$. | ||
| + | By this, they do not include the voltage drop in the force leads or contacts. \\ | ||
| + | Since almost no current flows to the measuring instrument, the voltage drop in the sense leads is negligible. | ||
| + | This method can be a practical tool for finding poor connections or unexpected resistance in an electrical circuit. | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{url> | ||
| + | </ | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| Zeile 500: | Zeile 529: | ||
| The resistors can be connected in series: | The resistors can be connected in series: | ||
| \begin{equation*} | \begin{equation*} | ||
| - | R_{series} = 3\cdot R = 3\cdot20~k\Omega | + | R_{\rm series} = 3\cdot R = 3\cdot20~k\Omega |
| \end{equation*} | \end{equation*} | ||
| The resistors can also be connected in parallel: | The resistors can also be connected in parallel: | ||
| \begin{equation*} | \begin{equation*} | ||
| - | R_{parallel} = \frac{R}{3} = \frac{20~k\Omega}{3} | + | R_{\rm parallel} = \frac{R}{3} = \frac{20~k\Omega}{3} |
| \end{equation*} | \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: | 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*} | \begin{equation*} | ||
| - | R_{total} = R+\frac{R\cdot R}{R+R} = \frac{3}{2}R = \frac{3}{2} \cdot 20~k\Omega | + | R_{\rm res} = R + \frac{R\cdot R}{R+R} = \frac{3}{2}R = \frac{3}{2} \cdot 20~k\Omega |
| \end{equation*} | \end{equation*} | ||
| </ | </ | ||
| <button size=" | <button size=" | ||
| \begin{equation*} | \begin{equation*} | ||
| - | R_{series} = 60~k\Omega\qquad R_{parallel} = 6.7~k\Omega\qquad R_{total} = 30~k\Omega | + | R_{series} = 60~k\Omega\qquad R_{\rm parallel} = 6.7~k\Omega\qquad R_{\rm res} = 30~k\Omega |
| \end{equation*} | \end{equation*} | ||
| </ | </ | ||
| Zeile 566: | Zeile 595: | ||
| < | < | ||
| </ | </ | ||
| - | {{url> | + | {{url> |
| </ | </ | ||
| - | In the simulation in <imgref BildNr81> | + | In the simulation in <imgref BildNr81> |
| - | - What voltage $U_{out}$ would you expect if the switch were closed? After thinking about your result, you can check it by closing the switch. | + | - 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 (" | - First, think about what would happen if you would change the distribution of the resistors by moving the wiper (" | ||
| - | - At which position do you get a $U_{out} = 3.5~V$? | + | - At which position do you get a $U_{\rm O} = 3.5~\rm V$? |
| </ | </ | ||
| Zeile 578: | Zeile 607: | ||
| ==== The loaded Voltage Divider ==== | ==== The loaded Voltage Divider ==== | ||
| - | If - in contrast to the abovementioned, | + | If - in contrast to the abovementioned, |
| < | < | ||
| Zeile 590: | Zeile 619: | ||
| $ U_1 = \LARGE{{U} \over {1 + {{R_2}\over{R_L}} + {{R_2}\over{R_1}} }}$ | $ 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 $R_s = R_1 + R_2$: | + | 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_s}\over{R_L}} }}$ | + | $ U_1 = \LARGE{{k \cdot U} \over { 1 + k \cdot (1-k) \cdot{{R_{\rm s}}\over{R_{\rm L}}} }}$ |
| <imgref BildNr65> | <imgref BildNr65> | ||
| - | In principle, this is similar to <imgref BildNr14>, | + | In principle, this is similar to <imgref BildNr14>, |
| < | < | ||
| Zeile 603: | Zeile 632: | ||
| </ | </ | ||
| - | What does this diagram tell us? This shall be investigated by an example. First, assume an unloaded voltage divider with $R_2 = 4.0 ~k\Omega$ and $R_1 = 6.0 ~k\Omega$, and an input voltage of $10~V$. Thus $k = 0.60$, $R_s = 10~k\Omega$ and $U_1 = 6.0~V$. | + | 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 $R_L = R_1 = 10 ~k\Omega$, $k$ reduces to about $0.48$ and $U_1$ reduces to $4.8~V$ - so the output voltage drops. For $R_L = 4.0~k\Omega$, | + | 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: | What is the practical use of the (loaded) voltage divider? \\ Here are some examples: | ||
| Zeile 614: | Zeile 643: | ||
| <panel type=" | <panel type=" | ||
| - | Determine from the circuit in <imgref BildNr15> | + | Determine from the circuit in <imgref BildNr15> |
| <button size=" | <button size=" | ||
| According to the voltage division rule, the loaded voltage is | According to the voltage division rule, the loaded voltage is | ||
| \begin{align*} | \begin{align*} | ||
| - | U_1 & | + | U_1 & |
| - | & | + | & |
| - | & | + | & |
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| - | The divided resistor $R_1$ and $R_2$ are put together to form $R_S=R_1 + R_2$. | + | The divided resistor $R_1$ and $R_2$ are put together to form $R_{\rm s}=R_1 + R_2$. |
| \begin{equation*} | \begin{equation*} | ||
| - | U_1=\frac{R_1 | + | U_1=\frac{R_1 |
| \end{equation*} | \end{equation*} | ||
| - | With the equations given there is also $R_1=k(R_1+R_2)=k | + | With the equations given there is also $R_1=k(R_1+R_2)=k |
| \begin{equation*} | \begin{equation*} | ||
| - | U_1=\frac{k | + | U_1=\frac{k |
| \end{equation*} | \end{equation*} | ||
| - | Dividing the numerator and denominator by $R_S R_L$ yields to | + | Dividing the numerator and denominator by $R_{\rm s} R_{\rm L}$ yields to |
| \begin{equation*} | \begin{equation*} | ||
| - | U_1=\frac{k}{k(1-k)\frac{R_S}{R_L}+1}U | + | U_1=\frac{k}{k(1-k)\frac{R_{\rm s}}{R_{\rm L}}+1}U |
| \end{equation*} | \end{equation*} | ||
| </ | </ | ||
| Zeile 641: | Zeile 670: | ||
| <panel type=" | <panel type=" | ||
| - | In the simulation in <imgref BildNr82> | + | In the simulation in <imgref BildNr82> |
| - | - What voltage '' | + | - What voltage '' |
| - | - At which position of the wiper do you get $3.50~V$ 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. | + | - 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. |
| < | < | ||
| < | < | ||
| </ | </ | ||
| - | {{url> | + | {{url> |
| </ | </ | ||
| </ | </ | ||
| Zeile 660: | Zeile 689: | ||
| </ | </ | ||
| - | You wanted to test a micromotor for a small robot. Using the maximum current and the internal resistance ($R_M = 5~\Omega$) you calculate that this can be operated with a maximum of $U_{M, max}=4~V$. A colleague said that you can get $4~V$ using the setup in <imgref BildNr16> | + | 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> |
| - | - First, calculate the maximum current $I_{M,max}$ of the motor. | + | - 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. | - 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_{max}=M(I_{M, | + | - 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: |
| - | - What might a setup with a potentiometer look like that would actually allow you to set a voltage between $0.5~V$ to $4~V$ on the motor? What resistance value should the potentiometer have? | + | - 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: | - Build and test your circuit in the simulation below. For an introduction to online simulation, see: [[circuit_design: | ||
| - Routing connections can be activated via the menu: '' | - Routing connections can be activated via the menu: '' | ||
| Zeile 704: | Zeile 733: | ||
| </ | </ | ||
| - | At the beginning of the chapter, an example of a network was shown (<imgref BildNr91> | + | At the beginning of the chapter, an example of a network was shown (<imgref BildNr91> |
| + | It is visible, that there are many $\Delta$-shaped (or triangle-shaped) loops resp. $\rm Y$-shaped (or star-shaped) nodes, see <imgref BildNr98> | ||
| + | A method to calculate these will be discussed in more detail now. | ||
| < | < | ||
| Zeile 725: | Zeile 756: | ||
| In other words: The resistances measured between two terminals must be identical for the black box and for the known circuit. | In other words: The resistances measured between two terminals must be identical for the black box and for the known circuit. | ||
| - | For this purpose, the different resistances between the individual nodes $a$, $b$, and $c$ are now to be considered, see <imgref BildNr18> | + | For this purpose, the different resistances between the individual nodes $\rm a$, $\rm b$, and $\rm c$ are now to be considered, see <imgref BildNr18> |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| Zeile 741: | Zeile 772: | ||
| ==== Delta Circuit ==== | ==== Delta Circuit ==== | ||
| - | In the delta circuit, the 3 resistors $R_{ab}^1$, $R_{bc}^1$, and $R_{ca}^1$ are connected in a loop. At the connection of the resistors, an additional terminal is implemented. \\ | + | In the delta circuit, the 3 resistors $R_{\rm ab}^1$, $R_{\rm bc}^1$, and $R_{\rm ca}^1$ are connected in a loop. At the connection of the resistors, an additional terminal is implemented. \\ |
| The labeling with a superscript $\square^1$ refers to the three resistors in the next paragraphs. | The labeling with a superscript $\square^1$ refers to the three resistors in the next paragraphs. | ||
| - | For the measurable resistance between two terminals (e.g. $R_{ab}$ between $a$ and $b$), the third terminal (here: $c$) is considered as not connected to anything outside. This results in a parallel circuit of the direct delta resistor $R_{ab}^1$ with the series connection of the other two delta resistors $R_{ca}^1 + R_{bc}^1$: | + | For the measurable resistance between two terminals (e.g. $R_{\rm ab}$ between $\rm a$ and $\rm b$), the third terminal (here: $\rm c$) is considered as not connected to anything outside. This results in a parallel circuit of the direct delta resistor $R_{\rm ab}^1$ with the series connection of the other two delta resistors $R_{\rm ca}^1 + R_{\rm bc}^1$: |
| < | < | ||
| Zeile 751: | Zeile 782: | ||
| </ | </ | ||
| - | $R_{ab} = R_{ab}^1 || (R_{ca}^1 + R_{bc}^1) $ \\ | + | $R_{\rm ab} = R_{\rm ab}^1 || (R_{\rm ca}^1 + R_{\rm 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}} $ \\ | + | $R_{\rm ab} = {{R_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)}\over{R_{\rm ab}^1 + (R_{\rm ca}^1 + R_{\rm bc}^1)}} = {{R_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} $ \\ |
| The same applies to the other connections. This results in: | The same applies to the other connections. This results in: | ||
| \begin{align*} | \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_{\rm ab} = {{R_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm 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_{\rm bc} = {{R_{\rm bc}^1 \cdot (R_{\rm ab}^1 + R_{\rm ca}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm 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} | + | R_{\rm ca} = {{R_{\rm ca}^1 \cdot (R_{\rm bc}^1 + R_{\rm ab}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \tag{2.6.1} |
| ==== Star Circuit ==== | ==== Star Circuit ==== | ||
| - | Given the idea, that the star circuit shall behave equally to the delta circuit, the resistance measured between the terminals must be similar. Also in the star circuit, 3 resistors are connected, but now in a star (or $Y$) shape. The star resistors are all connected with another node $0$ in the middle: $R_{a0}^1$, $R_{b0}^1$ and $R_{c0}^1$. | + | Given the idea, that the star circuit shall behave equally to the delta circuit, the resistance measured between the terminals must be similar. |
| + | Also in the star circuit, 3 resistors are connected, but now in a star (or $\rm Y$) shape. The star resistors are all connected with another node $0$ in the middle: | ||
| + | $R_{\rm a0}^1$, $R_{\rm b0}^1$ and $R_{\rm c0}^1$. | ||
| - | 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 to be 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: | + | Again, the procedure is the same as for the delta connection: the resistance between two terminals (e.g. $\rm a$ and $\rm b$) is determined, and the further terminal ($\rm c$) is considered to be open. |
| + | The resistance of the further terminal ($R_{\rm c0}^1$) is only connected at one node. Therefore, no current flows through it - it is thus not to be considered. It results in: | ||
| \begin{align*} | \begin{align*} | ||
| - | R_{ab} = R_{a0}^1 + R_{b0}^1 | + | R_{\rm ab} = R_{\rm a0}^1 + R_{\rm b0}^1 \\ |
| - | R_{bc} = R_{b0}^1 + R_{c0}^1 | + | R_{\rm bc} = R_{\rm b0}^1 + R_{\rm c0}^1 \\ |
| - | R_{ca} = R_{c0}^1 + R_{a0}^1 | + | R_{\rm ca} = R_{\rm c0}^1 + R_{\rm a0}^1 \tag{2.6.2} |
| \end{align*} | \end{align*} | ||
| From equations $(2.6.1)$ and $(2.6.2)$ we get: | 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_{\rm ab} = {{R_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} = R_{\rm a0}^1 + R_{\rm 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_{\rm bc} = {{R_{\rm bc}^1 \cdot (R_{\rm ab}^1 + R_{\rm ca}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} = R_{\rm b0}^1 + R_{\rm 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} | + | \begin{align} R_{\rm ca} = {{R_{\rm ca}^1 \cdot (R_{\rm bc}^1 + R_{\rm ab}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} = R_{\rm c0}^1 + R_{\rm a0}^1 \tag{2.6.5} \end{align} |
| Equations $(2.6.3)$ to $(2.6.5)$ can now be cleverly combined so that there is only one resistor on one side. \\ | Equations $(2.6.3)$ to $(2.6.5)$ can now be cleverly combined so that there is only one resistor on one side. \\ | ||
| - | A variation is to write the formulas as ${{1}\over{2}} \cdot \left( (2.6.3) + (2.6.4) - (2.6.5) \right)$ or ${{1}\over{2}} \cdot \left(R_{ab} + R_{bc} - R_{ca}\right)$ to combine. This gives $R_{b0}^1$ \\ | + | A variation is to write the formulas as ${{1}\over{2}} \cdot \left( (2.6.3) + (2.6.4) - (2.6.5) \right)$ or ${{1}\over{2}} \cdot \left(R_{\rm ab} + R_{\rm bc} - R_{\rm ca}\right)$ to combine. This gives $R_{\rm b0}^1$ \\ |
| \begin{align*} | \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_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} |
| + | + {{R_{\rm bc}^1 \cdot (R_{\rm ab}^1 + R_{\rm ca}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} | ||
| + | - {{R_{\rm ca}^1 \cdot (R_{\rm bc}^1 + R_{\rm ab}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \right) & | ||
| + | {{1}\over{2}} \cdot \left( | ||
| - | {{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 | + | {{1}\over{2}} \cdot \left( {{R_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)} + {R_{\rm bc}^1 \cdot (R_{\rm ab}^1 + R_{\rm ca}^1)} |
| + | | ||
| + | | ||
| - | {{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) & | + | {{1}\over{2}} \cdot \left( {{R_{\rm ab}^1 R_{\rm ca}^1 + R_{\rm ab}^1 R_{\rm bc}^1 + R_{\rm bc}^1 R_{\rm ab}^1 + R_{\rm bc}^1 R_{\rm ca}^1 - R_{\rm ca}^1 R_{\rm bc}^1 - R_{\rm ca}^1 R_{\rm ab}^1}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \right) & |
| - | {{1}\over{2}} \cdot \left( {{ 2 \cdot R_{ab}^1 R_{bc}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) & | + | {{1}\over{2}} \cdot \left( {{ 2 \cdot R_{\rm ab}^1 R_{\rm bc}^1 }\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \right) & |
| - | {{ R_{ab}^1 R_{bc}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} & | + | {{ R_{\rm ab}^1 R_{\rm bc}^1 }\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} & |
| \end{align*} | \end{align*} | ||
| - | Similarly, one can resolve to $R_{a0}^1$ and $R_{c0}^1$, and with a slightly modified approach to $R_{ab}^1$, $R_{bc}^1$ and $R_{ca}^1$. | + | Similarly, one can resolve to $R_{\rm a0}^1$ and $R_{\rm c0}^1$, and with a slightly modified approach to $R_{\rm ab}^1$, $R_{\rm bc}^1$ and $R_{\rm ca}^1$. |
| ==== Y-Δ-Transformation | ==== Y-Δ-Transformation | ||
| Zeile 811: | Zeile 850: | ||
| \text{therefore: | \text{therefore: | ||
| - | R_{a0}^1 &= {{ R_{ca}^1 \cdot R_{ab}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \\ | + | R_{\rm a0}^1 &= {{ R_{\rm ca}^1 \cdot R_{\rm ab}^1 }\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \\ |
| - | R_{b0}^1 &= {{ R_{ab}^1 \cdot R_{bc}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \\ | + | R_{\rm b0}^1 &= {{ R_{\rm ab}^1 \cdot R_{\rm bc}^1 }\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \\ |
| - | R_{c0}^1 &= {{ R_{bc}^1 \cdot R_{ca}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} | + | R_{\rm c0}^1 &= {{ R_{\rm bc}^1 \cdot R_{\rm ca}^1 }\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} |
| \end{align*} | \end{align*} | ||
| Zeile 826: | Zeile 865: | ||
| \text{therefore: | \text{therefore: | ||
| - | R_{ab}^1 &= {{ R_{a0}^1 \cdot R_{b0}^1 +R_{b0}^1 \cdot R_{c0}^1 +R_{c0}^1 \cdot R_{a0}^1 }\over{ R_{c0}^1}} \\ | + | R_{\rm ab}^1 &= {{ R_{\rm a0}^1 \cdot R_{\rm b0}^1 +R_{\rm b0}^1 \cdot R_{\rm c0}^1 +R_{\rm c0}^1 \cdot R_{\rm a0}^1 }\over{ R_{\rm c0}^1}} \\ |
| - | R_{bc}^1 &= {{ R_{a0}^1 \cdot R_{b0}^1 +R_{b0}^1 \cdot R_{c0}^1 +R_{c0}^1 \cdot R_{a0}^1 }\over{ R_{a0}^1}} \\ | + | R_{\rm bc}^1 &= {{ R_{\rm a0}^1 \cdot R_{\rm b0}^1 +R_{\rm b0}^1 \cdot R_{\rm c0}^1 +R_{\rm c0}^1 \cdot R_{\rm a0}^1 }\over{ R_{\rm a0}^1}} \\ |
| - | R_{ca}^1 &= {{ R_{a0}^1 \cdot R_{b0}^1 +R_{b0}^1 \cdot R_{c0}^1 +R_{c0}^1 \cdot R_{a0}^1 }\over{ R_{b0}^1}} | + | R_{\rm ca}^1 &= {{ R_{\rm a0}^1 \cdot R_{\rm b0}^1 +R_{\rm b0}^1 \cdot R_{\rm c0}^1 +R_{\rm c0}^1 \cdot R_{\rm a0}^1 }\over{ R_{\rm b0}^1}} |
| \end{align*} | \end{align*} | ||
| Zeile 866: | Zeile 905: | ||
| ==== Simple Example ==== | ==== Simple Example ==== | ||
| - | An example of such a circuit is given in <imgref imageNo89> | + | An example of such a circuit is given in <imgref imageNo89> |
| + | 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 $R_{\rm ab}$. | ||
| < | < | ||
| Zeile 874: | Zeile 914: | ||
| </ | </ | ||
| - | As already described in the previous subchapters, | + | As already described in the previous subchapters, |
| + | It is important to note that these partial circuits for conversion into equivalent resistors may only ever have two connections (= two nodes to the " | ||
| Zeile 884: | Zeile 925: | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | <imgref imageNo88 > shows the step-by-step conversion of the equivalent resistors in this example. \\ As a result of the equivalent resistance one gets: | + | <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*} | \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_{\rm |
| &= {{ {{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}\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)}} \\ | ||
| Zeile 923: | Zeile 965: | ||
| \begin{align*} | \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 | + | R_{\rm eq} = 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*} | \end{align*} | ||
| Zeile 982: | Zeile 1024: | ||
| </ | </ | ||
| - | <panel type=" | ||
| - | More German exercises can be found online on the pages of [[https:// | ||
| - | </ | ||