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 18:45] – 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 481: | Zeile 484: | ||
| Using Kirchhoff' | Using Kirchhoff' | ||
| - | $U_1 + U_2 + ... + U_n = U_{\m res}$ | + | $U_1 + U_2 + ... + U_n = U_{\rm res}$ |
| $R_1 \cdot I_1 + R_2 \cdot I_2 + ... + R_n \cdot I_n = R_{\rm eq} \cdot I $ | $R_1 \cdot I_1 + R_2 \cdot I_2 + ... + R_n \cdot I_n = R_{\rm eq} \cdot I $ | ||
| Zeile 490: | Zeile 493: | ||
| __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 566: | Zeile 595: | ||
| < | < | ||
| </ | </ | ||
| - | {{url> | + | {{url> |
| </ | </ | ||
| - | In the simulation in <imgref BildNr81> | + | In the simulation in <imgref BildNr81> |
| - | - What voltage $U_{\rm | + | - What voltage $U_{\rm |
| - 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_{\rm | + | - At which position do you get a $U_{\rm |
| </ | </ | ||
| 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 $10~\rm V$. Thus $k = 0.60$, $R_s = 10~k\Omega$ and $U_1 = 6.0~\rm V$. | + | 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 $R_{\rm L} = R_1 = 10 ~k\Omega$, $k$ reduces to about $0.48$ and $U_1$ reduces to $4.8~\rm V$ - so the output voltage drops. For $R_{\rm L} = 4.0~k\Omega$, | + | 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 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 $3.50~\rm V$ 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. | - 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. | ||
| Zeile 648: | Zeile 677: | ||
| < | < | ||
| </ | </ | ||
| - | {{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 $U_{\rm M, max}=4~\rm V$. A colleague said that you can get $4~\rm V$ 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 $I_{\rm M,max}$ 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. | ||
| Zeile 753: | Zeile 782: | ||
| </ | </ | ||
| - | $R_{/rm ab} = R_{/rm ab}^1 || (R_{/rm ca}^1 + R_{/rm bc}^1) $ \\ | + | $R_{\rm ab} = R_{\rm ab}^1 || (R_{\rm ca}^1 + R_{\rm 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}} $ \\ | + | $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_{/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} = {{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} = {{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 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 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} | + | 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. | 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: | + | 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 $R_{/rm c0}^1$. | + | $R_{\rm a0}^1$, $R_{\rm b0}^1$ and $R_{\rm c0}^1$. |
| - | Again, the procedure is the same as for the delta connection: the resistance between two terminals (e.g. $/rm a$ and $/rm b$) is determined, and the further terminal ($/rm c$) is considered to be open. | + | 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: | + | 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_{/rm ab} = R_{/rm a0}^1 + R_{/rm b0}^1 \\ | + | R_{\rm ab} = R_{\rm a0}^1 + R_{\rm b0}^1 \\ |
| - | R_{/rm bc} = R_{/rm b0}^1 + R_{/rm c0}^1 \\ | + | R_{\rm bc} = R_{\rm b0}^1 + R_{\rm c0}^1 \\ |
| - | R_{/rm ca} = R_{/rm c0}^1 + R_{/rm a0}^1 \tag{2.6.2} | + | 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_{/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_{\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_{/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_{\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_{/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} | + | \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 ${{1}\over{2}} \cdot \left(R_{/rm ab} + R_{/rm bc} - R_{/rm ca}\right)$ to combine. This gives $R_{/rm b0}^1$ \\ | + | 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_{/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}} | + | {{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 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) &= | + | - {{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( |
| - | {{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_{\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)} |
| - | - {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) &= | + | - {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( 2 \cdot R_{/rm b0}^1 \right) \\ | + | {{1}\over{2}} \cdot \left( 2 \cdot R_{\rm b0}^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( {{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_{/rm ab}^1 R_{/rm bc}^1 }\over{R_{/rm ab}^1 + R_{/rm ca}^1 + R_{/rm 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_{/rm ab}^1 R_{/rm bc}^1 }\over{R_{/rm ab}^1 + R_{/rm ca}^1 + R_{/rm 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 $R_{/rm a0}^1$ and $R_{/rm c0}^1$, and with a slightly modified approach to $R_{/rm ab}^1$, $R_{/rm bc}^1$ and $R_{/rm ca}^1$. | + | 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 821: | Zeile 850: | ||
| \text{therefore: | \text{therefore: | ||
| - | 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_{\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_{/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_{\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_{/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}} | + | 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 836: | Zeile 865: | ||
| \text{therefore: | \text{therefore: | ||
| - | 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_{\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_{/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_{\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_{/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}} | + | 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 877: | Zeile 906: | ||
| 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 $R_{/rm ab}$. | + | 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 900: | Zeile 929: | ||
| \begin{align*} | \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_{\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}\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 936: | 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 995: | Zeile 1024: | ||
| </ | </ | ||
| - | <panel type=" | ||
| - | More German exercises can be found online on the pages of [[https:// | ||
| - | </ | ||