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electrical_engineering_1:simple_circuits [2023/03/19 18:46] – mexleadminelectrical_engineering_1:simple_circuits [2024/10/24 08:13] (aktuell) – mexleadmin
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-====== 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. \\ 
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 {{youtube>VojwBoSHc8U}} {{youtube>VojwBoSHc8U}}
 </WRAP> </WRAP>
 +\\ \\
 +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's current law. \\ +$\large{{I_1}\over{I_{\rm res}} = {{G_1}\over{G_{\rm res}}}$ 
-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 by 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 also $R_1 \cdot I_1 = R_2 \cdot I_2 = ... = R_{eq} \cdot I_{res}$. \\ +The rule also be derived from Kirchhoff's current law: \\ 
-Therefore, we get with the conductance: ${{I_1} \over {G_1}} = {{I_2} \over {G_2}}= ... = {{I_{eq}} \over {G_{res}}}$+  - 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~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
  
 +<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%> <panel type="info" title="Exercise 2.4.1 Current divider"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
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 In the simulation in <imgref BildNr85> a current divider can be seen. The resistances are just inversely proportional to the currents flowing through it. 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 $5V$ to $3.3V$? __After__ thinking about your result, you can adjust the ''Voltage'' (bottom right of the simulation) accordingly by moving the slider.+  - 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?   - 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?
  
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 Using Kirchhoff's voltage law, the total resistance of a series circuit (in German: //Reihenschaltung//, see <imgref BildNr13>) can be easily determined: Using Kirchhoff's voltage law, the total resistance of a series circuit (in German: //Reihenschaltung//, see <imgref BildNr13>) can be easily determined:
  
-$U_1 + U_2 + ... + U_n = U_{\m res}$+$U_1 + U_2 + ... + U_n = U_{\rm res}$
  
 $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 $
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 __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>Four-terminal sensing}} or 4 wire sensing, is used.
 +This is a method of measuring electrical resistance avoiding errors caused by wire resistances. \\
 +The simulation in <imgref BildNr005> shows such a setup.
 +
 +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.
 +
 +<WRAP>
 +<imgcaption BildNr005 | Example of a circuit>
 +</imgcaption> \\
 +{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsAsYBsAOdAmSYBOdAdkwUkxRB2JCRQGY6BTAWjDACgAnEIkFJnBgcAoZGRdeOBOmGiZcsCioSp86hT5y84yZwDuGwdVljqh00pVXN4y4rsbdFo444KUE-pAdnlVDhe4OgSvgBuIOzEOlrRciZhAmqiSTAIlqyYEi78Lr4ADlHZ5lk5WkxhmWAxTmVOONW1JvUmjUXecmqhIJVQPNqDHoNh0FxG-D4jA-wYaiIhaWOW3Z2L-UatQvyJDsFzGge+bsGzCz6cAB5RHIEoogy0OJi0VGBCAGoA9gA2AC4AW2Yf2Y3AAOgBnACWADtIWCwTCcKwDFDuMxIRDmDDoTCAOaQgCGMIAJgiYSgUWiMRCsTjYXirnwIEEFBxqOgEuAhKj0RDIai-gALeGI3k0vlQiF-YkAY2YTJwDCYrI5BGoKHo3JAABEAKoAFU4jxyZja-gWECqeJutRc8WoQSg-WuCCQDCQJF6QjeQgASkz0OqGOrKAJRG9RLKflxrkHNBIw3gmJGQNHGq6kIQ5F7COrfSAA3HQ7QwwQJKnwjGmW6QJgmF7sAJtUW6FmTLmEBAC6342AEKIw-2tQsqxmOUJ+yqSv2ue9C4rMKJ+09MJOeqmIZwgA noborder}}
 +</WRAP>
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
Zeile 566: Zeile 595:
 <imgcaption BildNr81| unloaded voltage divider> <imgcaption BildNr81| unloaded voltage divider>
 </imgcaption> \\ </imgcaption> \\
-{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsAsB2ATAZgGw4QhmimJGOiApJSClgFABu4O1RGL1YaOUfK1JNWHQE9bihBZIk7rywY2CXhFJxkSAEoBTAM4BLXQBcAhgDsAxtvoB3EIoAc9zFJnOOkW68nspDny6edgpsGJJY-vZhUPQADn4+0REBHBDCcfaQTnKZTr5pMXZgrOA8uaW8QZzu5b6eAObeNdI+Dk6eimgJUZICob0gADq6w6O6AKoA+gD2AK5GXo4VPctVLcv5ZZ66KzmbleAgAGYmADa61kA noborder}}+{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsAsB2ATAZgGw4QhmimJGOiApJSClpQKYC0YYAUAG7g7VEbfUwaHFFEpqSalOgI2QlCCyQFQkVgy8EIiKTjIkAJQYBnAJbGALgEMAdgGMGbAO4gNADleZFyz-0jPvBT5FNyCvfxd1XgwFLFDXGKg2AAcQoMS4sP4IKRTXSA9VfI9gnKSXMB5wYWLqkQiBX1rg-wBzQKalILcPXI00NISFEhEMRIUAHWMpmeMAVQB9AHkA9zqh9Yau9dKa-2MNot368BAAMysAG2NHIA noborder}}
 </WRAP> </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 $1~k\Omega$. In the configuration shown, this is divided into $500 ~\Omega$ and $500 ~\Omega$. +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 $U_{\rm our}$ 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 ("intermediate terminal"). \\ You can check your assumption by using the slider at the bottom right of the simulation.   - 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 $U_{\rm out} = 3.5~\rm V$?+  - At which position do you get a $U_{\rm O} = 3.5~\rm V$?
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
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 </WRAP> </WRAP>
  
-What does this diagram tell us? This shall be investigated by an example. First, assume an unloaded voltage divider with $R_2 = 4.0 ~k\Omega$ 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 $R_{\rm L} = 4.0~k\Omega$, $k$ 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.+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: 
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 <panel type="info" title="Exercise 2.5.3 loaded voltage divider II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <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 $R_{\rm L} = 1.00 ~k\Omega$. +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_OUT'' 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.+  - 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.   - 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.
  
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 <imgcaption BildNr82| loaded voltage divider> <imgcaption BildNr82| loaded voltage divider>
 </imgcaption> \\ </imgcaption> \\
-{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAN3BppExRbbA0Kn5pUkVEdAT0ebXN14hsadgj4Qw8KGJAAlAKYBnAJa6ALgEMAdgGNt9AO7t87DJ2mPOkW6-CyXPPu7suHHIInEH+cgooKFIh7NFQ9AAOcnhsUVKpnhAi9ABOwaHx2LFhyHAexaFOKWnV4ZVxGWnx4WCsnmh4nq3tvvZdfe7JnV58I0HZCXYjfemjCbqN8xOyVBAAZiYANrrWaOTYmXOCVMcgADq6l9e6AKoA+gD2AK5G9ADmcpC1zt9fbO4gA noborder}}+{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3cGmkTFbrzAZCUMTSpIq06AjbDeuISJDY0fBKIhh4UWSABKDAM4BLYwBcAhgDsAxgzYB3Pvj4YBS9wMjPv4FS9hUV8XL35VBAEI0NV1FBRFKL5EqDYAB1U8XgTFbP8IaTYAJ0jo1OxkmOQ4P0rojyycxtj6lLyc1NiwHn80PH9u3uDXAZHfTP6A0SmIwrSXKZHc6bTjdtW5lSoIADMrABtjRzRybHyVmjAZ1N4AHWMHp+MAVQB9AHsAVws2AHNVJBmp4gYDeEUgA noborder}}
 </WRAP> </WRAP>
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
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 </WRAP> </WRAP>
  
-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> from a $9~\rm V$ block battery.+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.   - 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.
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 \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*}
  
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="other Exercises"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
  
-More German exercises can be found online on the pages of [[https://www.eit.hs-karlsruhe.de/hertz/teil-b-gleichstromtechnik/zusammenschaltung-von-widerstaenden-und-idealen-quellen/uebungsaufgaben-zusammenschaltung-von-widerstaenden/berechnung-von-ersatzwiderstaenden.html|HErTZ]] (selection on the left in the menu). 
-</WRAP></WRAP></panel>