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electrical_engineering_and_electronics_2:block12 [2026/06/02 03:27] – mexleadminelectrical_engineering_and_electronics_2:block12 [2026/06/10 03:06] (current) – mexleadmin
Line 86: Line 86:
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm E}+U_{\rm I}
 = =
 U_R+U_{\rm D}. U_R+U_{\rm D}.
Line 106: Line 106:
 I_{\rm D} I_{\rm D}
 \approx \approx
-\frac{U_{\rm E}-U_{\rm TO}}{R}.+\frac{U_{\rm I}-U_{\rm TO}}{R}.
 \end{align*} \end{align*}
 \] \]
Line 120: Line 120:
 The required forward voltage depends on the semiconductor material and therefore on the color. The required forward voltage depends on the semiconductor material and therefore on the color.
  
-For a supply voltage \(U_{\rm E}\), an LED forward voltage \(U_{\rm F}\), and a desired LED current \(I_{\rm F}\), the series resistor is+For a supply voltage \(U_{\rm I}\), an LED forward voltage \(U_{\rm F}\), and a desired LED current \(I_{\rm F}\), the series resistor is
  
 \[ \[
 \begin{align*} \begin{align*}
 \boxed{ \boxed{
-R_{\rm V} = \frac{U_{\rm E}-U_{\rm F}}{I_{\rm F}}+R_{\rm V} = \frac{U_{\rm I}-U_{\rm F}}{I_{\rm F}}
 } }
 \end{align*} \end{align*}
Line 139: Line 139:
 \[ \[
 \begin{align*} \begin{align*}
-P_R = (U_{\rm E}-U_{\rm F})I_{\rm F} = R_{\rm V}I_{\rm F}^2 \quad \overset{!}{<} \quad P_{R, \rm max}+P_R = (U_{\rm I}-U_{\rm F})I_{\rm F} = R_{\rm V}I_{\rm F}^2 \quad \overset{!}{<} \quad P_{R, \rm max}
 \end{align*} \end{align*}
 \] \]
Line 240: Line 240:
 \[ \[
 \begin{align*} \begin{align*}
-{\color{blue }{I_{\rm V}}} &= \frac{U_{\rm E}-U_{\rm Z}}{R_{\rm V}} &&\text{current supplied through the series resistor},+{\color{blue }{I_{\rm V}}} &= \frac{U_{\rm I}-U_{\rm Z}}{R_{\rm V}} &&\text{current supplied through the series resistor},
 \\ \\
 {\color{green}{I_{\rm L}}} &= \frac{U_{\rm Z}}{R_{\rm L}}           &&\text{useful load current}, {\color{green}{I_{\rm L}}} &= \frac{U_{\rm Z}}{R_{\rm L}}           &&\text{useful load current},
Line 539: Line 539:
 ===== Exercises ===== ===== Exercises =====
  
-<panel type="info" title="Exercise 2.1.6 Circuit with multiple diodes"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Circuit with multiple diodes: which lamps light up? 
 +#@TaskText_HTML@#
  
-The following simulation includes multiple diodes. The shown lambs light bright, when a voltage of $5~\rm V$ or more drops over them. \\  +The following simulation includes multiple diodes and several lamps.   
-Which lambs will light up, when the switch is closed?+A lamp lights brightly when a voltage of approximately 
 + 
 +\[ 
 +\begin{align*} 
 +U_{\rm lamp}\geq 5~{\rm V} 
 +\end{align*} 
 +\] 
 + 
 +drops across it. 
 + 
 +Close the switch in the simulation. 
 + 
 +  * Which lamps light up brightly? 
 +  * Which lamps remain dark? 
 +  * Explain the result using the idea of diode bypass paths.
  
 {{url>http://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxG3KQBZsQEBTAWjDACgBzEa4wkTFN14U0UKGwAmQvgPAYZGQYIkMAZgEMArgBsALmwBK4MILDFBeSOHNir1K7ltRoCSf3vuHhPJ-4gVGjr6AO4U3r7Y4WaCkGyhKB4JVknWMWxgeBC0pjY8fNFiuFYQSDBwEGWQ7KFg8vyKcvk2saG41KkUkO0mPi2NHbX5KL1ubeDD-T1+AVp6o1ET2eM+ymqzIdzYOYJLU7EAzsbbk83gIBra+wzpmWE+BbunRWelsFXO5TcQKQVjBQ5wF4fd6VdgZCB-GyRe5PQElYEVN5g26DDo-WHFegIhFxaT1HbCf646EdEl7NhAA noborder}} {{url>http://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxG3KQBZsQEBTAWjDACgBzEa4wkTFN14U0UKGwAmQvgPAYZGQYIkMAZgEMArgBsALmwBK4MILDFBeSOHNir1K7ltRoCSf3vuHhPJ-4gVGjr6AO4U3r7Y4WaCkGyhKB4JVknWMWxgeBC0pjY8fNFiuFYQSDBwEGWQ7KFg8vyKcvk2saG41KkUkO0mPi2NHbX5KL1ubeDD-T1+AVp6o1ET2eM+ymqzIdzYOYJLU7EAzsbbk83gIBra+wzpmWE+BbunRWelsFXO5TcQKQVjBQ5wF4fd6VdgZCB-GyRe5PQElYEVN5g26DDo-WHFegIhFxaT1HbCf646EdEl7NhAA noborder}}
  
-</WRAP></WRAP></panel> 
  
-<panel type="info" title="Exercise 2.1.7 Circuit with multiple diodes II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+1. Determine which lamps light up brightly.
  
-The following simulation includes multiple diodes. Assume a simple diode model (the forward voltage drop is $V_F=0.6~\rm V$ and constant). The source voltage shall be $U0 = 4~\rm V$.+<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12001~@# 
 +<WRAP leftalign> 
 +Number the lamps from left to right:
  
-Calculate the currents through $D1$, $R1$, and $R2$.+\[ 
 +\begin{align*} 
 +L_1,\;L_2,\;L_3,\;L_4,\;L_5. 
 +\end{align*} 
 +\] 
 + 
 +After closing the switch, check the voltage across each lamp in the simulation. 
 + 
 +A lamp is assumed to light brightly if 
 + 
 +\[ 
 +\begin{align*} 
 +U_{\rm lamp}\geq 5~{\rm V}. 
 +\end{align*} 
 +\] 
 + 
 +The simulation shows that the outer lamps have a sufficiently large voltage across them, while the inner lamps are bypassed by conducting diodes. 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 + 
 +<WRAP half column> 
 +#@ResultBegin_HTML~12001~@# 
 +The lamps 
 + 
 +\[ 
 +\begin{align*} 
 +L_1 
 +\quad \text{and} \quad 
 +L_5 
 +\end{align*} 
 +\] 
 + 
 +light up brightly. 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +2. Determine which lamps remain dark. 
 + 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12002~@# 
 +<WRAP leftalign> 
 +The inner lamps are connected in parts of the circuit that are bypassed by forward-biased diodes. 
 + 
 +A forward-biased diode has only a small voltage drop. Therefore, a lamp in parallel with such a diode path receives only a small voltage. 
 + 
 +If 
 + 
 +\[ 
 +\begin{align*} 
 +U_{\rm lamp}<5~{\rm V}, 
 +\end{align*} 
 +\] 
 + 
 +the lamp does not light brightly. 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 + 
 +<WRAP half column> 
 +#@ResultBegin_HTML~12002~@# 
 +The lamps 
 + 
 +\[ 
 +\begin{align*} 
 +L_2,\;L_3,\;L_4 
 +\end{align*} 
 +\] 
 + 
 +remain dark or almost dark. 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +#@TaskEnd_HTML@# 
 + 
 + 
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Circuit with multiple diodes II: current calculation 
 +#@TaskText_HTML@# 
 + 
 +The following simulation includes two diodes and two resistors. 
 + 
 +Assume a simple constant-voltage diode model: 
 + 
 +\[ 
 +\begin{align*} 
 +U_{\rm F}=0.6~{\rm V}. 
 +\end{align*} 
 +\] 
 + 
 +The source voltage is 
 + 
 +\[ 
 +\begin{align*} 
 +U_0=4.0~{\rm V}. 
 +\end{align*} 
 +\] 
 + 
 +The resistors are 
 + 
 +\[ 
 +\begin{align*} 
 +R_1=200~\Omega, 
 +\qquad 
 +R_2=100~\Omega. 
 +\end{align*} 
 +\] 
 + 
 +Calculate the currents through 
 + 
 +  * \(D_1\), 
 +  * \(R_1\), 
 +  * \(R_2\).
  
 {{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxABZykLsQEBTAWjDACgA3EFFC7iyN17gUeKOIGVxgmAjYAnENjQixywbxnc4CpXj5hRevpvFgdAd2P9B6m1DZW7pnicmQ2AD24IU4c9wE-nRuIAAi7N7Y2EistoSxYCH84SheSmTgGH4UmFk0KQBKaVG4WXTYhIJgGIRSwoWRQmI1dCgILbX1fACqHgAmQgZGdoZifv0MAGYAhgCuADYALmyDoyP6qtwgk7OLK0A noborder}} {{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxABZykLsQEBTAWjDACgA3EFFC7iyN17gUeKOIGVxgmAjYAnENjQixywbxnc4CpXj5hRevpvFgdAd2P9B6m1DZW7pnicmQ2AD24IU4c9wE-nRuIAAi7N7Y2EistoSxYCH84SheSmTgGH4UmFk0KQBKaVG4WXTYhIJgGIRSwoWRQmI1dCgILbX1fACqHgAmQgZGdoZifv0MAGYAhgCuADYALmyDoyP6qtwgk7OLK0A noborder}}
  
-</WRAP></WRAP></panel>+1. Calculate the current through \(R_1\).
  
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~12004~@#
 +<WRAP leftalign>
 +The current through \(R_1\) passes through one forward-biased diode.
  
-<panel type="info" title="Exercise 2.1.8 Circuit with multiple diodes III"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+Therefore the voltage across \(R_1\) is
  
-The following simulation includes multiple diodes. Assume a simple diode model (the forward voltage drop is $V_{\rm F} = 0.7~\rm V$ and constant). The source voltage shall be $U0 = 5~\rm V$.+\[ 
 +\begin{align*} 
 +U_{R1} 
 += 
 +U_0-U_{\rm F}. 
 +\end{align*} 
 +\]
  
-Calculate the currents through $R1$, $D1$, and $D2$ depending on the switch state S.+Insert the values: 
 + 
 +\[ 
 +\begin{align*} 
 +U_{R1} 
 +&= 
 +4.0~{\rm V}-0.6~{\rm V} 
 +\\ 
 +&= 
 +3.4~{\rm V}. 
 +\end{align*} 
 +\] 
 + 
 +Now apply Ohm's law: 
 + 
 +\[ 
 +\begin{align*} 
 +I_{R1} 
 += 
 +\frac{U_{R1}}{R_1} 
 += 
 +\frac{3.4~{\rm V}}{200~\Omega} 
 += 
 +17~{\rm mA}. 
 +\end{align*} 
 +\] 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 + 
 +<WRAP half column> 
 +#@ResultBegin_HTML~12004~@# 
 +\[ 
 +\begin{align*} 
 +I_{R1}=17~{\rm mA} 
 +\end{align*} 
 +\] 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +2. Calculate the current through \(R_2\). 
 + 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12005~@# 
 +<WRAP leftalign> 
 +The current through \(R_2\) passes through two forward-biased diodes. 
 + 
 +Therefore the voltage across \(R_2\) is 
 + 
 +\[ 
 +\begin{align*} 
 +U_{R2} 
 += 
 +U_0-2U_{\rm F}. 
 +\end{align*} 
 +\] 
 + 
 +Insert the values: 
 + 
 +\[ 
 +\begin{align*} 
 +U_{R2} 
 +&= 
 +4.0~{\rm V}-2\cdot 0.6~{\rm V} 
 +\\ 
 +&= 
 +2.8~{\rm V}. 
 +\end{align*} 
 +\] 
 + 
 +Now apply Ohm's law: 
 + 
 +\[ 
 +\begin{align*} 
 +I_{R2} 
 += 
 +\frac{U_{R2}}{R_2} 
 += 
 +\frac{2.8~{\rm V}}{100~\Omega} 
 += 
 +28~{\rm mA}. 
 +\end{align*} 
 +\] 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 + 
 +<WRAP half column> 
 +#@ResultBegin_HTML~12005~@# 
 +\[ 
 +\begin{align*} 
 +I_{R2}=28~{\rm mA} 
 +\end{align*} 
 +\] 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +3. Calculate the current through \(D_1\). 
 + 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12006~@# 
 +<WRAP leftalign> 
 +The diode \(D_1\) supplies both current paths. 
 + 
 +Therefore, by Kirchhoff's current law, 
 + 
 +\[ 
 +\begin{align*} 
 +I_{D1} 
 += 
 +I_{R1}+I_{R2}. 
 +\end{align*} 
 +\] 
 + 
 +Insert the values: 
 + 
 +\[ 
 +\begin{align*} 
 +I_{D1} 
 +&= 
 +17~{\rm mA}+28~{\rm mA} 
 +\\ 
 +&= 
 +45~{\rm mA}. 
 +\end{align*} 
 +\] 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 + 
 +<WRAP half column> 
 +#@ResultBegin_HTML~12006~@# 
 +\[ 
 +\begin{align*} 
 +I_{D1}=45~{\rm mA} 
 +\end{align*} 
 +\] 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +#@TaskEnd_HTML@# 
 + 
 + 
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Circuit with multiple diodes III: switch-dependent currents 
 +#@TaskText_HTML@# 
 + 
 +The following simulation includes two diodes and a switch. 
 + 
 +Assume a simple constant-voltage diode model: 
 + 
 +\[ 
 +\begin{align*} 
 +U_{\rm F}=0.7~{\rm V}. 
 +\end{align*} 
 +\] 
 + 
 +The source voltage is 
 + 
 +\[ 
 +\begin{align*} 
 +U_0=5.0~{\rm V}. 
 +\end{align*} 
 +\] 
 + 
 +The resistor is 
 + 
 +\[ 
 +\begin{align*} 
 +R_1=1.0~{\rm k}\Omega. 
 +\end{align*} 
 +\] 
 + 
 +Calculate the currents through 
 + 
 +  * \(R_1\), 
 +  * \(D_1\), 
 +  * \(D_2\), 
 + 
 +depending on the switch state \(S\).
  
 {{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWKswDZ0A4BMYDMAWfAdiIE59UFcRVIQl9qEBTAWjDACgA3ELLfH3x1+gsFgxQpw+lLowEnAE5C64ybkhiJUsPE4B3EIyyqQuVJIHzD5y2dFnInACZ3J69w-AA5fGHwMTgAPcwxqMCJBfwiSYyEQABEuUKwENUhSPgw1cXiBEAAlFL4dSOo0jyJUfMEAVWc3E3AdZus+X39AkONicCjjOMiiWqSsTgBnL09mz3kQADMAQwAbCeZOXCI6TW0NCxaPOVtHT3a521wDzwsPHWdQ3HJwTOM9cDzBAoBlTiA noborder}} {{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWKswDZ0A4BMYDMAWfAdiIE59UFcRVIQl9qEBTAWjDACgA3ELLfH3x1+gsFgxQpw+lLowEnAE5C64ybkhiJUsPE4B3EIyyqQuVJIHzD5y2dFnInACZ3J69w-AA5fGHwMTgAPcwxqMCJBfwiSYyEQABEuUKwENUhSPgw1cXiBEAAlFL4dSOo0jyJUfMEAVWc3E3AdZus+X39AkONicCjjOMiiWqSsTgBnL09mz3kQADMAQwAbCeZOXCI6TW0NCxaPOVtHT3a521wDzwsPHWdQ3HJwTOM9cDzBAoBlTiA noborder}}
  
-</WRAP></WRAP></panel>+1. Calculate the currents for open switch \(S\). 
 + 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12007~@# 
 +<WRAP leftalign> 
 +With the switch open, only \(D_1\) is connected to the resistor path. 
 + 
 +The conducting diode clamps the node voltage to approximately 
 + 
 +\[ 
 +\begin{align*} 
 +U_{\rm node}\approx U_{\rm F}=0.7~{\rm V}. 
 +\end{align*} 
 +\] 
 + 
 +The resistor current is therefore 
 + 
 +\[ 
 +\begin{align*} 
 +I_{R1} 
 +&= 
 +\frac{U_0-U_{\rm F}}{R_1} 
 +\\ 
 +&= 
 +\frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega} 
 +\\ 
 +&= 
 +4.3~{\rm mA}. 
 +\end{align*} 
 +\] 
 + 
 +Since only \(D_1\) conducts, 
 + 
 +\[ 
 +\begin{align*} 
 +I_{D1}=I_{R1}, 
 +\qquad 
 +I_{D2}=0. 
 +\end{align*} 
 +\] 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 + 
 +<WRAP half column> 
 +#@ResultBegin_HTML~12007~@# 
 +For open switch: 
 + 
 +\[ 
 +\begin{align*} 
 +I_{R1}&=4.3~{\rm mA}, 
 +\\ 
 +I_{D1}&=4.3~{\rm mA}, 
 +\\ 
 +I_{D2}&=0. 
 +\end{align*} 
 +\] 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +2. Calculate the currents for closed switch \(S\). 
 + 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12008~@# 
 +<WRAP leftalign> 
 +With the switch closed, \(D_1\) and \(D_2\) are connected in parallel. 
 + 
 +The resistor current is still determined by the source voltage, the forward diode voltage, and \(R_1\): 
 + 
 +\[ 
 +\begin{align*} 
 +I_{R1} 
 +&= 
 +\frac{U_0-U_{\rm F}}{R_1} 
 +\\ 
 +&= 
 +\frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega} 
 +\\ 
 +&= 
 +4.3~{\rm mA}. 
 +\end{align*} 
 +\] 
 + 
 +Kirchhoff's current law gives 
 + 
 +\[ 
 +\begin{align*} 
 +I_{D1}+I_{D2}=I_{R1}. 
 +\end{align*} 
 +\] 
 + 
 +With the ideal constant-voltage diode model, the individual currents through two parallel diodes are not uniquely determined. 
 + 
 +If both real diodes are approximately identical, the current splits approximately equally: 
 + 
 +\[ 
 +\begin{align*} 
 +I_{D1} 
 +\approx 
 +I_{D2} 
 +\approx 
 +\frac{4.3~{\rm mA}}{2} 
 += 
 +2.15~{\rm mA}. 
 +\end{align*} 
 +\] 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 + 
 +<WRAP half column> 
 +#@ResultBegin_HTML~12008~@# 
 +For closed switch: 
 + 
 +\[ 
 +\begin{align*} 
 +I_{R1}=4.3~{\rm mA} 
 +\end{align*} 
 +\] 
 + 
 +and 
 + 
 +\[ 
 +\begin{align*} 
 +I_{D1}+I_{D2}=4.3~{\rm mA}. 
 +\end{align*} 
 +\] 
 + 
 +For approximately identical real diodes: 
 + 
 +\[ 
 +\begin{align*} 
 +I_{D1} 
 +\approx 
 +I_{D2} 
 +\approx 
 +2.15~{\rm mA}. 
 +\end{align*} 
 +\] 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +3. Explain why the current sharing is not unique in the simple model. 
 + 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12009~@# 
 +<WRAP leftalign> 
 +The constant-voltage diode model assumes that each conducting diode has exactly the same voltage drop: 
 + 
 +\[ 
 +\begin{align*} 
 +U_{\rm D}=U_{\rm F}. 
 +\end{align*} 
 +\] 
 + 
 +For two parallel diodes, this condition is true for many possible current distributions. 
 + 
 +Therefore, the model only determines the sum 
 + 
 +\[ 
 +\begin{align*} 
 +I_{D1}+I_{D2}, 
 +\end{align*} 
 +\] 
 + 
 +not the individual diode currents. 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 + 
 +<WRAP half column> 
 +#@ResultBegin_HTML~12009~@# 
 +The constant-voltage diode model determines only 
 + 
 +\[ 
 +\begin{align*} 
 +I_{D1}+I_{D2}=4.3~{\rm mA}. 
 +\end{align*} 
 +\] 
 + 
 +It does not uniquely determine \(I_{D1}\) and \(I_{D2}\) separately. 
 + 
 +<callout type="warning" icon="true"> 
 +Parallel diodes are sensitive to small differences in real diode characteristics.   
 +Current sharing should not be assumed to be perfect without checking the design. 
 +</callout> 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +#@TaskEnd_HTML@# 
  
  
Line 578: Line 1100:
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm E}=24~{\rm V}.+U_{\rm I}=24~{\rm V}.
 \end{align*} \end{align*}
 \] \]
Line 604: Line 1126:
 R_{\rm V} R_{\rm V}
 &= &=
-\frac{U_{\rm E}-U_{\rm F}}{I_{\rm F}}+\frac{U_{\rm I}-U_{\rm F}}{I_{\rm F}}
 \\ \\
 &= &=
Line 628: Line 1150:
 P_R P_R
 &= &=
-(U_{\rm E}-U_{\rm F})I_{\rm F}+(U_{\rm I}-U_{\rm F})I_{\rm F}
 \\ \\
 &= &=
Line 659: Line 1181:
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm E}=12~{\rm V}.+U_{\rm I}=12~{\rm V}.
 \end{align*} \end{align*}
 \] \]
Line 692: Line 1214:
 I_{\rm V} I_{\rm V}
 &= &=
-\frac{U_{\rm E}-U_{\rm Z}}{R_{\rm V}}+\frac{U_{\rm I}-U_{\rm Z}}{R_{\rm V}}
 \\ \\
 &= &=