SuperminarenandClaude Sonnet 5.5
f7fa8cfbc1
- Sections and slides rewritten in Swedish; German law and institutions replaced by Swedish/CEPT/ITU equivalents (PTS, SSA, PTSFS 2025:1, SSMFS 2008:18), unverified statements marked "% TODO VERIFY" - New toc: INSTEG, HAREC, UPPGRADERING; section idents are Swedish (see ident-map.tsv) - Exam-question tags, BNetzA question pool, solutions and the German SWL course removed - Static pages, snippets, applets, READMEs and image alt-texts translated - TikZ labels translated; SVG drawings regenerated (drawing 689 still fails to build upstream: it references a missing photo) - Changed from the original: not endorsed by DARC e.V., SSA or PTS Co-Authored-By: Claude Sonnet 5.5 <[email protected]>
106 lines
3.7 KiB
Markdown
106 lines
3.7 KiB
Markdown
## Maskregeln och nodregeln
|
|
|
|
*Maskregeln*: I varje sluten krets (maska) är summan av spänningarna lika med noll.
|
|
*Nodregeln*: I varje nod är summan av inflödande strömmar lika med summan av utflödande strömmar.
|
|
|
|
<note>
|
|
Kirchhoffs lagar
|
|
</note>
|
|
---
|
|
#### Lösningsgång
|
|
* givet: $R_1 = R_2 = R_3 = \qty{10}{\kilo\ohm}$
|
|
* givet: $I_3 = I_2 = \qty{1}{\milli\ampere}$
|
|
* sökt: $U$
|
|
|
|
<fragment>
|
|
$R_{\mathrm{tot}} = R_1 + \frac{R_2 \cdot R_3}{R_2 + R_3} = \qty{10}{\kilo\ohm} + \frac{\qty{10}{\kilo\ohm} \cdot \qty{10}{\kilo\ohm}}{\qty{10}{\kilo\ohm} + \qty{10}{\kilo\ohm}} = \qty{15}{\kilo\ohm}$
|
|
</fragment>
|
|
<fragment>
|
|
$I = I_2 + I_3 = \qty{1}{\milli\ampere} + \qty{1}{\milli\ampere} = \qty{2}{\milli\ampere}$
|
|
</fragment>
|
|
<fragment>
|
|
$U = R_{\mathrm{tot}} \cdot I = \qty{15}{\kilo\ohm} \cdot \qty{2}{\milli\ampere} = \qty{30}{\volt}$
|
|
</fragment>
|
|
|
|
---
|
|
|
|
#### Lösningsgång
|
|
* givet: $R_1 = R_2 = R_3 = \qty{10}{\kilo\ohm}$
|
|
* givet: $U=\qty{15}{\volt}$
|
|
* sökt: $I_3$
|
|
|
|
<fragment>
|
|
$R_{\mathrm{tot}} = R_1 + \frac{R_2 \cdot R_3}{R_2 + R_3} = \qty{10}{\kilo\ohm} + \frac{\qty{10}{\kilo\ohm} \cdot \qty{10}{\kilo\ohm}}{\qty{10}{\kilo\ohm} + \qty{10}{\kilo\ohm}} = \qty{15}{\kilo\ohm}$
|
|
</fragment>
|
|
|
|
<fragment>
|
|
$\frac{U_3}{U} = \frac{R_2 \parallel R_3}{R_{\mathrm{tot}}} \Rightarrow U_3 = \frac{R_2 \parallel R_3}{R_{\mathrm{tot}}} \cdot U = \frac{\qty{5}{\kilo\ohm}}{\qty{15}{\kilo\ohm}} \cdot \qty{15}{\volt} = \qty{5}{\volt}$
|
|
</fragment>
|
|
|
|
<fragment>
|
|
$I_3 = \frac{U_3}{R_3} = \frac{\qty{5}{\volt}}{\qty{10}{\kilo\ohm}} = \qty{0,5}{\milli\ampere}$
|
|
</fragment>
|
|
|
|
---
|
|
#### Lösningsgång
|
|
* givet: $R_1 = R_2 = R_3 = \qty{10}{\kilo\ohm}$
|
|
* givet: $U=\qty{15}{\volt}$
|
|
* sökt: $P_2$
|
|
|
|
<fragment>
|
|
$\frac{U_2}{U} = \frac{R_2 \parallel R_3}{R_{\mathrm{tot}}} \Rightarrow U_2 = \frac{R_2 \parallel R_3}{R_{\mathrm{tot}}} \cdot U = \frac{\qty{5}{\kilo\ohm}}{\qty{15}{\kilo\ohm}} \cdot \qty{15}{\volt} = \qty{5}{\volt}$
|
|
</fragment>
|
|
|
|
<fragment>
|
|
$P_2 = \frac{U_2^2}{R_2} = \frac{(\qty{5}{\volt})^2}{\qty{10}{\kilo\ohm}} = \qty{2,5}{\milli\watt}$
|
|
</fragment>
|
|
--- style="font-size: 0.7em;"
|
|
#### Lösningsgång
|
|
|
|
<left>
|
|
* givet: $R = \qtyrange{0}{1}{\kilo\ohm}$
|
|
* givet: $R_1 = \qty{200}{\ohm}$
|
|
* givet: $R_2 = \qty{100}{\ohm}$
|
|
* givet: $R_3 = \qty{200}{\ohm}$
|
|
</left>
|
|
|
|
<right>
|
|
$R_{\mathrm{tot}} = R_1 + \frac{R_2 \cdot (R_3 + R)}{R_2 + (R_3 + R)}$
|
|
</right>
|
|
|
|
<fragment>
|
|
Vid $R = \qty{0}{\ohm}$:
|
|
$R_{\mathrm{tot}} = \qty{200}{\ohm} + \frac{\qty{100}{\ohm} \cdot (\qty{200}{\ohm} + \qty{0}{\ohm})}{\qty{100}{\ohm} + \qty{200}{\ohm} + \qty{0}{\ohm}} \approx \qty{267}{\ohm}$
|
|
</fragment>
|
|
|
|
<fragment>
|
|
Vid $R = \qty{1}{\kilo\ohm}$:
|
|
$R_{\mathrm{tot}} = \qty{200}{\ohm} + \frac{\qty{100}{\ohm} \cdot (\qty{200}{\ohm} + \qty{1}{\kilo\ohm})}{\qty{100}{\ohm} + \qty{200}{\ohm} +\qty{1}{\kilo\ohm}} \approx \qty{292}{\ohm}$
|
|
</fragment>
|
|
|
|
---
|
|
#### Lösningsgång
|
|
* givet: $R_1 = R_3 = \qty{2,2}{\kilo\ohm}$
|
|
* givet: $R_2 = R_4 = \qty{220}{\ohm}$
|
|
* sökt: $R_{\mathrm{tot}}$
|
|
|
|
<fragment>
|
|
$R_1 || R_3 + R_2 || R_4 = \qty{1100}{\ohm} + \qty{110}{\ohm} = \qty{1210}{\ohm}$
|
|
</fragment>
|
|
--- style="font-size: 0.7em;"
|
|
#### Lösningsgång
|
|
* givet: $R_1 = \qty{10}{\kilo\ohm}$
|
|
* givet: $R_2 = \qty{2,2}{\kilo\ohm}$
|
|
* givet: $R_L = \qty{8,2}{\kilo\ohm}$
|
|
* givet: $U_B = \qty{12}{\volt}$
|
|
* sökt: $U_2$
|
|
|
|
<fragment>
|
|
$\frac{U_2}{U_B} = \frac{R_2 \parallel R_L}{R_{\mathrm{tot}}}$
|
|
$R_2 \parallel R_L = \frac{R_2 \cdot R_L}{R_2 + R_L} = \frac{\qty{2,2}{\kilo\ohm} \cdot \qty{8,2}{\kilo\ohm}}{\qty{2,2}{\kilo\ohm} + \qty{8,2}{\kilo\ohm}} = \qty{1,74}{\kilo\ohm}$
|
|
$R_{\mathrm{tot}} = R_1 + R_2 \parallel R_L = \qty{10}{\kilo\ohm} + \qty{1,74}{\kilo\ohm} = \qty{11,74}{\kilo\ohm}$
|
|
</fragment>
|
|
<fragment>
|
|
$U_2 = \frac{R_2 \parallel R_L}{R_{\mathrm{tot}}} \cdot U_B = \frac{\qty{1,74}{\kilo\ohm}}{\qty{11,74}{\kilo\ohm}} \cdot \qty{12}{\volt} \approx \qty{1,8}{\volt}$
|
|
</fragment>
|