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Lösning 4.2:3 - Versionshistorik
2026-04-07T19:57:11Z
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Louwah den 19 mars 2018 kl. 12.46
2018-03-19T12:46:09Z
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<td colspan='2' style="background-color: white; color:black;">← Äldre version</td>
<td colspan='2' style="background-color: white; color:black;">Versionen från 19 mars 2018 kl. 12.46</td>
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<tr><td colspan="2" class="diff-lineno">Rad 3:</td>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div><math>E-R_i \cdot 4,0/0,8–4,0 = 0</math><br\> </div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div><math>E-R_i \cdot 4,0/0,8–4,0 = 0</math><br\> </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div><math>E-R_i \cdot 3,0/0,2–3,0 = 0 </math><br\></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div><math>E-R_i \cdot 3,0/0,2–3,0 = 0 </math><br\></div></td></tr>
<tr><td class='diff-marker'>-</td><td style="background: #ffa; color:black; font-size: smaller;"><div><math>E = 4,5 V</math><br\> </div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div><math>E = 4,5 <ins style="color: red; font-weight: bold; text-decoration: none;">\,\mathrm{</ins>V<ins style="color: red; font-weight: bold; text-decoration: none;">}</ins></math><br\> </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div><math>R_i = 0,1\Omega </math><br\></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div><math>R_i = 0,1\Omega </math><br\></div></td></tr>
<tr><td class='diff-marker'>-</td><td style="background: #ffa; color:black; font-size: smaller;"><div><math>I=<del style="color: red; font-weight: bold; text-decoration: none;">5A</del></math> vid <math>0,8\Omega</math><br\> </div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div><math>I=<ins style="color: red; font-weight: bold; text-decoration: none;">5 \,\mathrm{A}</ins></math> vid <math>0,8\Omega</math><br\> </div></td></tr>
<tr><td class='diff-marker'>-</td><td style="background: #ffa; color:black; font-size: smaller;"><div><math>I=<del style="color: red; font-weight: bold; text-decoration: none;">15A</del></math> vid <math>0,2 \Omega</math><br\></div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div><math>I=<ins style="color: red; font-weight: bold; text-decoration: none;">15 \,\mathrm{A}</ins></math> vid <math>0,2 \Omega</math><br\></div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>[[Bild:losning_4_2_3.gif|center]]</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>[[Bild:losning_4_2_3.gif|center]]</div></td></tr>
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Louwah
http://wiki.math.se/wikis/forberedandefysik/index.php?title=L%C3%B6sning_4.2:3&diff=2143&oldid=prev
Lena Chytraeus: Ny sida: Kirchhoffs lag ger <math>E-R_i \cdot I–U = 0</math> och Ohms lag <math>I=U/R</math> ger med insatta värden<br\> <math>E-R_i \cdot 4,0/0,8–4,0 = 0</math><br\> <math>E-R_i \cdot 3,0/0,...
2010-01-21T14:22:32Z
<p>Ny sida: Kirchhoffs lag ger <math>E-R_i \cdot I–U = 0</math> och Ohms lag <math>I=U/R</math> ger med insatta värden<br\> <math>E-R_i \cdot 4,0/0,8–4,0 = 0</math><br\> <math>E-R_i \cdot 3,0/0,...</p>
<p><b>Ny sida</b></p><div>Kirchhoffs lag ger <math>E-R_i \cdot I–U = 0</math> och Ohms lag <math>I=U/R</math> ger med insatta värden<br\><br />
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<math>E-R_i \cdot 4,0/0,8–4,0 = 0</math><br\> <br />
<math>E-R_i \cdot 3,0/0,2–3,0 = 0 </math><br\><br />
<math>E = 4,5 V</math><br\> <br />
<math>R_i = 0,1\Omega </math><br\><br />
<math>I=5A</math> vid <math>0,8\Omega</math><br\> <br />
<math>I=15A</math> vid <math>0,2 \Omega</math><br\><br />
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[[Bild:losning_4_2_3.gif|center]]</div>
Lena Chytraeus