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<title>Vurdering i elektrisitet</title>
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<body>
<div id="content">
<h1 class="title">Vurdering i elektrisitet</h1>
<div id="table-of-contents">
<h2>Innhold</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#orge0251d5">Oppgave 1</a></li>
<li><a href="#orgaccac91">Oppgave 2</a></li>
<li><a href="#orgdf64da0">Oppgave 3</a></li>
</ul>
</div>
</div>
<div id="outline-container-orge0251d5" class="outline-2">
<h2 id="orge0251d5">Oppgave 1</h2>
<div class="outline-text-2" id="text-orge0251d5">
<p>
Et batteri med en spenning på 12,0 V er koblet i serie med en motstand med resistansen 2,0 Ω og en parallellkobling av to motstander, begge med resistans 3,0 Ω.
</p>
</div>
<div id="outline-container-org35bae4f" class="outline-3">
<h3 id="org35bae4f">a</h3>
<div class="outline-text-3" id="text-org35bae4f">
<p>
Lag koblingsskjema
</p>
</div>
<div id="outline-container-org325384f" class="outline-4">
<h4 id="org325384f">Løsning</h4>
<div class="outline-text-4" id="text-org325384f">
<p>
<img src="figurer/vurdering_elektrisitet_koblingsskjema_1a.png" alt="vurdering_elektrisitet_koblingsskjema_1a.png"/>
</p>
</div>
</div>
</div>
<div id="outline-container-org90782a7" class="outline-3">
<h3 id="org90782a7">b</h3>
<div class="outline-text-3" id="text-org90782a7">
<p>
Finn resultantresistansen i kretsen.
</p>
</div>
<div id="outline-container-org425530c" class="outline-4">
<h4 id="org425530c">Løsning</h4>
<div class="outline-text-4" id="text-org425530c">
<p>
De to parallelle motstandene har sammen en motstand på \[R_P = \left(3,0^{-1}+3,0^{-1}\right)^{-1}\,\textrm{Ω} = 1,5\,\textrm{Ω}.\]Lagt sammen med seriemotstanden på 2,0 Ω får vi da en resultantmotstand på \[R=2,0\,\textrm{Ω}+1,5\,\textrm{Ω}=3,5\,\textrm{Ω}.\]
</p>
</div>
</div>
</div>
<div id="outline-container-orge1785f7" class="outline-3">
<h3 id="orge1785f7">c</h3>
<div class="outline-text-3" id="text-orge1785f7">
<p>
Finn strømmen gjennom batteriet.
</p>
</div>
<div id="outline-container-org6ddeabe" class="outline-4">
<h4 id="org6ddeabe">Løsning</h4>
<div class="outline-text-4" id="text-org6ddeabe">
<p>
Ved å bruke Ohms lov får vi at strømmen gjennom batteriet er på \[I=\frac{U}{R}=\frac{12,0\,\textrm{V}}{3,5\,\textrm{Ω}}=3,4\,\textrm{A}.\]
</p>
</div>
</div>
</div>
</div>
<div id="outline-container-orgaccac91" class="outline-2">
<h2 id="orgaccac91">Oppgave 2</h2>
<div class="outline-text-2" id="text-orgaccac91">
<p>
Tre like motstander er koblet i parallell. Se figuren under. Strømmen gjennom amperemeter A3 er 0,60 A. Bestem strømmen gjennom amperemetrene 1, 2, 4, og 5.
</p>
<p>
<img src="figurer/vurdering_elektrisitet_koblingsskjema.png" alt="vurdering_elektrisitet_koblingsskjema.png"/>
</p>
</div>
<div id="outline-container-org2aa62a8" class="outline-4">
<h4 id="org2aa62a8">Løsning</h4>
<div class="outline-text-4" id="text-org2aa62a8">
<p>
Siden alle de tre motstandene er like store, vil det gå en like stor strøm gjennom dem. A3 viser strømmen gjennom én motstand, det samme gjør A4, altså er A4 = A3 = 0,60 A. Videre kan vi bruke Kirchhoffs første lov og få at strømmen gjennom A1 skal deles over tre motstander, altså er \(I_1 = 3\cdot I_3 = 1,8\,\textrm{A}\), dersom \(I_n\) angir strømmen gjennom amperemeter A<sub>n</sub>. Videre er \(I_2 = I_3 + I_4 = 1,2\,\textrm{A}\). Vi har altså
</p>
\begin{align*}
I_1 &= I_5 = 1,8\,\textrm{A} \\
I_2 &= 1,2\,\textrm{A} \\
I_3 &= I_4 = 0,6\,\textrm{A}
\end{align*}
</div>
</div>
</div>
<div id="outline-container-orgdf64da0" class="outline-2">
<h2 id="orgdf64da0">Oppgave 3</h2>
<div class="outline-text-2" id="text-orgdf64da0">
</div>
<div id="outline-container-orgc3291e4" class="outline-3">
<h3 id="orgc3291e4">a</h3>
<div class="outline-text-3" id="text-orgc3291e4">
<p>
En lyspære står på 60 W i én time. Hvor stor ladning har passert et tverrsnitt av glødetråden i løpet av denne tiden? Anta at lyspæren er koblet til nettspenningen som er 230 V.
</p>
</div>
<div id="outline-container-org19f7cfe" class="outline-4">
<h4 id="org19f7cfe">Løsning</h4>
<div class="outline-text-4" id="text-org19f7cfe">
<p>
Vi har \(I=Q/t\), hvor \(Q\) er antall ladninger som passerer et tverrsnitt over tida \(t\), ved å bruke \(P=UI\) kan vi da finne ladningen \(Q\) som
</p>
\begin{align*}
P &= UI \\
I &= \frac{P}{U} \\
\frac{Q}{t} &= \frac{P}{U} \\
Q &= \frac{P}{U}\cdot t = \frac{60\,\textrm{W}}{230\,\textrm{V}}\cdot 3600\,\textrm{s} \\
Q &= 940\,\textrm{C}
\end{align*}
</div>
</div>
</div>
<div id="outline-container-org84e4956" class="outline-3">
<h3 id="org84e4956">b</h3>
<div class="outline-text-3" id="text-org84e4956">
<p>
Et varmekabelanlegg består av to like motstandstråder. Ved å koble én eller begge trådene til nettet på forskjellige måter kan vi oppnå tre ulike verdier for avgitt effekt. Tegn koblingsskjema for hvert av de tre tilfellene.
</p>
</div>
<div id="outline-container-org203cc5f" class="outline-4">
<h4 id="org203cc5f">Løsning</h4>
<div class="outline-text-4" id="text-org203cc5f">
<p>
En hvor de er koblet i serie, en hvor kun en motstandstråd er tilkoblet, og en hvor de er koblet i parallell.
</p>
</div>
</div>
</div>
<div id="outline-container-org73c7b74" class="outline-3">
<h3 id="org73c7b74">c</h3>
<div class="outline-text-3" id="text-org73c7b74">
<p>
Varmekabelanlegget har en bryter med trinnene 0, 1, 2 og 3, der trinnene fra 1 til 3 representerer de tre koblingene i oppgave b. På trinn 1 (laveste effekt) avgir anlegget effekten 330 W. Nettspenningen er 230 V. Finn strømmen.
</p>
</div>
<div id="outline-container-orgdeabb52" class="outline-4">
<h4 id="orgdeabb52">Løsning</h4>
<div class="outline-text-4" id="text-orgdeabb52">
<p>
Vi har at \(P=UI\) og vi kan da finne strømmen som \[I = \frac{P}{U}=\frac{330\,\textrm{W}}{230\,\textrm{V}}=1,4\,\textrm{A}.\]
</p>
</div>
</div>
</div>
<div id="outline-container-org61ad771" class="outline-3">
<h3 id="org61ad771">d</h3>
<div class="outline-text-3" id="text-org61ad771">
<p>
Hvilken effekt avgir anlegget når bryteren står på trinn 3?
</p>
</div>
<div id="outline-container-org71549b5" class="outline-4">
<h4 id="org71549b5">Løsning</h4>
<div class="outline-text-4" id="text-org71549b5">
<p>
Om man skal koble sammen opptil to like motstander med en gitt spenning over begge og få lavest mulig effekttap (varme fra varmeanlegget), må disse to motstandene kobles i serie, siden totalmotstanden da er størst.
</p>
<p>
Om vi setter motstanden i én tråd til å være \(R\), kan vi med svaret fra oppgave c finne denne via Ohms lov \(U=2RI\) (\(2R\) siden motstandene er koblet i serie). Vi får da \[R=\frac{U}{2I}=\frac{230\,\textrm{V}}{2\cdot1,4\,\textrm{A}}=80\,\textrm{Ω}.\]
</p>
<p>
Totalmotstanden når kablene er koblet i parallell er da \[R_\textrm{tot}=\left(2\cdot 80^{-1}\right)^{-1}\,\textrm{Ω}=40\,\textrm{Ω}.\] Vi kan da finne effekten til anlegget som
</p>
\begin{align*}
P &= UI \\
P &= \frac{U^2}{R_\textrm{tot}} \\
P &= \frac{(230\,\textrm{V})^2}{40\,\textrm{Ω}} = 1300\,\textrm{W}.
\end{align*}
</div>
</div>
</div>
</div>
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</body>
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