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The following quantities appear in the problem: Länge \(\ell\) / elektrische Stromstärke \(I\) / elektrische Spannung \(U\) / elektrischer Widerstand \(R\) / Fläche \(A\) / Radius \(r\) / spezifischer elektrischer Widerstand \(\rho\) /
The following formulas must be used to solve the exercise: \(A = \pi r^2 \quad \) \(U=RI \quad \) \(R = \varrho \dfrac{\ell}{A} \quad \)
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Exercise:
Compute the voltage drop along a m length of household no. copper Formelbuch.ohmmeter wire used in A circuits. The wire has diameter .mm and carries A current.

Solution:
NewQtyelm NewQtyre.ohmmeter NewQtyIpA NewQtyd.m NewQtyIA % Geg ell el rhoel re I' Ip d .mm  d I I % GesSpannungU siV % Der Kupferdraht hat einen Querschnitt von SolQtyAfracd^pidX***pi/squaremeter al A AF fracqtyd^ pi A. % Sein Widerstand ist folglich SolQtyRrhoel fracelld^pireX*elX/AXohm al R rhoel fracellA RF re frac elqtyd^ pi R. % Beim angegebenen Strom ist damit der Spannungsverlust SolQtyUrhoel fracell Id^piRX*IXV al U RI UF re frac el Iqtyd^ pi U. % U UF US
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Exercise:
Compute the voltage drop along a m length of household no. copper Formelbuch.ohmmeter wire used in A circuits. The wire has diameter .mm and carries A current.

Solution:
NewQtyelm NewQtyre.ohmmeter NewQtyIpA NewQtyd.m NewQtyIA % Geg ell el rhoel re I' Ip d .mm  d I I % GesSpannungU siV % Der Kupferdraht hat einen Querschnitt von SolQtyAfracd^pidX***pi/squaremeter al A AF fracqtyd^ pi A. % Sein Widerstand ist folglich SolQtyRrhoel fracelld^pireX*elX/AXohm al R rhoel fracellA RF re frac elqtyd^ pi R. % Beim angegebenen Strom ist damit der Spannungsverlust SolQtyUrhoel fracell Id^piRX*IXV al U RI UF re frac el Iqtyd^ pi U. % U UF US
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Topic
Tags
copper, current, electric, electricity, physics, resistance, resistivity, wire
Difficulty
(3, default)
Points
2 (default)
Language
ENG (English)
Type
Calculative / Quantity
Decoration
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