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SuperminarenandClaude Sonnet 5.5 f7fa8cfbc1 Swedish adaptation: Instegscertifikat and HAREC content
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Co-Authored-By: Claude Sonnet 5.5 <[email protected]>
2026-10-10 15:03:03 +02:00

708 B

Effektberäkning

Vi känner redan

P = U\cdot I = \dfrac{U^2}{R} = I^2\cdot R

Utlöst för U: $U = \dfrac{P}{I} = \sqrt{P \cdot R}$ Utlöst för I: $I = \dfrac{P}{U} = \sqrt{\dfrac{P}{R}}$

Effekt vid växelspänning

  • Vid växelspänning räknar man med effektivvärdet
  • U_\text{eff} = \dfrac{\hat{U}}{\sqrt{2}}

Exempel

  • U_\text{eff} = \qty{100}{\volt} över \qty{50}{\ohm}: P = \dfrac{U^2}{R} = \qty{200}{\watt}
  • \qty{2}{\ampere} genom \qty{50}{\ohm}: P = I^2 \cdot R = \qty{200}{\watt}
  • \qty{10}{\volt} över \qty{100}{\ohm}: P = \qty{1}{\watt}
  • Konstlast av 11 parallella motstånd á \qty{5}{\watt}: \qty{55}{\watt}