Add equations for Peak-Peak values
Add \hat{I} to equation
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@@ -4,13 +4,15 @@
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<right>
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Berechnung mit Effektivwerten
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<fragment>
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$U_{\text{eff}} = \frac{\hat{U}}{\sqrt{2}}$
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$U_{\mathrm{eff}} = \frac{\hat{U}}{\sqrt{2}} = \frac{U_\mathrm{SS}}{2\sqrt{2}}$
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</fragment>
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<fragment>
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$I_{\text{eff}} = \frac{\hat{I}}{\sqrt{2}}$
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$I_{\mathrm{eff}} = \frac{\hat{I}}{\sqrt{2}} = \frac{I_\mathrm{SS}}{2\sqrt{2}}$
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</fragment>
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<fragment>
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$P = U_{\text{eff}} \cdot I_{\text{eff}}$
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$P = U_{\mathrm{eff}} \cdot I_{\mathrm{eff}}$
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</fragment>
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</right>
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@@ -18,7 +20,7 @@ $P = U_{\text{eff}} \cdot I_{\text{eff}}$
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[question:AB301]
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---
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### Lösungsweg
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* gegeben: $I_{\mathrm{max}} = \qty{0,5}{\ampere}$
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* gegeben: $I_{\mathrm{max}} = \hat{I} = \qty{0,5}{\ampere}$
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* gegeben: $R = \qty{20}{\ohm}$
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* gesucht: $P$
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