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Q1: A current of I = 3 +0.2 A is applied to a resistor R of 10 +012 (with zero uncertainty so you can consider R as a constan
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Given I = 3to.z А. AI=0.2 R = 10 de AR = O a) V = IR = 3x10 = 30 Uncertanity ein v RAI + IAR = (1080.2) + (3x0) - Z y V = 30 R = 10 to.5 2 AR = 0.5 - V=IR = 3x10 = 30v AV IAR + RAI = (3x0.5) + (10x0.2) N/a 3.5 volti = 1.5 + 2 = - 30 + 3.5) notte

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