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#1: A high-voltage AC power supply (Emax = 311 V) is connected to an inductor. A maximum current of 1.79 A is measured in the

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Answer #1

! I-V Pk Emax = 3/10, Imax = 1-79A So inductive reacheunce XL= 311 -143-72 I 1-79 2 Frequency current = 2.8 minutes . = 2.6X6

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