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When an AC source is connected across a 23.0 12 resistor, the output voltage is given by AV = (160 V)sin(70st). Determine the

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Answer #1

Peak voltage = 160 volt @ 6 vous = 160 = 113-137 volt @ 6 A Npeak peek 6.956 SEE 6.8 56 sin ( 7067X0.00us) = s.a14 Amp @

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