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Electric Currents

  1. Mar 8, 2005 #1
    What is the maximum instantaneous power dissipated by, and maximum current passing through, a 3.0-hp pump connected to a 240-V ac power source?

    ANSWER: 6 hp, 13 A

    Given:

    P = 3 hp = 2238 W (1 hp = 246 W)
    V = 240 V

    I don't know which formula to use. At first, I attempted to use P = I*V, but, then, I got flustered over the provided formulas in my book. Peak current is mentioned, and there are different variations of P. Did I start off correctly? How am I supposed to execute this problem?

    Thanks.
     
  2. jcsd
  3. Mar 8, 2005 #2

    xanthym

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    Science Advisor

    Standard electrical specifications for AC Current and AC Voltage are expressed with "RMS" ("Root Mean Square") values of the AC sine wave quantities. Standard AC Power specifications are expressed with "Average Power" given by {(RMS Voltage)*(RMS Current)}. Following equations convert these standard specifications to alternate specifications concerning other properties of the AC sine wave:
    {Peak AC Voltage} = (1.414)*{RMS AC Voltage}
    {Peak AC Current} = (1.414)*{RMS AC Current}
    {Peak AC Power} = {Peak AC Voltage}*{Peak AC Current} =
    = (1.414)*{RMS AC Voltage}*(1.414)*{RMS AC Current} =
    = 2*{Average AC Power}

    For this problem, following values are given:
    {RMS AC Voltage} = (240 V)
    {Average AC Power} = (3 hp) = (2238 W)
    from which can be derived:
    {RMS AC Current} = {Average AC Power}/{RMS AC Voltage} = (2238)/(240) = (9.33 A)
    {Peak AC Current} = (1.414)*{RMS AC Current} = (1.414)*(9.33 A) = (13.2 A)
    {Peak AC Power} = (2)*{Average AC Power} = (2)*(3 hp) = (6 hp)



    ~~
     
    Last edited: Mar 8, 2005
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