A hair dryer draws 1 200 W, a curling iron draws 800 W, and an electric light fixture draws 500 W. If all three of these appliances are oper

Question

A hair dryer draws 1 200 W, a curling iron draws 800 W, and an electric light fixture draws 500 W. If all three of these appliances are operating in parallel on a 120-V circuit, what is the total current drawn?

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Yến Oanh 1 year 2021-08-31T21:38:09+00:00 1 Answers 5 views 0

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    2021-08-31T21:39:09+00:00

    Answer:

    Therefore,

    The Total current drawn is 20.83 Ampere.

    Explanation:

    Given:

    Voltage, V = 120 V

    Power of Appliances,

    P1 = 1200-W Hair dryer,

    P2 = 800-W Curling iron,

    P3 = 500-W Light

    To Find:

    Current  flowing in the line, I = ?

    Solution:

    We have Power formula,

    [tex]Power = Voltage\times Current\\P=V\times I[/tex]

    Current for hair dryer,

    [tex]P_{1}=V\times I_{1}[/tex]

    Substituting the values we get

    [tex]1200=120\times I_{1}\\I_{1}=\dfrac{1200}{120}=10\ Ampere[/tex]

    Similarly for curling iron,

    [tex]P_{2}=V\times I_{2}[/tex]

    [tex]800=120\times I_{2}\\I_{2}=\dfrac{800}{120}=6.6667\ Ampere[/tex]

    Similarly for electric light

    [tex]P_{3}=V\times I_{3}[/tex]

    [tex]500=120\times I_{3}\\I_{3}=\dfrac{500}{120}=4.1667\ Ampere[/tex]

    As all three of these appliances are operating in parallel on a 120-V circuit,  the total current will be given as

    [tex]I=I_{1}+I_{2}+I_{3}[/tex]

    Substituting the values we get

    [tex]I=10+6.6667+4.1667=20.8334\approx 20.83\ Ampere[/tex]

    Therefore,

    The Total current drawn is 20.83 Ampere.

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