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Electrical Circuits and Power

  1. Jan 31, 2016 #1
    1. The problem statement, all variables and given/known data
    Questions 10&11 in pictures.

    2. Relevant equations
    P=IV
    W=VIt
    W=Pt
    3. The attempt at a solution
    For 10a) P=IV
    P=70*14= 980W.
    10b) P=IV
    1500=12*I
    I=125A (which is massive so probably wrong)
    10c)P=IV
    60=I*12
    I=5A
    The 8A fuse because the current flowing through each headlamp should be 5A. Anymore than that could be dangerous.
    10d) W=VIt
    W=12*1*62
    W=744J (which is quite low??)
    10e)
    W=Pt
    It just doesn't give me the right answer as the sidelights appear to use more than the whole battery capacity if left on for 12 hours, which makes me believe I've done the whole of Q10 wrong.
    Q11)
    Maybe you could work out the work done / GPE then you could use P=W/t to work out how much power the car needs to go up the hill? (Using an hour as t from the 90km/h?)

    I would greatly appreciate any help.
     

    Attached Files:

  2. jcsd
  3. Jan 31, 2016 #2

    SteamKing

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    Your answer for 10b) is correct. Starter motors take a large amount of current to turn the engine over. They do so for only a short time, however, lest the battery be drained completely.
     
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