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Please help, one question about electric circuits

  1. Jan 16, 2007 #1
    1. The problem statement, all variables and given/known data
    I have to give the power being absorbed by the circuit element. There's a picture. This is an easy problem. They give me the current flow in a function of time, the voltage is also in function of time. The time in which I require to give the power is at 2ms. The function for the current is 8 cos (1000t) mA , and the one for voltage is 4 sin (1000t) V. MY QUESTION IS: Do I insert 2 as the variable for time or do I change 2ms into seconds (.002s) and put it in the function?

    Yeah I know this has almost nothing to do with circuits but I hope someone helps me. HW is due tomorrow...

    2. Relevant equations
    P=Vi


    3. The attempt at a solution
     
  2. jcsd
  3. Jan 16, 2007 #2

    chroot

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    Well, the SI standard unit of time is the second, not the millisecond, so I presume that the current and voltage are given in terms of seconds. It is possible, though unlikely, that t is in some other unit instead.

    - Warren
     
  4. Jan 16, 2007 #3
    yes, I always change everything to SI standard units. BUT, when a variable is inside the argument of a cosine o sine, what do you do? I had this same debate while doing wave function problems (sin (kx - wt)) . So what would you do?
     
  5. Jan 16, 2007 #4

    chroot

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    It depends on the units chosen by the person who wrote the problem. Normally, any unit of time is expressed in seconds, but this is just a convention. Being part of the argument to a trig function makes no difference; t is conventionally in seconds.

    I would use all times in seconds.

    - Warren
     
  6. Jan 16, 2007 #5
    thx chroot.

    I say this about trig functions because when you have something like sin (4t). If you put say 3s into "t", then you have the sine of 12 seconds, then when you do the trig. operation, the unit seconds is eliminated, or at least that's what I think.
    The angle measures 12 s (even if that doesn't make sense), but sine is not the angle but rather the side of the triangle oposing the angle 12s over the hypothenuse, so sin (12s) is no measured in seconds.

    This is very weird lol, but I've been having this confusion forever.
     
  7. Jan 16, 2007 #6

    chroot

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    The argument of a trig function is an angle, usually measured in radians. If you have an expression like sin(4t), where t is in seconds, then "4" is actually in units of radians per second. It's unfortunately quite common to omit the "radians per second" unit, but it's always there.

    No units are "eliminated." The resulting argument (4t) is in radians.

    [itex]
    \frac{{4\,{\text{radians}}}}
    {{{\text{second}}}} \cdot t\,{\text{seconds}}\,{\text{ = 4}}t{\text{ radians}}
    [/itex]

    The number "4" in your example is sometimes even called an "angular frequency," since it has units of angle per unit time, and is often represented in variable form by the greek letter [itex]\omega[/itex]. This is the greek letter conventionally used to describe angular frequencies, which you called "w" in post #3.

    - Warren
     
  8. Jan 16, 2007 #7

    Meir Achuz

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    The t in 1000t should be in seconds.
     
  9. Jan 17, 2007 #8
    Thanks guys.
    I'll receive my homework back in a couple of days, should be right...
     
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