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    I have been trying to derive P=IV.

    However, I can't find anything online. So, using the definition, I tried to do it myself. I'm not completely sure how close I am. (Probably no where near it lol) - But, if someone could please help, it would appreciated.
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    (Original post by Higherdude)
    I have been trying to derive P=IV.

    However, I can't find anything online. So, using the definition, I tried to do it myself. I'm not completely sure how close I am. (Probably no where near it lol) - But, if someone could please help, it would appreciated.
    Differentiate line 2 in your working. Just simplify and don't multiply by dt.
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    (Original post by morgan8002)
    Differentiate line 2 in your working. Just simplify and don't multiply by dt.
    What do you mean?

    So you skip to QV/T?
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    (Original post by Higherdude)
    What do you mean?

    So you skip to QV/T?
    P = \dfrac{d(QV)}{dt}, then simplify.

    No. What's T?
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    (Original post by morgan8002)
    P = \dfrac{d(QV)}{dt}, then simplify.

    No. What's T?
    Mistype (was meant to be t) - Aw, I see it now! Thank you so much!
    Also, can you do it the way i did it?
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    (Original post by Higherdude)
    Mistype (was meant to be t) - Aw, I see it now! Thank you so much!
    Also, can you do it the way i did it?
    The way you did it is pointless  P=d(QV)/dt = V(dQ/dt)+Q(dV/dt)
    if the voltage is constant then dV/dt=0. The definition of I is dQ/dt. So P=IV
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    (Original post by Higherdude)
    Mistype (was meant to be t) - Aw, I see it now! Thank you so much!
    Also, can you do it the way i did it?
    Assume P and I are constant. We can pull P out of the integral. We can also make the integrals definite rather than indefinite.
    This gives P\displaystyle\int_0^{\Delta t} dt = \displaystyle\int_0^{QV}d(QV)'.
    Then P\Delta t = QV.
    So P = \dfrac{QV}{\Delta t} = IV.

    This argument isn't as general though since we had to assume that I and P were constant. It's also more work. Algebra and/or differentiation are often the way to go.
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    (Original post by morgan8002)
    P = \dfrac{d(QV)}{dt}, then simplify.
    Has anyone done Advanced Higher Physics and could confirm this would be enough for the mathematics part of my underlying physics (in my project report) for this individual experiment.
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    (Original post by Higherdude)
    Has anyone done Advanced Higher Physics and could confirm this would be enough for the mathematics part of my underlying physics (in my project report).
    No I don't know, sorry. Hopefully someone who does will reply. There is a Scottish qualifications forum if you don't get any answers here.
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    (Original post by morgan8002)
    No I don't know, sorry. Hopefully someone who does will reply. There is a Scottish qualifications forum if you don't get any answers here.
    Ye, I used it frequently last year haha.
 
 
 
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