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Proof that current is the rate of change of charge with respect to time Watch

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    I've been looking for a proof of this but its impossible to find one.
    Ive started with the fact that E=QV
    therefore dQ/dt=d(E/V)/dt
    dE/dt= power = P= IV
    but I'm stuck with derivative of voltage.
    I know quotient rule will help.

    thanks
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    (Original post by kennz)
    I've been looking for a proof of this but its impossible to find one.
    Ive started with the fact that E=QV
    therefore dQ/dt=d(E/V)/dt
    dE/dt= power = P= IV
    but I'm stuck with derivative of voltage.
    I know quotient rule will help.

    thanks
    Instead of using calculus, you could just use normal equations:
    E=QV and \frac{E}{t}=P=IV
    Equate the two equations for E:

    \frac{E}{t}=\frac{QV}{t}=P=IV

    Cancel the voltages off both sides after combining:
    \frac{QV}{t}=IV \rightarrow \frac{Q}{t}=I

    So current is change in charge, respect to time.
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    (Original post by The-Spartan)
    Instead of using calculus, you could just use normal equations:
    E=QV and \frac{E}{t}=P=IV
    Equate the two equations for E:

    \frac{E}{t}=\frac{QV}{t}=P=IV

    Cancel the voltages off both sides after combining:
    \frac{QV}{t}=IV \rightarrow \frac{Q}{t}=I

    So current is change in charge, respect to time.
    Thankk you for that, is there not a way to do it with calculus though?
 
 
 
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