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The power $P$ (watts) of an electric circuit is related to the circuit's resistance $R$ (ohms) and current $I$ (amperes) by equation $P=RI^2$. How is $\frac{dR}{dt}$ related to $\frac{dI}{dt}$ if $P=P_0$ is constant?

First, I relate the equation $P=RI^2$ to $\frac{dP}{dt} = I^2 \frac{dR}{dt} + R \frac{dI}{dt}$.

If $P=P_0$ is constant can I say that $\frac{dP}{dt}=0$, therefore $\frac{dR}{dt} = \frac{-R}{I^2} \frac{dI}{dt}$.

Gary
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user307640
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1 Answers1

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$0=\frac{d}{dt}(RI^2)=I^2\frac{dR}{dt}+R\frac{d}{dt}(I^2)=I^2\frac{dR}{dt}+R(2I\frac{dI}{dt})\implies I\frac{dR}{dt}+2R\frac{dI}{dt}=0$

If $I(t)\neq 0$. you have:

$\frac{dR}{dt}=-2\frac{R}{I}\frac{dI}{dt}$

It may be interesting to find $R=R(I)$:

$\frac{dR}{R}=-2\frac{dI}{I}\implies \int_{R_0}^{R}\frac{dR}{R}=-2\int_{I_0}^{I}\frac{dI}{I}\implies \log|R|_{R_0}^{R}=-2\log|I|_{I_0}^{I}\implies R(I)=R(I_0)( \frac{I_0}{I} )^2$

Where, $R(I)=R(I_0)( \frac{I_0}{I} )^2=\frac{P_0}{I^2}$, as expected. So the differential relationship is validated.

Gino
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