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Thread: Calculating rate of voltage change through a circuit.

  1. #1
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    Calculating rate of voltage change through a circuit.

    Here's the question:
    Voltage across a circuit is given by Ohm's law, V=IR, where I=the current & R= resistance.
    If we place two circuits with resistance R1 and R2 in parallel their combined resistance, R, is given as
    $\displaystyle \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

    Suppose the current is 2 amps and increasing at $\displaystyle 10^(-2)$ amp/sec and $\displaystyle R_1$ is 3 ohms and increasing at 0.5 ohm/sec, while $\displaystyle R_2$ is 5 ohms and decreasing at 0.1 ohm/sec.
    Calculate the rate at which the voltage is changing.

    This is what I've got:
    Need to find $\displaystyle \frac{dV}{dt}$
    Using the Chain Rule, I get this:
    $\displaystyle \frac{dV}{dt}=I\frac{dR}{dt}+R\frac{dI}{dt}$
    where
    $\displaystyle R_1=3$ and $\displaystyle \frac{dR_1}{dt}=0.5$

    $\displaystyle R_2=5$ and $\displaystyle \frac{dR_2}{dt}=-0.1$

    $\displaystyle I=2$ and $\displaystyle \frac{dI}{dt}=0.01$

    $\displaystyle \frac{1}{R}=\frac{1}{3}+\frac{1}{5} => R=\frac{15}{8}$

    $\displaystyle \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} = \frac{R_1R_2}{R_1+R_2}$

    Which all comes to:
    $\displaystyle \frac{dV}{dt}=2\frac{dR}{dt}+\frac{15}{8}*\frac{1} {100}$

    $\displaystyle \frac{dV}{dt}=2\frac{dR}{dt}+\frac{15}{800}$

    and that's as far as I've got! How do I find the change in resistance? I've tried a few ways but none of them seem satisfactory. any ideas?
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  2. #2
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    Since R depends on $\displaystyle R_1\text{ and }R_2$ which then depend on t, there are two parts to $\displaystyle \frac{dR}{dt}$:

    $\displaystyle \frac{dR}{dt}=\frac{dR}{dR_1}\frac{dR_1}{dt}+\frac {dR}{dR_2}\frac{dR_2}{dt}$

    The two expressions $\displaystyle \frac{dR}{dR_1}$ and $\displaystyle \frac{dR}{dR_2}$ are actually partial derivatives, by the way. You treat one resistance as a constant while taking the derivative with respect to the other.

    - Hollywood
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  3. #3
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    Thanks. I was thinking that might be the answer.
    Is this correct:
    $\displaystyle \frac{dR}{dR_1}=-\frac{R_2^2}{(R_1+R_2)^2}$
    with $\displaystyle dR_2$ being the same.
    Last edited by Dr Zoidburg; Apr 17th 2010 at 12:09 AM. Reason: made a mistake in differentiating
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  4. #4
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    I get the same without the minus sign:

    $\displaystyle \frac{d}{dR_1}\left(\frac{R_1R_2}{R_1+R_2}\right)= \frac{(R_1+R_2)\frac{d}{dR_1}(R_1R_2)-R_1R_2\frac{d}{dR_1}(R_1+R_2)}{(R_1+R_2)^2}$ $\displaystyle =\frac{(R_1+R_2)(R_2)-(R_1R_2)(1)}{(R_1+R_2)^2}=\frac{R_2^2}{(R_1+R_2)^2 }$

    - Hollywood
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  5. #5
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    Quote Originally Posted by Dr Zoidburg View Post
    $\displaystyle \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} = \frac{R_1R_2}{R_1+R_2}$
    This was probably just a typo but I wanted to make sure. This should be:

    $\displaystyle \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} = \frac{R_1+R_2}{R_1 R_2}$

    Which implies:

    $\displaystyle R = \frac{R_1 R_2}{R_1+R_2}$

    Which further implies:

    $\displaystyle \frac{dR}{dt} = \frac{(R_1+R_2)\left(\tfrac{dR_1}{dt}\cdot R_2 + R_1 \cdot \tfrac{dR_2}{dt}\right) - (R_1R_2) \left(\tfrac{dR_1}{dt}+ \tfrac{dR_2}{dt} \right) }{(R_1+R_2)^2}$

    You know all the other values so you can just substitute in and find $\displaystyle \frac{dR}{dt}$. This is an alternative approach to doing partial derivatives.
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  6. #6
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    thanks for that. All things considered I think I'll stick with the partial derivative - the other eqn looks very unsightly!
    drum: it was a typo, I meant to say thus R = ...
    holly: thanks for the heads-up on the sign.
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