# Thread: Calculating rate of voltage change through a circuit.

1. ## 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?

2. 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

3. 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.

4. 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

5. Originally Posted by Dr Zoidburg
$\displaystyle \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} = \frac{R_1R_2}{R_1+R_2}$
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.

6. 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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