# Voltage Change

• Oct 1st 2011, 05:28 PM
mgwalker
Voltage Change
The voltage, V, (in volts) across a circuit is given by Ohm's law: V = IR, where I is the current (in amps) flowing through the circuit and R is the resistance (in ohms). If we place two circuits, with resistance R1 and R2, in parallel, then their combined resistance, R, is given by 1/R=1/R1+1/R2. Suppose the current is 2 amps and increasing at 10-2 amp/sec and R1 is 4 ohms and increasing at 0.5 ohm/sec, while R2 is 5 ohms and decreasing at 0.1 ohm/sec. Calculate the rate at which the voltage is changing.
• Oct 2nd 2011, 04:10 AM
tom@ballooncalculus
Re: Voltage Change
I note that your more recent post shows slightly more willing to discuss your own thoughts about the problem. That's good, because it will make it easier to help you.

I'll assume you've worked out the present momentary resistance R. (?) Using the given formula.

Then you can substitute values into 5 of the six balloons on the bottom row here...

http://www.ballooncalculus.org/asy/d...n/rates29b.png

... as explained here, but a key also in this spoiler...

Spoiler:
http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case t), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

For the subsequent part:

http://www.ballooncalculus.org/asy/prod.png

... is the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.

Then solve for $\frac{\mathrm{d}R}{\mathrm{d}t}$.

Next apply the product rule to V = IR...

http://www.ballooncalculus.org/asy/diffProd/VIR.png

To sub into this you need to know, which perhaps you do, whether 10-2 means that or 10.2 or whatever.

I hope that helps.

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