How the current changes

C

Cpro1

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Apr 26, 2007
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The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasing as the resistor heats up. Use Ohm's Law, V = IR, to find how the current I is changing at the moment when R = 400 ohms, I = .08 A, dV/dt = -.01 V/s, and dR/dt = .03 ohms/s.

I think i am suppost to use the chain rule, an equation similar to:
dz/dt = dz/dx*dx/dt + dz/dy*dy/dt
(This problem is in the chain rule section of my book, so i'm pretty sure i'm suppost to use the chain rule)

I'm not sure how to set up the problem. I'm guessing something like:
dI/dt = dI/dV*dV/dt + dI/R*dR/dt
dI/dt = dI/dV*(-.01) + dI/R*(.03)
I'm not sure how to get dI/dV and dI/R.

However, i'm not sure i set up the equation correctly in the first place.
 
You just shuffled your formula and then got a little derivative happy on the product rule.

V(t) = I(t)*R(t)

So,

dV/dt = I*(dR/dt) + R*(dI/dt)
 
dV/dt = I*(dR/dt) + R*(dI/dt)

ok, so:
-.01 = .08(.03)+400(dI/dt)
-.0124 = 400(dI/dt)
-.0124/400 = dI/dt

so, the current is changing at a rate of -.0124/400 ?
 
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