Answer to Question #103908 in Trigonometry for Martin

Question

A voltage V across a resistance R generates a current I=V/R. If a constant voltage of 8 volts is put across a resistance that is increasing at a rate of 0.3 ohms per second when the resistance is 3 ohms, at what rate is the current changing? (Give units.)

Answer 1

Given

V = 8 volts ( constant )

Resistance is increasing at the rate of 0.3 ohms per second

i.e. $\frac{dR}{dt}=0.3\:$ ohms/sec

We need to find rate of change of current when resistance is 3 ohms

$\frac{dI}{dt}=?$ when R= 3 ohms

We know the relation

$I=\frac{V}{R}$

Differentiating with respect to time

$\frac{d}{dt}\left(I\right)=\frac{d}{dt}\left(\frac{V}{R}\right)$

since voltage is constant, it can be taken out.

$\frac{dI}{dt}=V\cdot \frac{d}{dt}\left(\frac{1}{R}\right)$

$\frac{dI}{dt}=V\cdot \:\left(-\frac{1}{R^2}\cdot \frac{dR}{dt}\right)$

substituting values

$\frac{dI}{dt}=8\cdot \:\left(-\frac{1}{3^2}\cdot 0.3\right)=-\frac{4}{15}$ amperes per second

Negative sign indicates current is decreasing with respect to time.

So current is decreasing at the rate of 4/15 amperes per second when R is 3 ohms.