Answer to Question #104853 in Calculus for Kassim

Question

A voltage V across a resistance R generates a current I =
V/R. If a constant voltage of 22 volts is put across a resistance
that is increasing at a rate of 0.2 ohms per second when the re￾sistance is 5 ohms, at what rate is the current changing?

Answer 1

The relation among voltage $(V)$ , resistance $(R)$ and current $(I)$ is given by the following equation:

$I = \frac{V}{R} \hspace{1 cm}$ (1)

We are given:

The constant votage is:

$V = 22$ volts

The resistance is:

$R = 5$ ohms.

And:

The increasing rate of resistance is:

$\frac{dR}{dt} = 0.2$ ohms per second

Now:

The derivative of equation(1) with respect to time $t$ is:

$\frac{dI}{dt} = V \frac{d}{dt} (\frac{1}{R})$ $\hspace{1 cm}$ [ Because V is a constant ]

$\,\,\,\,\,\,\,\,= V \frac{d}{dt} (R^{-1} )$

$\,\,\,\,\,\,\,\,= V \left( - \frac{1}{R^2} \frac{dR}{dt} \right)$ $\hspace{1 cm} \left[ \because \frac{d}{dx} (x^n) = nx^{n-1} \right]$

$\,\,\,\,\,\,\,\,= - \frac{22}{5^2} .(0.2)$ $\hspace{1cm} \left[ \because V = 22 \, , \, R = 5 \, and \, \frac{dR}{dt} = 0.2 \right]$

$\,\,\,\,\,\,\, \, = - 0.176$

Hence:

The decreasing rate of current is $0.176$ amperes per second