Question

The combined electrical resistance R of two resistors R_1 and R_2, connected in parallel, is given by 1/R = 1/R_1 + 1/R_2 where R, R_1, and R_2 are measured in ohms. R_1 and R_2 are increasing at rates of 1 and 1.5 ohms per second, respectively. At what rate is R changing when R_1=50 ohms and R_2=75 ohms?

Answer 1

Answer:

$0.6\Omega/s$

Explanation:

We are given that

$R_1=150\Omega$

$R_2=75\Omega$

$\frac{dR_1}{dt}=1\Omega/s$

$\frac{dR_2}{dt}=1.5\Omega/s$

We have to find the rate at which R is changing.

In parallel

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

Using the formula

$\frac{1}{R}=\frac{1}{50}+\frac{1}{75}=\frac{3+2}{150}=\frac{5}{150}=\frac{1}{30}$

$R=30\Omega$

$-\frac{1}{R^2}\frac{dR}{dt}=-\frac{1}{R^2_1}\frac{dR_1}{dt}-\frac{1}{R^2_2}\frac{dR_2}{dt}$

$\frac{1}{R^2}\frac{dR}{dt}=\frac{1}{R^2_1}\frac{dR_1}{dt}+\frac{1}{R^2_2}\frac{dR_2}{dt}$

Substitute the values

$\frac{1}{900}\frac{dR}{dt}=1\times\frac{1}{2500}+1.5\times\frac{1}{(75)^2}$

$\frac{dR}{dt}=900(\frac{1}{2500}+\frac{1.5}{5625}$

$\frac{dR}{dt}=0.6\Omega/s$