If two resistors with resistances R1 and R2 are connected in parallel, as in the figure below, then the total resistance R, measured in ohms (Ω), is given by 1/R = 1/R1 + 1/R2

Question

if R₁ and R₂ are increasing at rates of 0.3 Ω/s and 0.2 Ω/s, respectively, how fast is R changing when R₁ = 60 Ω and R₂ = 120 Ω? (Round your answer to three decimal places.)

Answer 1

Well we know the equation for the total resistance here is is called R

R= 1/(R_1) + 1/(R_2) where R_1 and R_2 are 2 parts of the total resistance R of this specific circuit

We also know the rate of R_1 and R_1 which you can think of as velocity and we know the values of R_1 and R_2 which you can think of as displacement

Remember from kinematics that velocity is the derivative of displacement

Well the same works here!

The rate of the total resistance R is the derivative of R

From the given equation R= (1/(R_1))+(1/(R_2)) = ((R_2)+(R_1))/((R_1)(R_2)) by rewriting the right as one term

we find its derivative with respect to time (in seconds) by differentiating using the product or quotient rule of 3 terms. the 3 terms here are R_1 , R_2 , and (((R_1)+ ((R_2)))^-1) since these 3 parts make the equation of the total resistance R

----Note: to make this problem easier I moved the denominator to the numerator which is the term ((R_1)+(R_2)) by giving it a (-1) power in the numerator which makes this problem a product rule differentiation instead of a quotient rule differentiation (much tougher)

Using product rule of 3 terms I get:

dR/dt = (d(R_1)/dt)(R_2)(((R_1)+(R_2))^-1) + (d(R_2)/dt)(R_1)(((R_1)+(R_2))^-1) - (d((R_1)+(R_2))/dt)((R_1)+R_2)^-2)(R_1)(R_2))

and plug in all the values given to get:

= 0.156 ohms per second !!

Answer 2

1/R = 1/R₁ + 1/R₂

R^-1 = R₁^-1 + R₂^-1

-R^-2(dR/dt) = -R₁^-2(dR₁/dt) - R₂^-2(dR₂/dt)

dR/dt = -R²[-(dR₁/dt) / R₁² - (dR₂/dt) / R₂²]

When R₁ = 60 and R₂ = 120, 1/R = 1/60 + 1/120 = 3/120 = 1/40. So, R = 40.

Therefore, dR/dt = (-1600)[-0.3/3600 - 0.2/14400] = 0.111 Ω/sec

R is increasing at the rate of 0.111 ohms/sec.