Question

Resistors connected in parallel If two resistors of R₁ and R₂ ohms are connected in parallel in an electric circuit to make an R-ohm resistor, the value of R can be found from the equation

1/R = 1/R₁ + 1/R₂

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If R₁ is decreasing at the rate of 1ohm/sec and R₂ is increasing at the rate of 0.5 ohm/sec, at what rate is R changing when R₁ = 75 ohms and R₂ = 50 ohms?

Accepted Answer

Start by understanding the given equation for resistors in parallel: $$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} . $$This equation relates the total resistance $$ R  to $$the individual resistances $$ R_1 $$ and $$ R_2 . $$Differentiate both sides of the equation with respect to time $$ t  to $$find the rate of change of $$ R . $$Use implicit differentiation: $$ \frac{d}{dt} \left( \frac{1}{R} \right) = \frac{d}{dt} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) . $$Apply the chain rule to differentiate: $$ -\frac{1}{R^2} \frac{dR}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} - \frac{1}{R_2^2} \frac{dR_2}{dt} . $$Substitute the given rates of change: $$ \frac{dR_1}{dt} = -1 $$ ohm/sec (since $$ R_1  is $$decreasing) and $$ \frac{dR_2}{dt} = 0.5 $$ ohm/sec (since $$ R_2  is $$increasing). Substitute the given values $$ R_1 = 75 $$ ohms and $$ R_2 = 50 $$ ohms into the differentiated equation to solve for $$ \frac{dR}{dt} $$, the rate at which $$ R  is $$changing.

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Key Concepts

Related Rates

Implicit Differentiation

Reciprocal Functions