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?

Verified step by step guidance

1. 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 \( R_2 \).
2. To find the rate of change of \( R \), differentiate both sides of the equation with respect to time \( t \). Use implicit differentiation: \( \frac{d}{dt} \left( \frac{1}{R} \right) = \frac{d}{dt} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \).
3. Apply the chain rule to differentiate each term: \( -\frac{1}{R^2} \frac{dR}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} - \frac{1}{R_2^2} \frac{dR_2}{dt} \).
4. 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). Also, substitute \( R_1 = 75 \) ohms and \( R_2 = 50 \) ohms into the differentiated equation.
5. Solve the resulting equation for \( \frac{dR}{dt} \) to find the rate at which \( R \) is changing. This will involve substituting the known values and solving for the unknown rate of change.