Part of a circuit has two resistors connected in parallel as shown in the figure. R(1) is a constant resistor of 10 ohm, while R(2) is a variable resistor with resistance that varies at a rate of 2 ohms per minute. The total, or effective, resistance, R, provided by this circuit is given by:
1/R=1/R(1)+1/R(2)
At what rate does the effective resistance change when R(2) is 100 ohms? Round off your answer to four decimal places.
20
To find the rate at which the effective resistance (R) changes when R(2) is 100 ohms, we need to differentiate the equation with respect to time. The given equation is:
1/R = 1/R(1) + 1/R(2)
Taking the reciprocal of both sides gives:
R = (R(1)*R(2))/(R(1) + R(2))
Now, let's differentiate this equation with respect to time, t:
dR/dt = [d/dt (R(1)*R(2)) * (R(1) + R(2)) - (R(1)*R(2)) * d/dt (R(1) + R(2))] / (R(1) + R(2))^2
The rate of change of R(1) is zero since it is a constant resistor. Therefore, we only need to consider the rate of change of R(2), which is given as 2 ohms per minute. So, d/dt (R(2)) = 2.
Substituting the given values and differentiating, we get:
dR/dt = [0 * (R(1) + 100) - (10 * 100) * 2]/(10 + 100)^2
Simplifying further:
dR/dt = (-2000)/110^2
Calculating the value:
dR/dt ≈ -0.1653 ohms per minute
Therefore, the effective resistance changes at a rate of approximately -0.1653 ohms per minute when R(2) is 100 ohms.