The rate of change of R with time in the given equation is 0.004 ohm/s
Given parameters:
[tex]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \\\\\frac{dR_1}{dt} = 0.3 \ ohm/s\\\\\frac{dR_2}{dt} = 0.2 \ ohm/s\\\\R_1 = 60 \ ohms\\\\R_2 = 80 \ ohms[/tex]
To find:
- The rate of change of R with time in the given equation.
First determine the value of R from the given equation;
[tex]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \\\\\frac{1}{R} = \frac{1}{60} + \frac{1}{80} \\\\\frac{1}{R} = \frac{4 + 3}{240} \\\\\frac{1}{R} = \frac{7}{240} \\\\R = \frac{240}{7} = 34.286 \ ohms[/tex]
Finally, to determine the rate of change of R, differentiate the given equation.
[tex]\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} \\\\\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}\\\\\frac{dR}{dt} = R^2(\frac{1}{R_1^2} \frac{dR_1}{dt} + \frac{1}{R_2^2} \frac{dR_2}{dt})[/tex]
[tex]\frac{dR}{dt} = 34.286(\frac{1}{(60)^2} \times 0.3 \ \ \ + \ \ \ \frac{1}{(80)^2} \times 0.2)\\\\\frac{dR}{dt} = 34.286(8.333 \times 10^{-5} \ \ \ + \ \ \ 3.125 \times 10^{-5})\\\\\frac{dR}{dt} = 34.286(11.458 \times 10^{-5})\\\\\frac{dR}{dt} = 0.00393\\\\\frac{dR}{dt} \approx 0.004 \ ohm/s[/tex]
Thus, from the given equation the rate of change of R with time is 0.004 ohm/s
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