What is the Rate of Change of Resistance in a Parallel Circuit?

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SUMMARY

The discussion focuses on calculating the rate of change of total resistance (R) in a parallel circuit consisting of two resistors, R1 and R2. The formula for total resistance in parallel is given by 1/R = 1/R1 + 1/R2. When R1 is 10Ω and R2 is 90Ω, with R1 increasing at 0.4Ω/s and R2 at 0.6Ω/s, the rate of change of R can be determined by differentiating the resistance formula with respect to time.

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  • Understanding of parallel resistor circuits
  • Knowledge of calculus, specifically differentiation
  • Familiarity with resistance formulas
  • Basic concepts of rates of change
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This discussion is beneficial for electrical engineering students, physics learners, and anyone interested in applying calculus to electrical circuits.

musad
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Im having trouble with this problem to do with applictions of calculus.

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

If R1 and R2 are increasing at rates of 0.4Ω/s and 0.6Ω/s respectively, how fast is R changing when R1=10Ω and R2=90Ω?
Thanks
 
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musad said:
Im having trouble with this problem to do with applictions of calculus.

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

If R1 and R2 are increasing at rates of 0.4Ω/s and 0.6Ω/s respectively, how fast is R changing when R1=10Ω and R2=90Ω?
Thanks

First consideration ... if R1 and R2 are connected in parallel, is $\displaystyle R = \frac{R_{1}\ R_{2}}{R_{1} + R_{2}} \ne R_{1} + R_{2}$, isn't it?...

Kind regards

$\chi$ $\sigma$
 
I suspect the OP meant to give:

$$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\tag{1}$$

So, use (1) to determine $R$, and differentiate this with respect to time $t$, then plug in the given and computed values. What do you find?
 

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