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

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In a parallel circuit, the total resistance R is calculated using the formula 1/R = 1/R1 + 1/R2. To determine how fast R is changing when R1 is 10Ω and R2 is 90Ω, the rates of change for R1 and R2, which are 0.4Ω/s and 0.6Ω/s respectively, need to be differentiated with respect to time. The correct approach involves differentiating the total resistance equation and substituting the known values. This method will yield the rate of change of resistance R at the specified conditions. Understanding these calculations is crucial for 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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