Increment value of equvalent resistance when the resistances are parallel

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SUMMARY

The discussion focuses on calculating the incremental change in equivalent resistance (R3) when two resistances (R1 and R2) are connected in parallel. The formula for equivalent resistance is given as R3 = R1 * R2 / (R1 + R2). The total differential of R3 is expressed as dR3 = (∂R3/∂R1) dR1 + (∂R3/∂R2) dR2. The relationship d(R3)/R3 = d(R1)/R1 + d(R2)/R2 + d(R1 + R2)/(R1 + R2) is derived from these principles, illustrating how changes in R1 and R2 affect R3.

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sudhir1962
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R1 & R2 are two resistances in parallel. d(R1) & d(R2) are the incremnet in resistance. If R3 is the equvalent resisitance such that R3=R1*R2/R1+R2, then if d(R3) is increment in equvalent resistance. What is the value of d(R3)/R3.
 
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Hmm, well, if this is anything like I'm seeing it to be, you're dealing with functions:
[tex]R_3:\mathbb{R}^2\to\mathbb{R}[/tex]
So the total differential/change of [tex]R_3[/tex] would be:
[tex]dR_3=\frac{\partial R_3}{\partial R_1} dR_1 + \frac{\partial R_3}{\partial R_2} dR_2[/tex]
 
Actually I found it in some place as follows

d(R3)/R3= d(R1)/R1+d(R2)/R2+d(R1+R2)/(R1+R2), please let me know how it is arrived.
 
Last edited:

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