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  • error in parallel resistance part is what I m not getting How is it derived

That error in parallel resistance part is what I m not getting....How is it derived?

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@ Anupam 

Since R1 and R2 are in parallel 

So \frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} ......(1)

Differtiate LHS we get \frac{\partial (\frac{1}{R_{p}} )}{\partial R_{p}}=\frac{-1}{R_{p}^2}

                                    \partial (\frac{1}{R_{p}} )=\frac{ -\partial R_{p}}{R_{p}^2}

                                    \Delta (\frac{1}{R_{p}} )=\frac{-\Delta R_{p}}{R_{p}^2} ....(2)

Similary We get for RHS   \frac{\partial (\frac{1}{R_{1}} )}{\partial R_{1}}=\frac{-1}{R_{1}^2}

                                            \partial (\frac{1}{R_{1}} )=\frac{ -\partial R_{1}}{R_{1}^2}

                                           \Delta (\frac{1}{R_{1}} )=\frac{-\Delta R_{1}}{R_{1}^2} ....(3)

&                                      \Delta (\frac{1}{R_{2}} )=\frac{-\Delta R_{2}}{R_{2}^2} ....(4)

So from equation 1,2 ,3 4 We get

                                     \frac{-\Delta R_{p}}{R_{p}^2} = \frac{-\Delta R_{1}}{R_{1}^2} +\frac{-\Delta R_{2}}{R_{2}^2}

                        So       {\Delta R_{p}} =\Delta R_{1} \frac{R_{p}^2}{R_{1}^2} +\Delta R_{2} \frac{R_{p}^2}{R_{2}^2}

                                

Posted by

avinash.dongre

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