Given `R_(1) = 5.0 +- 0.2 Omega, and R_(2) = 10.0 +- 0.1 Omega`. What is the total resistance in parallel with possible `%` error?

Given `R_(1) = 5.0 +- 0.2 Omega, and R_(2) = 10.0 +- 0.1 Omega`. What

1.	Given `R_(1) = 5.0 +- 0.2 Omega, and R_(2) = 10.0 +- 0.1 Omega`. What is the total resistance in parallel with possible `%` error?
Answer» In parallel , `R_(P) = (R_(1) R_(2))/( R_(1) + R_(2)) = ( 5.0 xx 10.0)/( 5.0 + 10.0) = ( 50)/(15) = 3.3 Omega` Also `(Delta R_(P))/(R_(P)) xx 100 = (Delta R_(1))/(R_(1))xx100+(Delta R_(2))/(R_(2))xx100+(Delta(R_(1)+R_(2)))/(R_(1)+R_(2))xx100` `= (0.2) /(5.0) xx 100 + (0.1)/(10.0) xx 100 + (0.3)/(15) xx 100 = 7%` `:. R_(P) = 3.3 Omega +- 7%`

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Given `R_(1) = 5.0 +- 0.2 Omega, and R_(2) = 10.0 +- 0.1 Omega`. What is the total resistance in parallel with possible `%` error?

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