Question

2) Two resistances R (16 + 0.310 and R: = (48 = 0.510, are connected in parallel. Find the
resistance of the combination and maximum percentage error.​

Answer 1

Here, `R_1 = 16 ohm, Delta R_1 = 0.3 Omega`

`R_2 = 48 ohm, DeltaR_2 =0.5 Omega, R_p = ?`

`(1)/(R_p) = (1)/(R_1) +(1)/(R_2) = (1)/(16) +(1)/(48) = (3+1)/(48) = (4)/(48) = (1)/(12)`

`R_p = 12 ohm`

On differentating, `(1)/(R_p) = (1)/(R_1) +(1)/(R_2)`, we get `(-DeltaR_p)/(R_p^2) = - (DeltaR_1)/(R_1^2) - (DeltaR_2)/(R_2^2)`

`:. DeltaR_p = DeltaR_1((R_p)/(R_1))^2 + DeltaR_2((R_p)/(R_2))^2 = 0.3((12)/(16))^2 +0.5((12)/(48))^2`

= `0.16875+0.03125 = 0.20 ohm`

`(DeltaR_p)/(R_p)xx10 = (0.20)/(12)xx100 = 1.6%`

Answer 2

Explanation:

I hope it helps u a lot

thanks