Question

2 resistance are expressed as R1=( 4 +- 0.5) ohm and R2= (12+-0.5)ohm. what is net resistance when they are connected in -

i- series ii- parallel ?​

Answer 1

Answer:

According to what I've learnt, in any expression of multiplication or division, the percentage errors of each term are added up to find the equivalent percentage error. That is, if

y=ABC

then

%error in y=%error in A+%error in B+%error in C

For the above problem, let Rs denote series combination. Then Rs=300±7 ohm.

Let Rp denote parallel combination.

∴Rp=R1R2R1+R2=R1R2Rs

Ignoring errors, we get Rp=2003 ohm =66.67 ohm

%error in R1=3, %error in R2=2, %error in Rs=73

Hence, %error in Rp=3+2+73=223

So, error in Rp will be 223% of 2003, which is approximately 4.89.

Hence, I got Rp=66.67±4.89 ohm.

However, the book used the formula described and proved here and arrived at the answer Rp=66.67±1.8 ohm.

Answer 2

Answer can be seen in the image provided