Question

how can i find the error from the resistanct

Answer 1

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.

HOPE IT HELPS YOU :)

Answer 2

There are two resistors with resistance values R1=100±3 ohm and R2=200±4 ohm. Find the equivalent resistance of parallel combination.

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.