Error in the equivalent resistance in parallel
R = R1 R2/(R1+R2). R is the equivalent resistance of R1 and R2 in parallel.
Error in a formula
Maximum error in R
1.78
R = 66.7 ± 1.82.7%
- R10.444 × 31.33
- R20.111 × 40.444
- Totaladd them1.78
This mixes a sum with a product, so it is worked out with the values: the sum of |slope| times error, the worst case.
R = 66.7 ± 1.8
ΔR = |∂R/∂R1| ΔR1 + |∂R/∂R2| ΔR2
This mixes a sum with a product, so it is worked out with the values: the sum of |slope| times error, the worst case. This mixes a sum with a product, so it is worked out with the values. The textbook answer is 66.7 ± 1.8 ohm.
Other formulas
Measured the same thing several times? Use repeated readings, or find the least count of your instrument.
Error in the equivalent resistance in parallel: quick answers
What is the formula for the error in the equivalent resistance in parallel?
R is the equivalent resistance of R1 and R2 in parallel. With R = R1 R2/(R1+R2), the maximum error is given by ΔR = |∂R/∂R1| ΔR1 + |∂R/∂R2| ΔR2. This mixes a sum with a product, so it is worked out with the values: the sum of |slope| times error, the worst case.
How do I work out the error with real numbers?
Take R1 = 100 ± 3 and R2 = 200 ± 4. The maximum error in R is 1.33 + 0.444 = 1.78, so R = 66.7 ± 1.8.
Why are the errors added and never subtracted?
The error shown is the largest possible one. In the worst case every measurement is off in the direction that pushes the result the same way, so the errors pile up. They are added even when the quantity is divided or subtracted.