Question

for two resistors R1 and R2 connected in parallel, find the relaive error in their equilavent resistor .if R1=(50+/-2)ohm and R2=(100+/-3)

Answer 1

Hi

Two resistors (R1 & R2) are connected in parallel.

We know,

1/re = 1/r₁ + 1/r₂

Then re = r₁r₂/r₁+r₂

Putting values:

=50×100/50+100

=5000/160

=100/3 ← Ans

Now, Parallel Connection error:

R_{error} = R1^2 dB +R2^2dA/ (R1+R2)^2

=50×50 (3)+100×100 (2)/150×150

=7500+20000/150×150

=11/9 ← (Ans)

Relative Error = \frac{R_{error}}{R}

=(11/9)/100/3

=0.03666