Question

For two resistors R1 and R2 connected in a parallel , find the relative error in their equivalent resistance , if R1 = (50+- 2) ohm and R2 = (100+- 30 ) ohm

Answer 1

Hey there!

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   ←(Ans)

Answer 2

Answer:  0.03666

Explanation:

Hey there!

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:

= 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 =  

                      =(11/9)/100/3

                      =0.03666   ←(Ans)