Question

Numarical The value of two registance are measured
in experimentare given by. R1= 5+-0.252
R =10+- 0.1 ohm calculate restance in series
and Parllel​

Answer 1

YOUR QUESTION:-

The value of two resistance measured in experiment are given by R1 = 5 ± 0.152 Ω and

R2 = 10 ± 0.1 Ω. Calculate resultant in series and parallel.

ANSWER:-

GIVEN:-

R1 = 5 ± 0.152 Ω

So, R1 = 5 Ω & ∆R1 = 0.152 Ω

R2 = 10 ± 0.1 Ω

So, R2 = 10Ω & ∆R2 = 0.1 Ω

CASE 1 :-

FOR SERIES COMBINATION:-

We know that in series combination, resultant resistance is the sum of the given resistances.

So, R = R1 + R2

& ∆R = ∆R1 + ∆R2

Now, putting the values , we get

R = (5 + 10 )Ω

R = 15 Ω

And, ∆R = (0.152 + 0.1) Ω

∆R = 0.252 Ω

hence , resistance with error is

R = 15 ± 0.252 Ω

CASE 2:-

FOR PARALLEL COMBINATION:-

We know that in parallel combination , reciprocal of resultant resistance is equal to the sum of the reciprocal of the given resistances.

So, 1/R = 1/R1 + 1/R2

R = (R1 × R2) / (R1 + R2)

and , ∆R = ∆R1 + ∆R2

Now, putting the values , we get

R = (5 × 10) / (10 + 5)

R = 50/15

R = 3.33 Ω

and, ∆R = 0.152 + 0.1

∆R = 0.252 Ω

Hence, resistance with error is

R = 3.33 ± 0.252 Ω