Question

If two resistances of values r1= (2.0 + 0.1) and r2= (12.3+ 0.2) are put (i) in parallel and (ii) in series, find the error in the equivalent resistance.

Answer 1

in parallel combination = (1.72 ± 0.78) Ω and in series combination = (14.3 ± 0.3) Ω

it is given that

R1 = (2 ± 0.1) and R2 = (12. ± 0.2)

in parallel combination,

1/Req = 1/R1 + 1/R2

⇒1/Req = 1/2 + 1/12.3

⇒Req = 2 × 12.3/(14.3) = 1.72 Ω

now error in the equivalent resistance, ∆Req

as ∆Req = ∆R1 (R1/R1 + R2)² + ∆R2 (R2/R1 + R2)²

= 0.1 × (12.3/14.3)² + 0.2(2/14.3)²

= 0.78 Ω

hence, equivalent resistance will be (1.72 ± 0.78)Ω

in series combination,

Req = R1 + R2

= (2 + 12.3) = 14.3 Ω

and error in the equivalent resistance, ∆Req

∆Req = ∆R1 + ∆R2

= 0.1 + 0.2 = 0.3

so, equivalent resistance is (14.3 ± 0.3)Ω

Answer 2

$\huge\bold\purple{Answer:-}$

it is given that

R1 = (2 ± 0.1) and R2 = (12. ± 0.2)

in parallel combination,

1/Req = 1/R1 + 1/R2

⇒1/Req = 1/2 + 1/12.3

⇒Req = 2 × 12.3/(14.3) = 1.72 Ω

now error in the equivalent resistance, ∆Req

as ∆Req = ∆R1 (R1/R1 + R2)² + ∆R2 (R2/R1 + R2)²

= 0.1 × (12.3/14.3)² + 0.2(2/14.3)²

= 0.78 Ω

hence, equivalent resistance will be (1.72 ± 0.78)Ω

in series combination,

Req = R1 + R2

= (2 + 12.3) = 14.3 Ω

and error in the equivalent resistance, ∆Req

∆Req = ∆R1 + ∆R2

= 0.1 + 0.2 = 0.3

so, equivalent resistance is (14.3 ± 0.3)Ω