Question

the value of two resistors are R1=(6+_0.3)kohm and R2=(10+_0.2)kohm. The % error in equivalent resistance when they are connected in parallel is?

Answer 1

% error in equivalent resistance when two resistors R₁ and R₂ are connected in parallel is given by
% error in Req = % error in R₁ + % error in R₂
= ∆R₁/R₁ × 100 + ∆R₂/R₂ × 100
Here, ∆R₁ = 0.3 , ∆R₂ = 0.2 , R₁ = 6 and R₂ = 10

∴ % error in Req = 0.3/6 × 100 + 0.2/10 × 100
= 30/6 + 20/10
= 5 + 2 = 7 %

Hence , % error in equivalent resistance = 7 %

Answer 2

Given conditions ⇒

Value of the Resistor R₁ = (6 +/- 0.3) Ω
R₁ = 6
∴ ΔR₁ = +/- 0.3 Ω

Now, % Error in R₁ = ΔR₁/R₁ × 100
= 0.3/6 × 100
= 5 %

Value of the Resistor R₂ = (10 +/- 0.2) Ω
R₂ = 10
∴ ΔR₂ = +/- 0.2 Ω

∴ % Error in R₂ = ΔR₂/R₂ × 100
= 0.2/10 × 100
= 2 %

A/c to the Question, Resistors are connected in Parallel, 
∴  Error % in eq. Resistance (R) = Error % in R₁ + Error % in R₂
∴  Error % in Eq.Resistance = 5% + 2 %
   = 7 %

Hence, the % Error in the Equivalent Resistance, when the Resistors are connected in parallel is 7 %.


Hope it helps.