Question

The resistances of two resistors are measured as R1 = (6 ± 0.3) Ω and R2 = (12 ± 0.2) Ω. The maximum percentage error in equivalent resistance when they are connected in parallel is


8.5%


3.9%


6.4%


2.1%

Answer 1

Given :-

The resistances of two resistors are measured

⇝       R₁ = (6 ± 0.3) Ω

⇝      R₂ = (12 ± 0.2) Ω

so,

↦    R₁ = 6 Ω   ( avoiding errors )

↦    R₂ = 12 Ω ( avoiding errors )

↦    Δ R₁ = 0.3 Ω

↦    Δ R₂ = 0.2 Ω

To find :-

→   Maximum percentage error in equivalent resistance when resistors

are connected in parallel .

Formulas used :-

→ Equivalent Resistance of two resistors in parallel combination is calculated as

$\boxed{\bf{\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}}}$

→ Formula to calculate Δ R_p

$\boxed{\bf{\frac{\triangle R_p}{(R_p)^2}=\frac{\triangle R_1}{(R_1)^2}+\frac{ \triangle R_2}{(R_2)^2}}}$

→ Formula to calculate maximum percentage error

$\boxed{\bf{max.\:percentage\:error=\frac{\triangle R_p}{R_p}\times100\%}}$

Solution :-

● Firstly we will calculate equivalent resistance in parallel avoiding errors

$\implies \rm{\frac{1}{R_p}=\frac{1}{6}+\frac{1}{12}}$

$\implies\rm{\frac{1}{R_p}=\frac{18}{72}}$

$\implies\rm{\purple{R_p=4\:\Omega}}$

● Calculating Δ R_p

using formula

$\implies\rm{\frac{\triangle R_p}{(R_p)^2}=\frac{\triangle R_1 }{(R_1)^2}+\frac{\triangle R_2 }{(R_2)^2} }$

$\implies\rm{\frac{\triangle R_p}{R_p}=\frac{\triangle R_1 \:R_p}{(R_1)^2}+\frac{\triangle R_2\: R_p}{(R_2)^2} }$

putting values in RHS

$\implies\rm{\frac{\triangle R_p}{R_p}=\frac{0.3\times4}{(6)^2} +\frac{0.2\times4}{(12)^2} }$

$\implies\rm{\purple{\frac{\triangle R_p}{R_p}=0.0389 }\:\:......eqn(1)}$

● Calculating maximum percentage error

Using formula

$\implies\rm{max. \:percentage\:error = \frac{\triangle R_p}{R_p}\times 100 \%}$

using eqn (1)

$\implies\rm{max.\:percentage\:error = 0.0389\times 100\%}$

$\boxed{\boxed{\large{\bf{\pink{max.\:percentage\:error=3.89\:\%}}}}}$

Hence,

the correct option is 3.9 % .