Mathematics in Physics

JEE Main level, single correct, medium

75 of 121

Question 75 of 121

Two resistance are given as R1=(10±0.5)Ω{R}_{1}=(10\pm 0.5)\Omega and R2=(15±0.5)Ω.{R}_{2}=(15\pm 0.5)\Omega . The percentage error in the measurement of equivalent resistance when they are connected in parallel is

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Answer: (D) 4.334.33

For a parallel combination the equivalent resistance is
1R=1R1+1R2...(i)\dfrac{1}{R}=\dfrac{1}{{R}_{1}}+\dfrac{1}{{R}_{2}}...(i)
A general expression for error can be written as x±∆xx\pm ∆x
Differentiating equation (i)
dRR2=dR1R12+dR2R22...(ii)\dfrac{dR}{{R}^{2}}=\dfrac{d{R}_{1}}{{R}_{1}^{2}}+\dfrac{d{R}_{2}}{{R}_{2}^{2}}...(ii)
The equivalent resistance with {R}_{1}=10\Omega &{R}_{2}=15\Omega is
1R=110+115\dfrac{1}{R}=\dfrac{1}{10}+\dfrac{1}{15}
⇒R=15025=6Ω\Rightarrow R=\dfrac{150}{25}=6\Omega
Substituting the above value in equation (ii)
dR=36(0.5100+0.515×15)=0.26dR=36\left(\dfrac{0.5}{100}+\dfrac{0.5}{15\times 15}\right)=0.26
⇒R=6±0.26\Rightarrow R=6\pm 0.26
Hence, the percentage error is
dRR×100=0.266×100\dfrac{dR}{R}\times 100=\dfrac{0.26}{6}\times 100=4.33=4.33%

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