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

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Units & Measurement

Two resistances are given as

R_{1} = (10 \pm 0.5) Ω

and

R_{2} = (15 \pm 0.5) Ω

. The percentage error in the measurement of equivalent resistance when they are connected in parallel is

Answer: (C)

For resistances in parallel:

\frac{1}{R} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

The equivalent resistance is:

R = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{10 \times 15}{10 + 15} = 6 Ω

Differentiating the parallel combination formula:

\frac{Δ R}{R ^{2}} = \frac{Δ R _{1}}{R _{1}^{2}} + \frac{Δ R _{2}}{R _{2}^{2}}

\frac{Δ R}{R} = (\frac{Δ R _{1}}{R _{1}^{2}} + \frac{Δ R _{2}}{R _{2}^{2}}) R = (\frac{0.5}{100} + \frac{0.5}{225}) \times 6

= \frac{6 \times 0.5}{25} (\frac{1}{4} + \frac{1}{9}) = \frac{3}{25} \times \frac{13}{36} = \frac{13}{300}

Percentage error:

\frac{Δ R}{R} \times 100 = \frac{13}{3} \approx 4.33%

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Single correct option