Two resistance are given as R 1 = ( 10 ± 0 . 5 ) Ω and R 2 = ( 15 ± 0 . 5 ) Ω . The percentage error in the measurement of equivalent resistance when they are connected in parallel is

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For a parallel combination the equivalent resistance is
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} . . . (i)$
A general expression for error can be written as $x \pm ∆ x$
Differentiating equation (i)
$\frac{d R}{R^{2}} = \frac{d R_{1}}{R_{1}^{2}} + \frac{d R_{2}}{R_{2}^{2}} . . . (i i)$
The equivalent resistance with $R_{1} = 10 Ω & R_{2} = 15 Ω$ is
$\frac{1}{R} = \frac{1}{10} + \frac{1}{15}$
$\Rightarrow R = \frac{150}{25} = 6 Ω$
Substituting the above value in equation (ii)
$d R = 36 (\frac{0.5}{100} + \frac{0.5}{15 \times 15}) = 0.26$
$\Rightarrow R = 6 \pm 0.26$
Hence, the percentage error is
$\frac{d R}{R} \times 100 = \frac{0.26}{6} \times 100$ $= 4.33 %$

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