Calculate equivalent resistance of two resistors $R_{1}$ and $R_{2}$ in parallel where, $R_{1} = (6 \pm 0.2) ohm$ and $R_{2} = (3 \pm 0.1) ohm$ .

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$(1 \pm 0.07) ohm$

$(5 \pm 0.17) ohm$

$(2 \pm 0.07) ohm$

$(2 \pm 0.7) ohm$

answer is C.

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Detailed Solution

In parallel,

$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$ …….. (i)

$R = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{(6) (3)}{6 + 3} = 2 ohm$

Differentiating Eq. (i), we have

$- \frac{d R}{R^{2}} = - \frac{d R_{1}}{R_{1}^{2}} - \frac{d R_{2}}{R_{2}^{2}}$

Therefore, maximum permissible error in equivalent resistance may be

$Δ R = (\frac{Δ R_{1}}{R_{1}^{2}} + \frac{Δ R_{2}}{R_{2}^{2}}) (R^{2})$

Substituting the values we get,

$Δ R = [\frac{0.2}{(6)^{2}} + \frac{0.1}{(3)^{2}}] (2)^{2}$

$= 0.07 ohm$

$R = (2 \pm 0.07) ohm$

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