First slide

Accuracy; precision of instruments and errors in measurements

Question

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

Moderate

A

$(1 \pm 0.07) ohm$
B

$(2 \pm 0.7) ohm$
C

$(2 \pm 0.07) ohm$
D

$(5 \pm 0.17) ohm$

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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