First slide

Accuracy; precision of instruments and errors in measurements

Question

Two resistors of resistances $R_{1} = 100 \pm 3 Ω and R_{2} = 200 \pm 4 Ω$ are connected in parallel, then the equivalent resistance in parallel is (in ohm)
$(Use \frac{1}{R^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} and \frac{{ΔR}^{'}}{{ΔR}^{' 2}} = \frac{{ΔR}_{1}}{R_{1}^{2}} + \frac{{ΔR}_{2}}{R_{2}^{2}})$

Moderate

A

$66.7 \pm 1.8$
B

$300 \pm 7$
C

$150.8 \pm 2$
D

$92.3 \pm 3$

Solution

The equivalent resistance of parallel combination is
$R^{'} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{200}{3} = 66.7 Ω$
From $\frac{1}{R^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}, we get$
$\frac{{ΔR}^{'}}{R^{' 2}} = \frac{{ΔR}_{1}}{R_{1}^{2}} + \frac{{ΔR}_{2}}{R_{2}^{2}}$
${ΔR}^{'} = (R^{' 2}) \frac{{ΔR}_{1}}{R_{1}^{2}} + (R^{' 2}) \frac{{ΔR}_{2}}{R_{2}^{2}}$
$= {(\frac{66.7}{100})}^{2} 3 + {(\frac{66.7}{200})}^{2} 4 = 1.8 Ω$
Hence, $R^{'} = (66.7 \pm 1.8) Ω$

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