The resistance R=VI where V=(200±5)V and I=(20±0.2)A, the percentage error in the measurement of R is :

R=V1According to error analysis dRR=dVV+dIIdRR=5200+0.220dRR=7200% error dRR×100=7200×100=3.5%

The resistance R=VI where V=(200±5)V and I=(20±0.2)A, the percentage error in the measurement of R is :

R=V1According to error analysis dRR=dVV+dIIdRR=5200+0.220dRR=7200% error dRR×100=7200×100=3.5%

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The resistance $R = \frac{V}{I} where V = (200 \pm 5) V$ and $I = (20 \pm 0.2) A$ , the percentage error in the
measurement of R is :

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Detailed Solution

$R = \frac{V}{1}$

According to error analysis

$\frac{dR}{R} = \frac{dV}{V} + \frac{dI}{I}$

$\frac{dR}{R} = \frac{5}{200} + \frac{0.2}{20}$

$\frac{dR}{R} = \frac{7}{200}$

$% error \frac{d R}{R} \times 100 = \frac{7}{200} \times 100 = 3.5 %$

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The resistance R=VI where V=(200±5)V and I=(20±0.2)A, the percentage error in the measurement of R is :

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The resistance $R = \frac{V}{I} where V = (200 \pm 5) V$ and $I = (20 \pm 0.2) A$ , the percentage error in the
measurement of R is :