Two resistances are given as R1 = (10 0.5) and R2 = (15 0.5). The… — Solution | JEE Physics

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Two resistances are given as $R_1 = (10 \pm 0.5)\Omega$ and $R_2 = (15 \pm 0.5)\Omega$ . The percentage error in the measurement of equivalent resistance when they are connected in parallel is:

A
$2.33\%$
B
$4.33\%$
C
$5.33\%$
D
$6.33\%$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $R_1 = (10 \pm 0.5)\Omega$ and $R_2 = (15 \pm 0.5)\Omega$ connected in parallel.

Find: Percentage error in the equivalent resistance.

For parallel combination,

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}

Therefore,

R_{eq} = \frac{R_1R_2}{R_1 + R_2} = \frac{150}{25} = 6\Omega

Using error propagation as shown,

\frac{dR_{eq}}{R_{eq}^2} = \frac{dR_1}{R_1^2} + \frac{dR_2}{R_2^2}

So,

\frac{dR_{eq}}{R_{eq}} = R_{eq}\left[\frac{dR_1}{R_1^2} + \frac{dR_2}{R_2^2}\right]

Substituting the given values,

\frac{dR_{eq}}{R_{eq}} = 6\left[\frac{0.5}{15^2} + \frac{0.5}{10^2}\right]

Hence percentage error is

\%\frac{dR_{eq}}{R_{eq}} = 6 \times 0.5 \times 100\left[\frac{1}{225} + \frac{1}{100}\right]

Now,

6 \times 0.5 \times 100\left[\frac{1}{225} + \frac{1}{100}\right] = 6 \times 0.5 \times 100 \times \frac{325}{225 \times 100} = 4.33

Therefore, the percentage error in equivalent resistance is $4.33\%$ . The correct option is B.

Common mistakes

Using the series combination formula instead of the parallel relation is incorrect because the equivalent resistance of parallel resistors is found from reciprocals. Start with $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$ .
Adding percentage errors of $R_1$ and $R_2$ directly is wrong here because $R_{eq}$ is not a simple product or quotient of measured quantities in the form used. First rewrite the parallel relation and then propagate error through the reciprocal expression.
Substituting $R_1 = 10\Omega$ and $R_2 = 15\Omega$ in the denominator terms in the wrong order gives an incorrect value. Follow the shown step carefully: $\frac{dR_{eq}}{R_{eq}} = 6\left[\frac{0.5}{15^2} + \frac{0.5}{10^2}\right]$ .

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Answer

Step-by-step solution

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