Question

two resistances are measured in ohm give as r1 is equal to 3 ohm + - 1% and r2 is equal to 6 minus 2% when they are connected in parallel the percentage errors and equivalent resisistance

Answer 1

4/3%

Answer 2

The equivalent resistance $R_p$ of two resistances $R_1$ and $R_2$ connected in parallel is given by:

$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$$ $$R_p = \frac{R_1 R_2}{R_1 + R_2}$$

The nominal values are $R_1 = 3 \, \Omega$ and $R_2 = 6 \, \Omega$. The nominal equivalent resistance is:

$$R_p = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 \, \Omega$$

The percentage errors are given as $1\%$ for $R_1$ and $2\%$ for $R_2$. The absolute error in $R_1$ is $\Delta R_1 = 1\% \text{ of } R_1 = 0.01 \times 3 = 0.03 \, \Omega$. The absolute error in $R_2$ is $\Delta R_2 = 2\% \text{ of } R_2 = 0.02 \times 6 = 0.12 \, \Omega$.

To find the error in the equivalent resistance $R_p$, we can use the formula for error propagation for parallel resistances, derived from $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$. Differentiating and considering maximum error:

$$\frac{|\Delta R_p|}{R_p^2} = \frac{|\Delta R_1|}{R_1^2} + \frac{|\Delta R_2|}{R_2^2}$$

We want to find the percentage error in $R_p$, which is $\frac{|\Delta R_p|}{R_p} \times 100$. From the error formula, we can write:

$$\frac{|\Delta R_p|}{R_p} = R_p \left( \frac{|\Delta R_1|}{R_1^2} + \frac{|\Delta R_2|}{R_2^2} \right)$$

Substitute the nominal values and absolute errors:

$$\frac{|\Delta R_p|}{R_p} = 2 \left( \frac{0.03}{3^2} + \frac{0.12}{6^2} \right)$$ $$\frac{|\Delta R_p|}{R_p} = 2 \left( \frac{0.03}{9} + \frac{0.12}{36} \right)$$ $$\frac{|\Delta R_p|}{R_p} = 2 \left( \frac{0.01}{3} + \frac{0.03}{9} \right)$$ $$\frac{|\Delta R_p|}{R_p} = 2 \left( \frac{0.01}{3} + \frac{0.01}{3} \right)$$ $$\frac{|\Delta R_p|}{R_p} = 2 \left( \frac{0.02}{3} \right) = \frac{0.04}{3}$$

The percentage error in $R_p$ is:

Percentage error $= \frac{|\Delta R_p|}{R_p} \times 100 = \frac{0.04}{3} \times 100 = \frac{4}{3} \%$

Alternatively, we can use the formula for absolute error in parallel combination derived from partial derivatives:

$$\Delta R_p = \frac{R_2^2 \Delta R_1 + R_1^2 \Delta R_2}{(R_1 + R_2)^2}$$ $$\Delta R_p = \frac{6^2 (0.03) + 3^2 (0.12)}{(3 + 6)^2} = \frac{36 \times 0.03 + 9 \times 0.12}{9^2}$$ $$\Delta R_p = \frac{1.08 + 1.08}{81} = \frac{2.16}{81}$$

Now calculate the percentage error:

Percentage error $= \frac{\Delta R_p}{R_p} \times 100 = \frac{2.16/81}{2} \times 100$

Percentage error $= \frac{2.16}{81 \times 2} \times 100 = \frac{2.16}{162} \times 100$

Percentage error $= \frac{216}{162} \% = \frac{36 \times 6}{27 \times 6} \% = \frac{36}{27} \% = \frac{4 \times 9}{3 \times 9} \% = \frac{4}{3} \%$

Both methods give the same result.