Question

Two resistances ${R_1} = \left( {16 \pm 0.3} \right)\Omega$ and ${R_2} = \left( {48 \pm 0.5} \right)\Omega$ are connected in parallel. Find the max $\%$ error.
(A) $3.2\%$
(B) $1.6\%$
(C) $0.8\%$
(D) $2\%$

Answer 1

To solve this question we first evaluate the equivalent resistance and using that equivalent resistance we will evaluate the total equivalent error. Hence after that using the maximum error formula and the obtained values, we will evaluate the maximum percentage error of the resistance provided in the question.
Formula used:
Equivalent resistance for parallel connection
$\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ....$
The formula for parallel equivalence error
$\Rightarrow \Delta {R_{eq}} = \Delta {R_1}\left( {\dfrac{{{\operatorname{R} _{eq}}}}{{{R_1}^2}}} \right) + \Delta {R_2}\left( {\dfrac{{{\operatorname{R} _{eq}}}}{{{R_2}^2}}} \right)$
Percentage error formula
$\Rightarrow \% error = \dfrac{{\Delta {R_{eq}}}}{{{R_{eq}}}} \times 100$

Complete Step-by-step solution
Here given that the resistance ${R_1}$ and ${R_2}$ are connected in parallel in a circuit. The value of the resistance is given with error including as ${R_1} = \left( {16 \pm 0.3} \right)\Omega$ and ${R_2} = \left( {48 \pm 0.5} \right)\Omega$ are connected in parallel. Hence for the parallel connection of resistance, the formula for equivalent resistance can be given as
$\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
$\Rightarrow {R_{eq}} = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$
Substituting the values of ${R_1} = 16\Omega$ and ${R_2} = 48\Omega$, hence
$\Rightarrow {R_{eq}} = \dfrac{{16 \times 48}}{{16 + 48}}\Omega$
$\therefore {R_{eq}} = 12\Omega$
Now we will evaluate the error for the parallel connections, which can be given by $\Delta {R_{eq}}$. The maximum percentage error can be obtained by carrying out the percentage of relative error.
$\Rightarrow \Delta {R_{eq}} = \Delta {R_1}\left( {\dfrac{{{\operatorname{R} _{eq}}}}{{{R_1}^2}}} \right) + \Delta {R_2}\left( {\dfrac{{{\operatorname{R} _{eq}}}}{{{R_2}^2}}} \right)$
Here substituting the values of $\Delta {R_1} = 0.3$, $\Delta {R_2} = 0.5$ and the values of resistance ${R_1} = 16\Omega$ and ${R_2} = 48\Omega$, hence
$\Rightarrow \Delta {R_{eq}} = \left( {0.3} \right)\left( {\dfrac{{12}}{{{{12}^2}}}} \right) + \left( {0.5} \right)\left( {\dfrac{{12}}{{{{48}^2}}}} \right)$
$\Rightarrow \Delta {R_{eq}} = 0.16875 + 0.03125$
$\therefore \Delta {R_{eq}} \approx 0.20\Omega$
Now using the maximum percentage error formula we will evaluate the max $\%$ error,
$\Rightarrow \% error = \dfrac{{\Delta {R_{eq}}}}{{{R_{eq}}}} \times 100$
Substituting the values of ${R_{eq}} = 12\Omega$ and $\Delta {R_{eq}} \approx 0.20\Omega$, in the equation results in
$\Rightarrow \% error = \dfrac{{0.20}}{{12}} \times 100$
$\therefore \% error = 1.6\%$
Hence resistances ${R_1} = \left( {16 \pm 0.3} \right)\Omega$ and ${R_2} = \left( {48 \pm 0.5} \right)\Omega$ are connected in parallel then the max $\%$ error of the combination of the resistance is given as $1.6\%$.

Therefore the option (B) is the correct answer.

Note: Here we have used the concept of percentage error where the maximum percentage error can be obtained by carrying out the percentage of relative error, where the relative error can be defined as the ratio of absolute error and the measured error.