Question

In an experiment the values of two resistances were measured to be ${R_1} = \left( {5.0 \pm 0.2} \right)\,\Omega$ and ${R_2} = \left( {10.0 \pm 0.1} \right)\,\Omega$. Then their combined resistance in series will be:
A. $\left( {15 \pm 0.1} \right)\,\Omega$
B. $\left( {15 \pm 0.2} \right)\,\Omega$
C. $\left( {15 \pm 2\% } \right)\,\Omega$
D. $\left( {15 \pm 3\% } \right)\,\Omega$

Answer 1

When the two resistors are connected in series, their equivalent resistance is the sum of the resistance of two resistors. Determine the percentage error in the equivalent resistance by taking the ratio of error in the resistance to the true value of the resistance and multiply it by 100.

Formula used:
Equivalent resistance in series combination, ${R_{eq}} = {R_1} + {R_2}$
where, ${R_1}$ and ${R_2}$ are the resistances of two resistors.

Complete step by step answer:
We know that when the two resistors are connected in series, their equivalent resistance is the sum of the resistance of two resistors. Therefore, we can express the equivalent resistance of the two resistances ${R_1}$ and ${R_2}$ as follows,
${R_{eq}} = {R_1} + {R_2}$
Substituting ${R_1} = \left( {5.0 \pm 0.2} \right)\,\Omega$ and ${R_2} = \left( {10.0 \pm 0.1} \right)\,\Omega$ in the above equation, we get,
${R_{eq}} = \left( {5.0 \pm 0.2} \right) + \left( {10.0 \pm 0.1} \right)$
$\Rightarrow {R_{eq}} = \left( {5.0 + 10.0} \right) \pm \left( {0.2 + 0.1} \right)$
$\Rightarrow {R_{eq}} = 15 \pm 0.3$
In the above equation, $15\,\Omega$ is the true value of the equivalent resistance and 0.3 is the error in the resistance.We can determine the percentage error in the equivalent resistance as follows,
$\% {\text{error}} = \dfrac{{\Delta R}}{R} \times 100$
Here, $\Delta R$ is the error in the resistance and R is the true value of the resistance.
Substituting $\Delta R = 0.3\,\Omega$ and $R = 15\,\Omega$ in the above equation, we get,
$\% {\text{error}} = \dfrac{{0.3}}{{15}} \times 100$
$\Rightarrow \% {\text{error}} = 2\%$
Therefore, we can express the equivalent resistance as,
$\therefore {R_{eq}} = \left( {15 \pm 2\% } \right)\,\Omega$

So, the correct answer is option C.

Note: Students must know the formula for equivalent resistance for series combination and parallel combination of resistances. In series, the resistance increases, therefore, the equivalent resistance is the sum of resistance of each resistor. To determine the percentage error in the resistance, you should take the ratio of error in the resistance to the true value of resistance and multiply it by 100.