Question

Two resistors of resistances ${{R}_{1}}=600\pm 18$and ${{R}_{2}}=300\pm 6$ ohm are connected in parallel. The equivalent resistance of the combination is:
$\begin{aligned} & \text{(1)900}\pm \text{24}\Omega \\\ & \text{(2)200}\pm \text{5}\Omega \\\ & \text{(3)200}\pm \text{15}\Omega \\\ & \text{(4)200}\pm \text{4}\text{.7}\Omega \\\ \end{aligned}$

Answer 1

While solving this question, two formulas are necessarily used. The first one is the formula of equivalent resistance of two resistors in parallel connection and the second is the error in the equivalent resistance of two resistors in parallel connection.

Formula used:
$\begin{aligned} & \dfrac{1}{{{R}_{eq}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}} \\\ & \\\ \end{aligned}$
$\dfrac{\Delta R}{R_{eq}^{2}}=\dfrac{\Delta {{R}_{1}}}{{{R}_{1}}}+\dfrac{\Delta {{R}_{2}}}{{{R}_{2}}}$

Complete step by step answer:
Given,
The values of the two resistors are given as:
$\begin{aligned} & {{R}_{1}}=600\Omega \\\ & {{R}_{2}}=300\Omega \\\ \end{aligned}$
The values of the errors in their resistances are given as:
$\begin{aligned} & \Delta {{R}_{1}}=18\Omega \\\ & \Delta {{R}_{2}}=6\Omega \\\ \end{aligned}$
For two resistors connected to each other in parallel combination, the equivalent resistance for the two resistors is given by the formula:
$\begin{aligned} & \dfrac{1}{{{R}_{eq}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}} \\\ & \\\ \end{aligned}$ ………… (1)
Where, ${{R}_{1}}$ and ${{R}_{2}}$ are the values of resistances of the two resistors connected in parallel.
${{R}_{eq}}$ is the equivalent resistance of the combination.
But, the solution of the problem does not end here,
Since the values of the errors in the resistances are also given in the question.
$\therefore$ The formula of equivalent resistance for the error contained in two resistors connected in parallel is given as:
$\dfrac{\Delta R}{R_{eq}^{2}}=\dfrac{\Delta {{R}_{1}}}{{{R}_{1}}}+\dfrac{\Delta {{R}_{2}}}{{{R}_{2}}}$ ………… (2)
Where,$\Delta R$ is the equivalent error in resistance
$\Delta {{R}_{1}}$ is the error in the value of resistance ${{R}_{1}}$
$\Delta {{R}_{2}}$ is the error in the value of resistance ${{R}_{2}}$
By making use of the two formulas the value of equivalent resistance and the error in equivalent resistance for parallel combination is calculated.
Plugging in the values of resistances in equation (1), we have
$\begin{aligned} & \dfrac{1}{{{R}_{eq}}}=\dfrac{1}{600}+\dfrac{1}{300} \\\ & \Rightarrow \dfrac{1}{{{R}_{eq}}}=\dfrac{1}{200} \\\ & \Rightarrow {{R}_{eq}}=200\Omega \\\ \end{aligned}$
Plugging in the values of errors of resistances in equation (2) along with the value of equivalent resistance, we have,
$\begin{aligned} & \dfrac{\Delta R}{{{(200)}^{2}}}=\dfrac{18}{{{(600)}^{2}}}+\dfrac{6}{{{(300)}^{2}}} \\\ & \Rightarrow \Delta R=2+2.67 \\\ & \Rightarrow \Delta R=4.67\Omega \\\ \end{aligned}$
Thus rounding off the value to the closed option, we get the final value of resistance as we get the final value of resistance as follows = $200\pm 4.7\Omega$
Therefore, the correct option is (4).
Additional Information: The formula when n no of resistors are connected in series is given as follows:
$\begin{aligned} & {{R}_{eq}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+.......+{{R}_{n}} \\\ & \\\ \end{aligned}$
And the formula for calculation of error in resistance is given as follows:
$\Delta R=\Delta {{R}_{1}}+\Delta {{R}_{2}}+\Delta {{R}_{3}}+.....+\Delta {{R}_{n}}$
Thus, even for series combination, the equivalent resistance and error in resistance can be calculated as shown above.

Note:
Both the formulas of equivalent resistance in parallel and series connection and the calculation of their errors must be kept in mind. For the calculation of error in parallel combination, the value of equivalent resistance in parallel connection needs to be calculated first. Moreover, the formula for parallel combination can also be extended for n number of resistors connected in parallel.