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Two resistors of resistances ${R_1} = 100 \pm 3$ ohm and ${R_2} = 200 \pm 4$ ohm are connected in parallel. The equivalent combination of the parallel connection is:
A) $66.7 \pm 1.8$ohm
B) $66.7 \pm 4$ohm
C) $66.7 \pm 3$ohm
D) $66.7 \pm 7$ohm

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Last updated date: 20th Jun 2024
Total views: 54k
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Answer
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Hint: In parallel combination of resistances the equivalent resistance is given as the ratio of the product of the two resistances and the error part will be solved using the formula given below:
$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$(Parallel combination formula)
$\dfrac{{dR}}{{{R^2}}} = \dfrac{{d{R_1}}}{{{R^2}_1}} + \dfrac{{d{R_2}}}{{{R^2}_2}}$ ($R_1$ and $R_2$ are the two resistances given to us, we have differentiated the parallel combination formula)
Using above relations we will find the parallel combination of two resistances and their error part.

Complete step by step solution:
Let’s have some discussion on parallel combination first and then we will proceed for the calculation part:
In parallel, the combination of resistors voltage remains the same in all the branches of the circuit be it AC or DC circuit. Current will have different values in all the branches and the magnitude of the current will depend on the impedance a particular branch has.
Let’s come to the calculation part now:
We will solve the magnitude part first:
$ \Rightarrow \dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$(We will substitute the value of both the resistances in the formula)
$ \Rightarrow \dfrac{1}{R} = \dfrac{1}{{100}} + \dfrac{1}{{200}}$(on taking LCM)
$ \Rightarrow R = \dfrac{{100 \times 200}}{{300}} $
$ \Rightarrow R = 66.67ohm $
(Value of magnitude part of the parallel combination)
Now, we will solve the error part of the two parallel resistors:
$ \Rightarrow \dfrac{{dR}}{{{R^2}}} = \dfrac{{d{R_1}}}{{{R^2}_1}} + \dfrac{{d{R_2}}}{{{R^2}_2}}$(We will substitute the numerical values in the given equation)
$ \Rightarrow \dfrac{{dR}}{{{{66.67}^2}}} = \dfrac{3}{{{{100}^2}}} + \dfrac{4}{{{{200}^2}}} $
$\Rightarrow \dfrac{{dR}}{{{{66.67}^2}}} = \dfrac{3}{{10000}} + \dfrac{4}{{40000}} $
(The error part was 3 and 4 respectively)
On multiplying $R_2$ term on RHS
$ \Rightarrow dR = {(66.67)^2}[{\dfrac{3}{10000} - \dfrac{4}{40000}}] $
$ \Rightarrow dR = ({(66.67)^2} \times .0003) - (.0004 \times {(66.67)^2}) $
(After simplification our equation becomes)
$dR = 1.78$ or $1.8$
We got the error value as 1.8 ohm.

Therefore, option (A) is correct.

Note: If the resistors are added in series combination instead of parallel then we would have added the magnitude of the resistors directly without doing any reciprocal of the magnitude. For the error part; we would have added the resistance value in the error part too again directly.