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Question: Given \[{R_1} = 5.0 + - 0.2\Omega\] and \[{R_2} = 10.0 + - 0.1\Omega \]. What is the total resistanc...

Given R1=5.0+−0.2Ω{R_1} = 5.0 + - 0.2\Omega and R2=10.0+−0.1Ω{R_2} = 10.0 + - 0.1\Omega . What is the total resistance in parallel with possible %\% error?

Explanation

Solution

In a parallel resistance, the inverse of the equivalent resistor is equal to the sum of the inverses of the two resistors connected in parallel. The error of the equivalent resistor can be obtained by differentiating the resistances in the formula of the equivalent resistor when resistors are connected in parallel.

Complete step by step answer:
Here, the two resistances are connected in parallel to each other. When such is the situation, the net resistance is lower than the lowest resistor present in the circuit. In addition to this, when the resistance is parallel, the voltage across the resistances is constant. Here, the resistances R1=5.0+0.2Ω{R_1} = 5.0 + 0.2\Omega and R2=10.0+0.1Ω{R_2} = 10.0 + 0.1\Omega are connected in parallel with equivalent resistance being R, and the primary resistance of Rp{R_p} is given as follows:

R=R1⋅R2R1+R2 ⇒R=5⋅105+10=3.33Ω R=\dfrac{{{R}_{1}}\centerdot {{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \\\ \Rightarrow R=\dfrac{5\centerdot 10}{5+10}=3.33\Omega \\\

Now, to find the error part of the equivalent resistance, we use the below formula;

\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} \\\ \Rightarrow \dfrac{{\Delta {R_p}}}{{{R_p}^2}} = \dfrac{{\Delta {R_1}}}{{{R_1}^2}} + \dfrac{{\Delta {R_2}}}{{{R_2}^2}} \\\\$$ Here, $$\Delta {R_p}$$ is the error in the measurement of the equivalent resistance which is obtained from the instrument, $$\Delta {R_1}$$ is the error in the measurement of the ${R_1}$ resistor and $$\Delta {R_2}$$ is the error in the measurement of the ${R_2}$ resistor. Thus, the error in the equivalent resistance is as below:

\Delta {R_p} = {(3.33)^2}\left[ {\dfrac{{0.2}}{{{5^2}}} + \dfrac{{0.1}}{{{{10}^2}}}} \right] \\
\Rightarrow \Delta {R_p} = {(3.33)^2}\left[ {\dfrac{{0.2}}{{25}} + \dfrac{{0.1}}{{100}}} \right] \\
\Rightarrow \Delta {R_p} = {(3.33)^2} \times 0.009 \\
\therefore \Delta {R_p} = 0.099 \\ ∗∗Thus **Thus\Delta {R_p} = 3.33 + - 0.099\Omega s$$ where 0.099 is the error in the measurement of the equivalent resistance which occurs due to instrumental error.**

Note: When the resistances are connected in parallel, the voltage remains constant and the net resistance is lower than the smallest resistance of the network, while when they are connected in series, the current remains constant and the net resistance is higher than the largest resistance of the network.