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  • Two resistances are given as <img alt="R_{1}=\left ( 10 \pm 0.5 \right )\Omega" src="https://entrancecorner.oncodecogs.com/gif.latex?R_%7B1%7D%3D%5Cleft%20%28%2010%20%5Cpm%200.5%20%5Cright%20%

Two resistances are given as R_{1}=\left ( 10 \pm 0.5 \right )\Omega  and R_{2}=\left ( 15 \pm 0.5 \right )\Omega  The percentage error in the measurement

of equivalent resistance when they are connected in parallel is -

Option: 1

2.33


Option: 2

4.33


Option: 3

5.33


Option: 4

6.33


Answers (1)

best_answer

In parallel combination, \frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}
\Rightarrow \frac{1}{R_{e q}}=\frac{1}{10}+\frac{1}{15}=\frac{5}{30}=\frac{1}{6}

Now, for error calculation,
\begin{aligned} & \frac{d R_{e q}}{R_{e q}^2}=\frac{d R_1}{R_1^2}+\frac{d R_2}{R_2^2} \\ & \Rightarrow \frac{d R_{e q}}{36}=\frac{0.5}{100}+\frac{0.5}{225} \\ & d R_{e q}=36 \times 0.5 \times\left(\frac{13}{900}\right)=18 \times \frac{13}{900}=\frac{26}{100}=0.26 \end{aligned}
Now, \frac{d R_{e q}}{R_{e q}} \times 100=\frac{0.26}{6} \times 100=\frac{26}{6}=4.33

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Anam Khan

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