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Question: For resistances \({R_1}\) and \({R_2}\), connected in parallel, find the relative error in their equ...

For resistances R1{R_1} and R2{R_2}, connected in parallel, find the relative error in their equivalent resistance, if R1{R_1}= (50±\pm 2) ohm and R2{R_2}=(100±\pm3).

Explanation

Solution

Hint : In this question, first we will see the definition of absolute error and relative error. After this, we will calculate the equivalent resistance of the parallel combination. Then we get the equivalent error of the parallel combination and finally calculate the relative error of the equivalent resistance.

Complete answer:
First let us see the definition of absolute error.
It is defined as the deviation in measured value from true value. Mathematically, we can write:
A=AtAm\vartriangle A = {A_{_t}} - {A_m}
Where, A\vartriangle A is the absolute error.At{A_t} is the true value and Am{A_m} is the measured value.
Relative error:
It is defined as the ratio of absolute error and true value.
Relative error = AAt\dfrac{{\vartriangle A}}{{{A_t}}} ……………... (1)
Now, we calculate the equivalent resistance of the parallel combination.
Let Rp{R_p}denote the equivalent resistance.
Rp=R1×R2R1+R2{R_p} = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}} .
Putting the value ofR1{R_1} and R2{R_2} in above equation, we get:
Rp=50×100150=1003{R_p} = \dfrac{{50 \times 100}}{{150}} = \dfrac{{100}}{3}ohm.
Now, we calculate the parallel connection error.
The parallel connection error is given as:
Req=R12R1+R22R2(R1+R2)2=502(2)+1002(3)1502=5000+30000150×150=149\vartriangle {R_{eq}} = \dfrac{{{R_1}^2\vartriangle {R_1} + {R_2}^2\vartriangle {R_2}}}{{{{({R_1} + {R_2})}^2}}} = \dfrac{{{{50}^2}(2) + {{100}^2}(3)}}{{{{150}^2}}} = \dfrac{{5000 + 30000}}{{150 \times 150}} = \dfrac{{14}}{9}
Putting the value of absolute error i.e. error in parallel connection and true value i.e. equivalent resistance in equation 1, we get:
Therefore, relative error in equivalent resistance is given as:
\therefore Relative error in equivalent combination =ReqReq=(14/9)(100/3)=0.0467\dfrac{{\vartriangle {\operatorname{R} _{eq}}}}{{{R_{eq}}}} = \dfrac{{(14/9)}}{{(100/3)}} = 0.0467.

Note: In such a type of question, you should know the formula to calculate relative error. To calculate the resultant error due to difference or sum of two quantities, we then add the absolute error due to each quantity. To calculate the resultant error for two quantities in a product is a summation of relative error of each.