QJEEM21Ph1594: JEE Main 2021 Physics - Class 12 JEEM21Ph1594 Solution: Two resistors R 1 = (4 0.8) and R 2 = (4 0.4) are connected in…

Question

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Two resistors R₁ = (4

\pm

0.8)

\Omega

and R₂ = (4

\pm

0.4)

\Omega

are connected in parallel. The equivalent resistance of their parallel combination will be :

Accepted Answer

Given,

R₁ = (4

\pm

0.8)

\Omega

R₂ = (4

\pm

0.4)

\Omega

Equivalent resistance when the resistors are connected in parallel is given by

{1 \over {{R_{eq}}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} \Rightarrow {1 \over {{R_{eq}}}} = {1 \over 4} + {1 \over 4}

{R_{eq}} = 2\,\Omega

Now,

{{\Delta {R_{eq}}} \over {R_{eq}^2}} = {{\Delta {R_1}} \over {R_1^2}} + {{\Delta {R_2}} \over {R_2^2}}

Substituting the values in the above equation, we get

{{\Delta {R_{eq}}} \over 4} = {{0.8} \over {16}} + {{0.4} \over {16}} \Rightarrow \Delta {R_{eq}} = 0.3\,\Omega

\therefore

The equivalent resistance in parallel combination is

{R_{eq}} = (2 \pm 0.3)\Omega

.

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