JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 9)
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 :
(4 $$\pm$$ 0.4) $$\Omega$$
(2 $$\pm$$ 0.4) $$\Omega$$
(2 $$\pm$$ 0.3) $$\Omega$$
(4 $$\pm$$ 0.3) $$\Omega$$
Explanation
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 $${R_{eq}} = (2 \pm 0.3)\Omega $$.
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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