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# Calculate the equivalent resistance between X and Y.A) $1\Omega$B) $2\Omega$C) $3\Omega$D) $4\Omega$

Last updated date: 01st Mar 2024
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Hint: Equivalent resistance of two resistances connected in series combination is equal sum of individual resistances. Multiplicative inverse of equivalent resistance of two resistances connected in parallel combination is equal sum of multiplicative inverse of individual resistance. Here to find equivalent resistance between X and Y, we divide the complete circuit in parts and solve the parts and combine them again.

Given, ${R_A} = 1\Omega$, ${R_B} = 3\Omega$, ${R_C} = 4\Omega$, ${R_D} = 2\Omega$, ${R_E} = 4\Omega$, ${R_F} = 2\Omega$ and ${R_G} = 1\Omega$.
First, we take ${R_A}$ and ${R_B}$, these are connected in series then their equivalent resistance is ${R_{AB}} = {R_A} + {R_B} = 1 + 3 = 4\Omega$.
Now, resistance ${R_{AB}}$ and ${R_C}$ are connected in parallel and their equivalent resistance is ${R_{ABC}} = \dfrac{{{R_{AB}} \times {R_C}}}{{{R_{AB}} + {R_C}}} = \dfrac{{4 \times 4}}{{4 + 4}} = 2\Omega$.
Now, ${R_{ABC}}$ and ${R_D}$ are connected in series and their equivalent resistance is ${R_{ABCD}} = {R_{ABC}} + {R_D} = 2 + 2 = 4\Omega$.
Now, resistance ${R_{ABCD}}$, ${R_E}$ and ${R_F}$ are connected in parallel combination and their equivalent resistance is ${R_{ABCDEF}} = \dfrac{1}{{\dfrac{1}{{{R_{ABCD}}}} + \dfrac{1}{{{R_E}}} + \dfrac{1}{{{R_F}}}}} = \dfrac{1}{{\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{2}}} = 1\Omega$.
Finally, we have two resistance ${R_{ABCDEF}}$ and ${R_G}$ connected in series and their equivalent resistance is ${R_{Eq}} = {R_{ABCDEF}} + {R_G} = 1 + 1 = 2\Omega$.
Hence equivalent resistance between X and Y is $2\Omega$ and the correct answer is option B.