Six resistors of 3 Ω each are connected along the sides of a hexagon and three resistors of 6 Ω each are connected along AC, AD and AE as shown in the figure. The equivalent resistance between A and B is equal to (a) 2 Ω (b) 6 Ω (c) 3 Ω (d) 9 Ω

Answer: (a). Start from F: AF + FE = 6 Ω, in parallel with AE = 6 Ω gives 3 Ω between A and E. Add ED: 3 + 3 = 6, parallel with AD = 6 gives 3 Ω between A and D. Add DC: 3 + 3 = 6, parallel with AC = 6 gives 3 Ω between A and C. Add CB: 3 + 3 = 6, parallel with AB = 3: (6 × 3)/(9) = 2 Ω So the answer is (a) 2 Ω.

Six resistors of 3 Ω each are connected along the sides of a hexagon… | NEET 1994 (AIPMT)

Q84NEET 1994 · AIPMTMedium

Combination of Resistors

Six resistors of $3\ \Omega$ each are connected along the sides of a hexagon and three resistors of $6\ \Omega$ each are connected along $AC, AD$ and $AE$ as shown in the figure. The equivalent resistance between $A$ and $B$ is equal to

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Answer: (a)

Start from F: $AF + FE = 6\ \Omega$ , in parallel with $AE = 6\ \Omega$ gives $3\ \Omega$ between A and E.
Add $ED$ : $3 + 3 = 6$ , parallel with $AD = 6$ gives $3\ \Omega$ between A and D.
Add $DC$ : $3 + 3 = 6$ , parallel with $AC = 6$ gives $3\ \Omega$ between A and C.
Add $CB$ : $3 + 3 = 6$ , parallel with $AB = 3$ : $\dfrac{6 \times 3}{9} = 2\ \Omega$
So the answer is (a) $2\ \Omega$ .

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