The net resistance of the circuit between A and B is (a) 8/3Ω (b) 14/3Ω (c) 16/3Ω (d) 22/3Ω

Answer: (b). Wheatstone bridge check: 3/4 = 6/8, balanced, so the 7 Ω carries no current. Upper arm 3 + 4 = 7 Ω, lower arm 6 + 8 = 14 Ω. RAB = (7 × 14)/(21) = 14/3 Ω. So the answer is (b).

The net resistance of the circuit between A and B is | NEET 2000 (AIPMT)

Q80NEET 2000 · AIPMTMedium

Combination of Resistors

The net resistance of the circuit between A and B is

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Answer: (b)

Wheatstone bridge check: $\dfrac{3}{4} = \dfrac{6}{8}$ , balanced, so the $7\ \Omega$ carries no current.
Upper arm $3 + 4 = 7\ \Omega$ , lower arm $6 + 8 = 14\ \Omega$ .
$R_{AB} = \dfrac{7 \times 14}{21} = \dfrac{14}{3}\ \Omega$ . So the answer is (b).

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