Current Electricity

JEE Main level, single correct, medium

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Question 1 of 380

In the given electrical circuit network between terminals AA and BB, resistors are R1=4 ΩR_1 = 4\,\Omega, R2=8 ΩR_2 = 8\,\Omega, R3=6 ΩR_3 = 6\,\Omega, and R4=12 ΩR_4 = 12\,\Omega. Determine the equivalent resistance RABR_{AB}.

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Answer: (B) 7.2 Ω7.2\,\Omega

Since R1R3=46=23=R2R4\frac{R_1}{R_3} = \frac{4}{6} = \frac{2}{3} = \frac{R_2}{R_4}, the Wheatstone bridge is balanced. The galvanometer branch carries zero current. Thus RAB=(4+8)(6+12)(4+8)+(6+12)=12×1830=7.2 ΩR_{AB} = \frac{(4+8)(6+12)}{(4+8)+(6+12)} = \frac{12 \times 18}{30} = 7.2\,\Omega.

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