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QJEEM24Phc319: JEE Main 2024 Physics - Class 12 JEEM24Phc319 Solution: Twelve wires each having resistance are joined to form a cube.…

Twelve wires each having resistance 2Ω2 \Omega are joined to form a cube. A battery of 6 V6 \mathrm{~V} emf is joined across point aa and cc . The voltage difference between ee and ff is ________ V.

JEE Main 2024 (Online) 4th April Morning Shift Physics - Current Electricity Question 27 English

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The circuit can be simplified as

JEE Main 2024 (Online) 4th April Morning Shift Physics - Current Electricity Question 27 English Explanation

amp;Rac=6×28=32Ωamp;amp;i=1Amp.amp;Vef=(i2)2amp;=1 V\begin{aligned}& R_{a c}=\frac{6 \times 2}{8}=\frac{3}{2} \Omega \\& \begin{aligned}& i=1 \mathrm{Amp} . \\& V_{e f}=\left(\frac{i}{2}\right)^2 \\& \quad=1 \mathrm{~V}\end{aligned}\end{aligned}

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