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JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Physics
Chapter: Current Electricity

JEE Main 2024 · 4 April · Morning Shift · Physics · Current Electricity

Twelve wires each having resistance 2 are joined to form a cube. A...

Twelve wires each having resistance

2 \Omega

are joined to form a cube. A battery of

6 \mathrm{~V}

emf is joined across point

a

and

c

. The voltage difference between

e

and

f

is ______V. Twelve wires each having resistance 2 are joined to form a cube. A battery of 6 ~V emf is joined across point a and c .

Twelve wires each having resistance 2 are joined to form a cube. A battery of 6 ~V emf is joined across point a and c .

Correct answer

0

Step-by-step solution

From symmetry, current through e-b & g-d =0

\begin{aligned} & \therefore \mathrm{R}_{\text{eq }}=\frac{3}{4} \times \mathrm{R}=\frac{3}{2} \Omega \\ & \therefore \text{ Current through battery }=\frac{6 \times 2}{3}=4 \mathrm{ A} \\ & \mathrm{i}_2=\frac{4}{8} \times 2=1 \mathrm{ A} \\ & \therefore \Delta \mathrm{V} \text{ across e-f }=\frac{\mathrm{i}_2}{2} \times \mathrm{R}=\frac{1}{2} \times 2=1 \mathrm{ V} \end{aligned}

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