JEE MAIN - Physics (2024 - 4th April Morning Shift - No. 29)
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.
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Answer
1
Explanation
The circuit can be simplified as
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$$\begin{aligned} & R_{a c}=\frac{6 \times 2}{8}=\frac{3}{2} \Omega \\ & \begin{aligned} & i=1 \mathrm{Amp} . \\ & V_{e f}=\left(\frac{i}{2}\right)^2 \\ & \quad=1 \mathrm{~V} \end{aligned} \end{aligned}$$
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