A straight wire of resistance $R$ is bent in the shape of a square. A cell of emf 12 V is connected between two adjacent corners of the square. The potential difference across any diagonal of the square is

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Aicharya · Accepted Answer

$\begin{aligned} & R_{A D}=\frac{\frac{3 R}{4} \times \frac{R}{4}}{R}=\frac{3 R}{16} \Omega \\ \therefore & I=\frac{V}{R_{A D}}=\frac{12}{\frac{3 R}{16}}=\frac{64}{R} \\ \therefore & I_1=\left(\frac{\frac{R}{4}}{R}\right) I \\ = & \frac{1}{4} \times \frac{64}{R}=\frac{16}{R} \\ \therefore \quad & V_{A C}=I_1\left(\frac{R}{2}\right)=\frac{16}{R} \times \frac{R}{2}=8 V\end{aligned}$

A straight wire of resistance $R$ is bent in the shape of a square. A cell of emf 12 V is connected between…

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