Question

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:

• A. \(8 \, V\)
• B. \(18 \, V\)
• C. \(6 \, V\)
• D. \(12 \, V\)

Hint:

In problems involving symmetric resistor networks, simplify the circuit using series and parallel combinations to find equivalent resistances and apply Ohm's law accordingly.

Answer 1

The Correct Option is A

When a square loop is formed from a wire with resistance \(R\) and a voltage of 12 V is applied between two adjacent corners, we can consider the square as two resistors in series along one side of the square and two in series along the other side, forming two parallel paths from one corner to the opposite corner. 
Step 1: Each side of the square has resistance \( \frac{R}{4} \). The total resistance between two adjacent corners (half the square) is: \[ R_{{half}} = \frac{R}{4} + \frac{R}{4} = \frac{R}{2} \] Step 2: The equivalent resistance between opposite corners (along the diagonal), with two such half resistances in parallel, is: \[ R_{{diag}} = \frac{R_{{half}} \times R_{{half}}}{R_{{half}} + R_{{half}}} = \frac{\frac{R}{2} \times \frac{R}{2}}{\frac{R}{2} + \frac{R}{2}} = \frac{R}{4} \] Step 3: The current through the circuit, using Ohm's law (\(I = \frac{V}{R}\)), is: \[ I = \frac{12 \, V}{\frac{R}{4}} = \frac{48}{R} \] Step 4: The voltage across the diagonal, where the resistance is \( \frac{R}{4} \), becomes: \[ V_{{diag}} = I \times R_{{diag}} = \frac{48}{R} \times \frac{R}{4} = 12 \, V \times \frac{2}{3} = 8 \, V \]