A 16 Ω wire is bent to form a square loop. A cell of 9 V is connected across one of its sides. The potential difference between the diagonals of the square loop is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 6 V
The wire is uniform, so each of the four sides has resistance \( 16/4 = 4\,\Omega \). Let the cell be across side AB. The potential difference across AB is the cell voltage, 9 V, and the other three sides (BC, CD, DA) form a series path of \( 12\,\Omega \) in parallel with AB.
\( \displaystyle I = \frac{9}{12} = 0.75\,\mathrm{A} \)
\( \displaystyle V_{\text{side}} = 0.75 \times 4 = 3\,\mathrm{V} \)
Each of the three sides carries the same current, so the potential falls by 3 V across each of them. Take B at 0 V, so A is at 9 V. Going from A through D and C to B, the potentials are D = 6 V, C = 3 V.
Across a diagonal, for example A and C (or B and D), the potential difference is that of two sides in series:
\( \displaystyle V_A - V_C = 9 - 3 = 6\,\mathrm{V} \qquad V_D - V_B = 6 - 0 = 6\,\mathrm{V} \)
Note: The wording ‘potential difference between the diagonals’ is read as the potential difference between the two ends of a diagonal, which is 6 V for either diagonal.