A metallic wire of resistance \( 12\,\Omega \) is bent to form a square loop. A source of \( 6\,\mathrm{V} \) is connected across one of its sides. The potential drop across its diagonal is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 4.0\,\mathrm{V} \)
The wire is uniform, so each side of the square has a resistance of \( 12\,\Omega/4 = 3\,\Omega \). Call the square ABCD with the source across side AB. The source is taken as ideal (no internal resistance is given), so side AB and the path A-D-C-B (three sides in series) both have 6 V across them.
\( \displaystyle I_{\mathrm{ADCB}} = \frac{6\,\mathrm{V}}{9\,\Omega} = \frac{2}{3}\,\mathrm{A} \)
The three equal resistances in series share the 6 V equally, 2 V each. Taking \( V_B = 0 \):
\( V_A = 6\,\mathrm{V},\ V_D = 4\,\mathrm{V},\ V_C = 2\,\mathrm{V},\ V_B = 0 \)
\( V_{AC} = 6 - 2 = 4\,\mathrm{V} \)
\( V_{DB} = 4 - 0 = 4\,\mathrm{V} \)
Both diagonals have a potential difference of 4 V.