Four resistors, AB = 2 Ω, BC = 3 Ω, CD = 2 Ω, and DA = 3 Ω are connected in the form of a closed loop. A current of 2 A is flowing through point A and leaving the loop from point C. The potential difference VB − VD will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → +1 V
Between A and C the loop offers two parallel paths: A→B→C and A→D→C. The current divides in the inverse ratio of their resistances.
\( R_{ABC} = 2\,\Omega + 3\,\Omega = 5\,\Omega \)
\( R_{ADC} = 3\,\Omega + 2\,\Omega = 5\,\Omega \)
The two paths are equal, so the 2 A splits equally:
\( I_{ABC} = I_{ADC} = 1\,\mathrm{A} \)
Potential falls in the direction of current (V = IR across each resistor):
\( V_A - V_B = (1\,\mathrm{A})(2\,\Omega) = 2\,\mathrm{V} \)
\( V_A - V_D = (1\,\mathrm{A})(3\,\Omega) = 3\,\mathrm{V} \)
Subtracting the first equation from the second:
\( V_B - V_D = 3\,\mathrm{V} - 2\,\mathrm{V} = +1\,\mathrm{V} \)