The current flowing through 8 Ω resistor is

The correct answer is Option (3) → 0.5 A

The \( 8\,\Omega \) resistor is in parallel with the branch of \( 1\,\Omega \) and \( 3\,\Omega \) in series, and both branches are connected directly across the 4 V cell (no internal resistance is given, so the cell is taken as ideal). The full 4 V therefore appears across the \( 8\,\Omega \) resistor.

    \( \displaystyle I_8 = \frac{V}{R} = \frac{4\,\mathrm{V}}{8\,\Omega} = 0.5\,\mathrm{A} \)

(The other branch, \( 1 + 3 = 4\,\Omega \), carries 1 A, so the cell supplies 1.5 A in all.)