The value of current through 5 Ω resistance is

The correct answer is Option (3) → 0.5 A

Treat both diodes as ideal: a forward-biased diode conducts like a plain wire, and a reverse-biased diode conducts nothing. The left terminal is at 5 V and the right terminal at 0 V, so current flows from left to right.

Diode D₁ has its arrow pointing from the 5 V side towards the right, so it is forward biased and acts as a wire.

Diode D₂ has its arrow pointing towards the left, against the direction in which the current would have to flow from the 5 V side to the 0 V side. It is reverse biased, so the branch with the 8 Ω resistor carries no current.

The circuit now has two paths from the 5 V terminal to the junction at the top of the 6 Ω resistor:

    Upper path: D₁ and \( 10\,\Omega \), so \( 10\,\Omega \)
    Lower path: \( 4\,\Omega \) and then the \( 6\,\Omega \) resistor, so \( 4 + 6 = 10\,\Omega \)

These two are in parallel:

    \( \displaystyle R_{\parallel} = \frac{10\times10}{10 + 10} = 5\,\Omega \)

This combination is in series with the \( 5\,\Omega \) resistor to the 0 V terminal, and the whole current passes through that \( 5\,\Omega \) resistor.

    \( \displaystyle I = \frac{5\,\mathrm{V}}{5\,\Omega + 5\,\Omega} = 0.5\,\mathrm{A} \)