In the given circuit, what is the value of the current I ?

The correct answer is Option (1) → 0.6 A

The four outer resistors form a Wheatstone bridge with the vertical 5 Ω resistor as the middle link. The bridge is balanced when the ratio of the two arms on one side equals the ratio on the other side; then no current flows through the middle link and it can be removed.

Readings from the figure (upper arms 10 Ω and 20 Ω, lower arms 5 Ω and 10 Ω):

    \( \displaystyle \frac{10\,\Omega}{20\,\Omega} = \frac{1}{2} \quad \text{and} \quad \frac{5\,\Omega}{10\,\Omega} = \frac{1}{2} \)

The bridge is balanced, so the upper path (10 + 20 = 30 Ω) and the lower path (5 + 10 = 15 Ω) are simply in parallel.

    \( \displaystyle R_{\mathrm{bridge}} = \frac{30 \times 15}{30 + 15}\,\Omega = 10\,\Omega \)

This is in series with the 10 Ω resistor next to the cell.

    \( R_{\mathrm{total}} = 10\,\Omega + 10\,\Omega = 20\,\Omega \)

    \( \displaystyle I = \frac{12\,\mathrm{V}}{20\,\Omega} = 0.6\,\mathrm{A} \)