The current I flowing in the network is:

The correct answer is Option (1) → 4 A

Every branch of the network joins the same two junctions, so the four branches are in parallel. In each crossing branch the two 10 Ω resistors are in series.

    \( R_1 = 20\,\Omega \)
    \( R_2 = 20\,\Omega \)
    \( R_3 = 10\,\Omega + 10\,\Omega = 20\,\Omega \)
    \( R_4 = 10\,\Omega + 10\,\Omega = 20\,\Omega \)

    \( \displaystyle \frac{1}{R_{\mathrm{eq}}} = \frac{1}{20} + \frac{1}{20} + \frac{1}{20} + \frac{1}{20} \)

    \( \displaystyle \frac{1}{R_{\mathrm{eq}}} = \frac{4}{20}\,\Omega^{-1} \)

    \( \displaystyle R_{\mathrm{eq}} = 5\,\Omega \)

    \( \displaystyle I = \frac{V}{R_{\mathrm{eq}}} = \frac{20\,\mathrm{V}}{5\,\Omega} \)

    \( \displaystyle I = 4\,\mathrm{A} \)