The current I in the circuit shown below is : (All diodes are ideal...

From the given circuit, the positive terminal of the $10\text{ V}$ battery is connected to the left side of the parallel branches.

Therefore, the left side is at a higher potential compared to the right side.

The diodes in the branches with $4\ \Omega$ and $2\ \Omega$ resistors have their anodes connected to the higher potential side, so they are forward-biased and act as short circuits (since they are ideal).

The diodes in the branches with $3\ \Omega$ and $5\ \Omega$ resistors have their cathodes connected to the higher potential side, so they are reverse-biased and act as open circuits.

The effective circuit consists of the $4\ \Omega$ and $2\ \Omega$ resistors connected in parallel across the $10\text{ V}$ battery.

The equivalent resistance of the circuit is:

$R_{eq} = \dfrac{4 \times 2}{4 + 2} = \dfrac{8}{6} = \dfrac{4}{3}\ \Omega$

The total current $I$ in the circuit is:

$I = \dfrac{V}{R_{eq}} = \dfrac{10}{\dfrac{4}{3}} = \dfrac{30}{4} = \dfrac{15}{2}\text{ A}$

Answer: $\dfrac{15}{2}$ A