The current \(I\) in the circuit shown below is :
(All diodes are ideal and identical)

(D) \(\frac{15}{2} A\)
From the given circuit, the positive terminal of the 10 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 V battery.
The equivalent resistance of the circuit is:
\(R_{e q}=\frac{4 \times 2}{4+2}=\frac{8}{6}=\frac{4}{3} \Omega\)\(B=\frac{-P}{\Delta V / V}=-P\left(\frac{V}{\Delta V}\right)\)
The total current \(I\) in the circuit is:
\(I=\frac{V}{R_{e q}}=\frac{10}{\frac{4}{3}}=\frac{30}{4}=\frac{15}{2} A\)