If the internal resistance of cell is negligible then current flowing through the circuit is

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From the given circuit, the positive terminal of the $5 V$ battery is on the left. Thus, current flows from left to right through the parallel branches.

Diode $D_{1}$ is forward-biased as its p-side is connected to the higher potential (left side).

Diode $D_{2}$ is reverse-biased as its n-side is connected to the higher potential. No current flows through the top branch.

The effective circuit is a series combination of the $20Ω$ resistor, the forward-biased diode $D_{1}$ , the $30Ω$ resistor, and the $5 V$ battery.

Total resistance, $R = 20Ω + 30Ω = 50Ω$ .

Current, $I = \frac{V}{R} = \frac{5}{50} = 0.10 A$ .

Answer: 0.10 A

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