Solution

Check the bias of each diode with respect to the battery.
The diode in the top (20 $\Omega$) branch is reverse biased, so it acts as an open switch and no current flows in that branch.
The diode in the middle (30 $\Omega$) branch is forward biased and acts as a closed switch (ideal diode).
So the equivalent circuit is the 5 V battery in series with 20 $\Omega$ and 30 $\Omega$.

$I = \dfrac{5\ V}{(30 + 20)\ \Omega} = \dfrac{5}{50}$ A

Concept Reverse biased branch carries no current

Semiconductor Electronics: Materials, Devices and Simple Circuits in past papers

106 questions from this chapter have appeared across 24 exam years.