Given, supply voltage, V=5V The circuit diagram, when positive terminal of the battery is connected to X is as shown below
Let I current is coming from battery. ∴D1 will act as closed circuit as forward biased and D2 will act as open circuit as reverse biased. Now, by using Kirchhoff's voltage law, ‌5−VD1−10I‌‌=0 ⇒‌‌5−0.7−10I‌‌=0‌‌‌(∵VD1=0.7V) ⇒‌‌4.3‌‌=101‌ ⇒‌‌I‌‌=0.43A