Semiconductor Electronics Part 1

Given, supply voltage,
V=5V
The circuit diagram, when positive terminal of the battery is connected to X is as shown belowLet 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