Q 12-14-138JEE MainJEE Main 2021 (25 Feb, Shift 1)Medium
A 5 V battery is connected across the points $X$ and $Y$. Assume $D_1$ and $D_2$ to be normal silicon diodes. Find the current supplied by the battery if the $+ve$ terminal of the battery is connected to point $X$.
Answer: (D) 0.43 A
With $X$ positive, current tends to flow from the left wire to the right wire through the branches. $D_1$ points from left to right, so it is forward biased; $D_2$ points the other way and is reverse biased (no current).
Taking 0.7 V across the forward-biased silicon diode:
$$I = \frac{5 - 0.7}{10} = 0.43\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics