Q 12-14-165JEE MainJEE Main 2020 (9 Jan, Shift 2)Medium
The current $i$ in the network is
Answer: (C) $0.3$ A
The current $i$ enters the network at the left node. Both diodes point against this flow (the one in the upper-left arm and the one in the lower-right arm), so they are reverse biased and those two arms carry no current.
The only path left is: left node $\to 10\ \Omega \to$ bottom node $\to 5\ \Omega$ (middle) $\to$ top node $\to 10\ \Omega \to$ right node.
Total resistance $= 5 + 10 + 5 + 10 = 30\ \Omega$:
$$i = \frac{9}{30} = 0.3\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics