JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 19)

The current i in the network is : JEE Main 2020 (Online) 9th January Evening Slot Physics - Semiconductor Question 142 English

0.6 A

0.3 A

0 A

0.2 A

Both the diodes are reverse biased. JEE Main 2020 (Online) 9th January Evening Slot Physics - Semiconductor Question 142 English Explanation

i = $${9 \over {5 + 10 + 5 + 10}}$$ = $${9 \over {30}}$$ A