The current i in the network is

The current $i$ in the network is

1. $0.2 \mathrm{A}$
2. $0.6 \mathrm{A}$
3. $0.3 \mathrm{A}$
4. $0 \mathrm{A}$

Solution

Both the diodes are in reverse biased.

$I=\frac{9}{30}=\frac{3}{10}\mathrm{A}=0.3 \mathrm{A}$

Asked in: JEE Main 2020 (09 Jan Shift 2)

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