The current i in the network is:


The current $i$ in the network is:

  1. (1) $0.2 \mathrm{~A}$

  2. (2) $0.6 \mathrm{~A}$

  3. (3) $0.3 \mathrm{~A}$

  4. (4) $0 \mathrm{~A}$

Correct Option: , 3


(3) Both the diodes are reverse biased, so, there is no flow of current through $5 \Omega$ and $20 \Omega$ resistances.

Now, two resistors of $10 \Omega$ and two resistors of $5 \Omega$ are in series.

Hence current $I$ through the network


$\Rightarrow I=\frac{9}{30}=0.3 \mathrm{~A}$

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