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The current i in the  network is :
Option: 1 0.3A
Option: 2 0\: A
Option: 3 0.2\: A
Option: 4 0.6\: A
 

Answers (1)

best_answer

Both diodes are in reverse biased 

 

So new circuits can be drawn as

 

 

So I=\frac{9}{30}=\frac{3}{10}=0.3\; A

Hence the correct option is (1).

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avinash.dongre

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