JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Evening)

  • question_answer
    The current i in the network is                               [JEE MAIN Held on 09-01-2020 Evening]

    A) 0.3A    

    B) 0.6 A

    C) 0 A

    D) 0.2 A

    Correct Answer: A

    Solution :

    (a) Both the diodes are reverse biased \[I=\frac{9}{30}=0.3A\]


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