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Wave Optics

Question
CBSEENPH12039791
Wired Faculty App

Consider the junction diode as ideal. The value of current flowing through AB is,

 

  • 10-2 A

  • 10-1 A

  • 10-3 A

  • 0 A

Solution

A.

10-2 A

Let, I be the current through the diode.
From the given condition,
straight I space equals space fraction numerator straight V subscript straight A space minus space straight V subscript straight B over denominator straight R end fraction space space equals fraction numerator 4 minus left parenthesis negative 6 right parenthesis over denominator 1 space kΩ end fraction space space equals space fraction numerator 10 over denominator 1 space straight x space 10 cubed end fraction space space equals space 10 to the power of negative 2 end exponent space straight A

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