A)
B)
C)
D)
Correct Answer: D
Solution :
Key Idea For negative potential the diode becomes reverse biased whereas for positive potential diode is forward biased. For \[=\frac{m{{d}^{2}}}{12}+\frac{m{{d}^{2}}}{4}\left( as\,{{I}_{EF}}=\frac{m{{d}^{2}}}{12} \right)\] the diode is reverse biased and hence offer infinite resistance, so circuit would be like as shown in Fig. (2) and \[{{I}_{AD}}=\frac{m{{d}^{2}}}{3}=4{{I}_{EF}}\] For \[{{V}_{i}}<0\], the diode is forward biased and circuit would be as shown in Fig. (3) and \[{{V}_{0}}=0\] Hence, the optical is correct.You need to login to perform this action.
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