\(\text{If diode }D_1 \text{ and }D_2 \text{ are made up of same material with the cut-in voltage }V_Y=0.7\text{ V, then the value of current I is equal to}\)
A
0.59 mA
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B
0.99 mA
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C
0.46 mA
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D
1.06 mA
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Solution
The correct option is D 1.06 mA When D2 is ON then the value of V0 will beV0=3−0.7V=2.3V Hence, D1 will be OFF. Thus, The current, I=2.3−(−3)5×10−3=5.35×10−3=1.06mA