Q 12-14-019NEETJEE MainEasy
Two identical silicon diodes (cut-in $0.7$ V each) are connected in series, both forward biased, with a $2.3\ \text{k}\Omega$ resistor across a $6$ V supply. The current is
Answer: (C) $2$ mA
Drop across the two diodes $= 1.4$ V, so the resistor has $6 - 1.4 = 4.6$ V.
$I = \dfrac{4.6}{2300} = 2$ mA.
Solution by Sreeraj P, M.Sc Physics