Q 12-14-018NEETJEE MainMedium
A germanium diode (cut-in $0.3$ V) and a silicon diode (cut-in $0.7$ V) are connected in parallel, both pointing in the forward direction. This pair is connected in series with a $10\ \text{k}\Omega$ resistor across a $12$ V battery. Which diode conducts, and what is the current drawn from the battery?
Answer: (B) Only the Ge diode; $1.17$ mA
The Ge diode switches on first and holds the voltage across the parallel pair at $0.3$ V. That is less than the $0.7$ V the Si diode needs, so the Si diode stays off.
$I = \dfrac{12 - 0.3}{10\,000} = 1.17$ mA.
Solution by Sreeraj P, M.Sc Physics