Discussion
Two identical silicon diodes in series feed \( 1 \ \mathrm{k\Omega} \) from \( 5 \ \mathrm{V} \). The current is
Answer: B
Two drops must be subtracted:
\[ I = \frac{5 - 1.4}{1000} = 3.6 \ \mathrm{mA} \]
so 3.6 mA. 4.3 mA counts one drop and 5 mA none. Halving the single-diode answer, which gives 1.8 mA, would be right for two resistors but not for two diodes, since a diode is a voltage drop rather than a resistance. Which model a component obeys decides whether series elements add voltages or resistances.
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