A 5 V DC source is connected in series with a 1 k resistor and two identical silicon diodes. The barrier potential for each silicon diode is 0.7 V . Consider the following two cases: Case 1: Both diodes are connected such that they are forward-biased. Case 2: One diode is forward-biased, while the other is reverse-biased. What is the potential drop across the resistor in Case 1 and Case 2, respectively?

Step-by-step solution

In Case 1, both silicon diodes are forward-biased. Each diode will have a potential drop equal to its barrier potential, which is 0.7 V . By Kirchhoff's Voltage Law, the total voltage drop across the diodes is 0.7 V + 0.7 V = 1.4 V . The remaining potential drops across the resistor. Potential drop across the resistor = 5 V - 1.4 V = 3.6 V . In Case 2, one diode is reverse-biased. A reverse-biased diode offers near-infinite resistance, effectively acting as an open circuit. As a result, the current flowing through