A circuit consists of a 5V source, a 100 ohm resistor R, and two diodes D1 and D2 in series. Calculate the voltage drops and current for two scenarios: a) The circuit is serviceable. b) D2 is shorted

Each diode has a voltage drop of $0.6V$. Since there are two identical diodes in series, the total voltage drop is:

$$V_{diodes} = V_{D1} + V_{D2} = 0.6V + 0.6V$$ $$V_{diodes} = \boxed{\text{1.2 V}}$$

Using Kirchhoff's Voltage Law, the sum of voltage drops in a series circuit equals the supply voltage.

$$V_{source} = V_{diodes} + V_R$$ $$5V = 1.2V + V_R$$ $$V_R = 5V - 1.2V$$ $$V_R = \boxed{\text{3.8 V}}$$

Using Ohm's Law, $I = \frac{V_R}{R}$:

$$I = \frac{3.8V}{100\Omega}$$ $$I = \boxed{\text{0.038 A}}$$

When D2 is shorted out, D1 is still in the circuit and forward-biased. Its voltage drop remains the same.

$$V_{D1} = \boxed{\text{0.6 V}}$$

Since D2 is shorted out, its voltage drop is $0V$. Using Kirchhoff's Voltage Law:

$$V_{source} = V_{D1} + V_{D2} + V_R$$ $$5V = 0.6V + 0V + V_R$$ $$V_R = 5V - 0.6V$$ $$V_R = \boxed{\text{4.4 V}}$$

Using Ohm's Law, $I = \frac{V_R}{R}$:

$$I = \frac{4.4V}{100\Omega}$$ $$I = \boxed{\text{0.044 A}}$$